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I know that there is no Call by reference in C language.
but, Some people say that there is a Call by reference in C.
I'm confused.
As far as I know, when handing over the factor to the function in C, I know that the value transferred to the function is received by making a local copy as a parameter.
However, in C++, "Call by reference" is possible because "the same element that differs only from the factor and name" is created by the reference "&". Is that true?
I know that there is no Call by reference in C language.
Correct. C always passes by value.
Some people say that there is a Call by reference in C. I'm confused.
They are wrong. This is very easy to test.
Let's start by looking at this small Perl program.
use 5.014;
sub f {
$_[0] = 456; # $_[0] is the first argument.
}
my $x = 123;
f( $x );
say $x; # 456
Changing the parameter changed the argument. This is an example of pass by reference. Perl arguments are passed by reference.
Now let's do the same thing in C.
#include <stdio.h>
void f( int x ) {
x = 456;
}
int main( void ) {
int x = 123;
f( x );
printf( "%d\n", x ); // 123
}
Changing the parameter had no effect on the argument. This is an example of pass by value. C's arguments are passed by value.
You can use pointers to achieve a similar result.
#include <stdio.h>
void f( int *xp ) {
*xp = 456;
}
int main( void ) {
int x = 123;
f( &x );
printf( "%d\n", x ); // 456
}
Note that the argument (the pointer) is still passed by value. Changing xp itself (as opposed to *xp) has no effect on the caller.
Same goes for arrays. The degenerate into a pointer which is passed by value.
#include <stdio.h>
void f( char a[] ) {
a = "def";
}
int main( void ) {
char a[] = "abc";
f( a );
printf( "%s\n", a ); // abc
}
This could be called passing a reference. It is not passing by reference, however.
However, in C++, "Call by reference" is possible
Correct.
C++ normally uses pass by value.
#include <iostream>
using namespace std;
void f( int x ) {
x = 456;
}
int main( void ) {
int x = 123;
f( x );
cout << x << endl; // 123
}
But pass by reference can be requested using &.
#include <iostream>
using namespace std;
void f( int &x ) {
x = 456;
}
int main( void ) {
int x = 123;
f( x );
cout << x << endl; // 456
}
I know that there is no Call by reference in C language. but, Some people say that there is a Call by reference in C. I'm confused.
Your confusion is, I'm afraid, inevitable, but it's not your fault.
People have been arguing about this question for a long time. (There's an FAQ list entry on the question that dates to the 1990's.) I didn't realize it was still a matter of debate, but evidently it is, because the same confusion you've experienced has been repeated (which is to say, validated) in the answers posted right here on this Stack Overflow question, where you hoped you'd get a definitive answer.
Depending on how you define your terms, all of the following statements are more or less true:
C does not have pass by reference. C always passes arguments by value.
C lets you simulate pass by reference, by passing a pointer instead. But the pointer is passed by value.
Arrays in C are passed by reference, because the array reference in the function call decays into a pointer to the array's first element. (But the pointer is passed by value.)
C++ has reference parameters, meaning that the programmer doesn't have to explicitly use the & operator in the call, or the * operator in the function. (But the implementation of that reference involves something very much like a pointer which is, again, passed by value.)
Or, in other words, for a sufficiently formal and restrictive definition of the term "pass by reference", C does not have it. But it has a couple of things that are pretty close, perhaps close enough to satisfy a less-formal definition, or to let you say, "You can get something a lot like call-by-reference in C, if you want."
C sometimes appears to pass by reference. The detail is that the argument goes through automatic conversions and that argument is passed by value. The effect is that the original argument experiences a pass-by-reference.
When an argument is a function, it may look/act like it is pass by reference. Functions are converted to the address of a function. The function sqrt is not passed by value. foo() receives the address of the function as a function pointer.
foo(sqrt);
An array looks like it is passed by reference in that bar() does not receive the value of the array. What happened under the table is that the array is converted to the type and address of the first element and bar() receives a pointer to a char. bar() may do things that change s[], exactly what pass by refence does in other languages, yet in C there is a technical under-the-table conversion that maintains the idea of pass only by value.
char s[1];
bar(s);
I know that there is no Call by reference in C language. but, Some
people say that there is a Call by reference in C. I'm confused.
The traditional definition of "pass by reference" is an aspect of subprogram calling semantics providing that the subprogram's parameters are bound to the same objects that are designated by the caller as the corresponding subprogram arguments.1,2 This has the effect that if the subprogram modifies the object identified by one of its parameters, including by assignment, then the caller can observe that modification (provided that the caller retains a way to examine that object). This is the typical implementation of Fortran's call semantics, among others.
For example, consider a program of this form (expressed in a polyglot pseudocode):
subprogram sub1(x)
x = 0
end
integer a
a = 42
call sub1(a)
print(a)
In a language with pass-by-reference semantics, the assignment to x in sub1 will modify the value of a in the caller, with the result that the program prints "0".
Pass-by-value is the main alternative: the names of subprogram parameters are not bound to the objects specified by the caller. They are instead bound to different objects with (initially) the same values as those presented by the caller. In a language with pass-by-value semantics, a program such as the above would be expected to print "42", as the subprogram's parameter x refers to a different object than the caller's a, therefore the subprogram's assignment to x is not visible to the caller.
As far as I know, when handing over the factor to the function in C, I
know that the value transferred to the function is received by making
a local copy as a parameter.
Yes, this is mandated by the C language specification:
An argument may be an expression of any complete object type. In
preparing for the call to a function, the arguments are evaluated, and
each parameter is assigned the value of the corresponding argument.
(C17 6.5.2.2/4; emphasis added)
As judged via the definitions above, this is unequivocally pass-by-value in all cases. However, there are a couple of cases that require special attention in this context:
One can pass the address of an object to a function -- for example, by means of the unary & operator. In that case, the function can modify the pointed-to object via the pointer it receives. Some people are inclined to characterize this as pass-by-reference, but it does not satisfy the definition above, because the argument was never the pointed-to object in the first place. Moreover, assignment to the received pointer itself does not modify the argument presented by the caller or the object to which it points.
The arguments presented by the caller are the results of evaluating the expressions presented, and C has some cases where the effect of that may be surprising to the uninitiated, especially
functions. Wherever the name of a function appears in a valid expression that is evaluated, it is automatically converted to a pointer.3 In particular, when the name of a function appears in the argument list of a function call, the corresponding parameter receives a pointer to the function. The called function can use that pointer to call the function it points to, even if that function's identifier is not in scope. But the function has not been passed by reference (by the above definition), for if the called function assigns a new value to the parameter, that does not modify the function it originally pointed to (nor the caller's copy of the function pointer, if it retains one).
arrays. C specifies that with only a few, narrow exceptions, expressions of array type are automatically converted to pointers. I think it's fair and consistent to describe that as the result of evaluating a (sub)expression of array type being a pointer to the first element of the array. The argument lists to function calls are no exception, so when you specify an array as a function argument, the corresponding function parameter receives a pointer to the first array element.
As a result, the called function can modify the array's elements via the pointer it receives. Some people describe that effect as the array having been passed by reference, but it doesn't actually satisfy the above definition. The parameter doesn't even have the same type as the caller's array, and moreover, if you assign a new value to the parameter itself then the effect is visible only in the function. In this sense, modifying array elements via a pointer received as a parameter is analogous to calling a function via a function pointer received as a parameter.
However, in C++, "Call by reference" is possible because "the same
element that differs only from the factor and name" is created by the
reference "&". Is that true?
Yes, one of the things that C++ has that C does not is references, and one of the major uses of references is providing pass-by-reference semantics that satisfy the above definition. It's not quite pass-by-reference in the Fortran sense because the parameter has a different type than the corresponding argument, but for most purposes, the parameter can be used in the same ways, and with the same effects, as the argument to which it is bound. In particular, assignment to a reference does affect the referenced object.
C++ references have some additional properties that differentiate them from pointers, among them:
A reference can be created only from another, valid object, either as a reference to that object or, if that object is itself a reference, as a copy of that reference (referring to the same object).
References cannot be rebound to different objects.
These play well with using C++ references for pass-by-reference.
Until now I have grounded my discussion in the definition given above, but it will be clear from the answers and comments given that there is a controversy here over whether that remains an appropriate definition. Some claim that the language has moved on, and in particular that in the context of C, the term "pass by reference" has become accepted as including passing a pointer. To be sure, some do use the term that way. On the other hand, "accepted" is clearly too strong a term, because plenty of others, including some voicing their opinions here, insist that it is is imprecise, sloppy, or simply wrong to describe passing a pointer to an object as passing that object by reference.
One thing to consider here is that the conventional meaning of these terms in context of most programming languages other than C has not appreciably moved from the traditional ones, even for much younger languages. C++ in particular is relevant because of the shared history and strong interoperability of these, and no C++ programmer would characterize passing a pointer as pass by reference. But the terminology is also well established in Java, which has only pass by value, including passing references by value, but not pass by reference. It also comes up in Python, which is like Java but more so, because all argument passing there is passing references by value. The distinction is important for explaining the semantics of those languages, as indeed it is for C, too.
Therefore, at present, if
you are engaging in comparative analysis of computer languages,
you want to express your ideas with maximum precision, or
you want to avoid, in many cases, earning a point of disrespect from a portion of your audience
then you will avoid conflating passing pointers with pass by reference. But if you do conflate the two then you can reasonably expect to be understood, at least in C-specific context.
1 Wikipedia definition
2 Strongly supported Stack Overflow definition
3 Including in their most common context, function calls: "The expression that denotes the called function shall have type pointer to function" (C17 6.5.2.2/1).
But, Some people say that there is a Call by reference in C. I'm confused.
This depends on how you define the term "call by reference":
Three years ago, I worked for a company in the automotive industry:
Pointer arguments that were pointing to a (single) variable (and not to an array) of the type someType were neither defined as "arguments of the type someType *" nor as "pointers to a someType" but as "'call by reference' arguments of the type someType".
You could also see this at the "[out]" or "[inout]" tag in the doxygen comments:
/*! \brief Example function (non-AUTOSAR)
* \param[in] foo Value of the "foo" force
* \param[out] bar Value of the "bar" speed
* \param[inout] foobar Current value of the "foobar" state
* \return True if calculating "bar" succeeded */
bool someFunction(someType foo, someType * bar, someType * foobar)
{
...
}
Microsoft's documentation of the Windows API seems to see this in a similar way - for example here:
BOOL GetExitCodeProcess(
[in] HANDLE hProcess,
[out] LPDWORD lpExitCode
);
“Pass by reference”1 is an old computer programming phrase that antedates the “reference” feature built into C++. It means providing a pointer to an object. The C standard says a pointer “provides a reference” (C 2018 6.2.5 20). References can be provided either manually by a programmer writing some explicit notation or by a feature built into a programming language.
When C++ developers named a new feature a “reference,” that did not change the prior usage of the term. When discussing “passing by reference” in languages other than C++ or others that provide built-in references, the phrase has its classic meaning.
C supports passing by reference (manually) but does not provide it as a built-in feature, except to the extent that arrays and functions undergo automatic conversions and adjustments that effect passing by reference. Programmers understand from context that “pass by reference” refers to passing a pointer in C and passing a built-in reference type in C++.
Footnote
1 The question uses the phrase “call by reference,” but this is an imprecise use of terminology. Whether we are discussing C++ or C, the issue is how arguments are passed, not how functions are called.
In C, all function arguments are passed by value, meaning each argument expression in the function call is fully evaluated and the result of that evaluation is passed to the function:
#include <stdio.h>
void swap( int a, int b )
{
int tmp = a;
a = b;
b = tmp;
}
int main( void )
{
int x = 10, y = 10;
printf( "Before swap: x = %d, y = %d\n" );
swap( x, y );
printf( " After swap: x = %d, y = %d\n" );
return 0;
}
The formal parameters a and b in the definition of swap are different objects in memory from the actual parameters x and y in main. The expressions x and y are fully evaluated, and the results of those evaluations (the values 10 and 20) are copied to a and b. Changing the values of a and b has no effect on x or y, and the output of the program will be
Before swap: x = 10, y = 20
After swap: x = 10, y = 20
We can fake pass-by-reference semantics by passing pointer values:
#include <stdio.h>
void swap( int *a, int *b )
{
int tmp = *a;
*a = *b;
*b = tmp;
}
int main( void )
{
int x = 10, y = 10;
printf( "Before swap: x = %d, y = %d\n" );
swap( &x, &y );
printf( " After swap: x = %d, y = %d\n" );
return 0;
}
Instead of passing the values of x and y, we're passing the results of the expressions &x and &y, which evaluate to the addresses of x and y.
a and b are still separate objects in memory from x and y, but instead of receiving the values of x (10) and y (20), they receive the addresses of x and y.
The expressions *a and *b can kinda-sorta be thought of as aliases for x and y, such that writing a new value to *a is the same as writing a new value to x and writing a new value to *b is the same as writing a new value to y. But this is not true pass-by-reference - it's passing pointers by value and manually dereferencing the pointers. The output of this program will be
Before swap: x = 10, y = 20
After swap: x = 20, y = 10
Switching to C++:
#include <iostream>
void swap( int &a, int &b )
{
int tmp = a;
a = b;
b = tmp;
}
int main( void )
{
int x = 10, y = 20;
std::cout << "Before swap: x = " << x << ", y = " << y << std::endl;
swap( x, y );
std::cout << " After swap: x = " << x << ", y = " << y << std::endl;
return 0;
}
The parameter declarations int &a and int &b declare a and b as references - a and b are not separate objects in memory, but rather they are alternate names or aliases for the objects designated by x and y1. You do not need to explicitly dereference a or b in the C++ code the way you have to in the C code. Like the second C example, the output will be:
Before swap: x = 10, y = 20
After swap: x = 20, y = 10
Logically speaking, anyway - the compiler may be using pointers under the hood to accomplish this, but that's hidden from you. As far as the behavior of the code is concerned, a and b are not independent objects from x and y.
What is the difference between
a parameter passed by reference
a parameter passed by value?
Could you give me some examples, please?
First and foremost, the "pass by value vs. pass by reference" distinction as defined in the CS theory is now obsolete because the technique originally defined as "pass by reference" has since fallen out of favor and is seldom used now.1
Newer languages2 tend to use a different (but similar) pair of techniques to achieve the same effects (see below) which is the primary source of confusion.
A secondary source of confusion is the fact that in "pass by reference", "reference" has a narrower meaning than the general term "reference" (because the phrase predates it).
Now, the authentic definition is:
When a parameter is passed by reference, the caller and the callee use the same variable for the parameter. If the callee modifies the parameter variable, the effect is visible to the caller's variable.
When a parameter is passed by value, the caller and callee have two independent variables with the same value. If the callee modifies the parameter variable, the effect is not visible to the caller.
Things to note in this definition are:
"Variable" here means the caller's (local or global) variable itself -- i.e. if I pass a local variable by reference and assign to it, I'll change the caller's variable itself, not e.g. whatever it is pointing to if it's a pointer.
This is now considered bad practice (as an implicit dependency). As such, virtually all newer languages are exclusively, or almost exclusively pass-by-value. Pass-by-reference is now chiefly used in the form of "output/inout arguments" in languages where a function cannot return more than one value.
The meaning of "reference" in "pass by reference". The difference with the general "reference" term is that this "reference" is temporary and implicit. What the callee basically gets is a "variable" that is somehow "the same" as the original one. How specifically this effect is achieved is irrelevant (e.g. the language may also expose some implementation details -- addresses, pointers, dereferencing -- this is all irrelevant; if the net effect is this, it's pass-by-reference).
Now, in modern languages, variables tend to be of "reference types" (another concept invented later than "pass by reference" and inspired by it), i.e. the actual object data is stored separately somewhere (usually, on the heap), and only "references" to it are ever held in variables and passed as parameters.3
Passing such a reference falls under pass-by-value because a variable's value is technically the reference itself, not the referred object. However, the net effect on the program can be the same as either pass-by-value or pass-by-reference:
If a reference is just taken from a caller's variable and passed as an argument, this has the same effect as pass-by-reference: if the referred object is mutated in the callee, the caller will see the change.
However, if a variable holding this reference is reassigned, it will stop pointing to that object, so any further operations on this variable will instead affect whatever it is pointing to now.
To have the same effect as pass-by-value, a copy of the object is made at some point. Options include:
The caller can just make a private copy before the call and give the callee a reference to that instead.
In some languages, some object types are "immutable": any operation on them that seems to alter the value actually creates a completely new object without affecting the original one. So, passing an object of such a type as an argument always has the effect of pass-by-value: a copy for the callee will be made automatically if and when it needs a change, and the caller's object will never be affected.
In functional languages, all objects are immutable.
As you may see, this pair of techniques is almost the same as those in the definition, only with a level of indirection: just replace "variable" with "referenced object".
There's no agreed-upon name for them, which leads to contorted explanations like "call by value where the value is a reference". In 1975, Barbara Liskov suggested the term "call-by-object-sharing" (or sometimes just "call-by-sharing") though it never quite caught on. Moreover, neither of these phrases draws a parallel with the original pair. No wonder the old terms ended up being reused in the absence of anything better, leading to confusion.4
(I would use the terms "new" or "indirect" pass-by-value/pass-by-reference for the new techniques.)
NOTE: For a long time, this answer used to say:
Say I want to share a web page with you. If I tell you the URL, I'm
passing by reference. You can use that URL to see the same web page I
can see. If that page is changed, we both see the changes. If you
delete the URL, all you're doing is destroying your reference to that
page - you're not deleting the actual page itself.
If I print out the page and give you the printout, I'm passing by
value. Your page is a disconnected copy of the original. You won't see
any subsequent changes, and any changes that you make (e.g. scribbling
on your printout) will not show up on the original page. If you
destroy the printout, you have actually destroyed your copy of the
object - but the original web page remains intact.
This is mostly correct except the narrower meaning of "reference" -- it being both temporary and implicit (it doesn't have to, but being explicit and/or persistent are additional features, not a part of the pass-by-reference semantic, as explained above). A closer analogy would be giving you a copy of a document vs inviting you to work on the original.
1Unless you are programming in Fortran or Visual Basic, it's not the default behavior, and in most languages in modern use, true call-by-reference is not even possible.
2A fair amount of older ones support it, too
3In several modern languages, all types are reference types. This approach was pioneered by the language CLU in 1975 and has since been adopted by many other languages, including Python and Ruby. And many more languages use a hybrid approach, where some types are "value types" and others are "reference types" -- among them are C#, Java, and JavaScript.
4There's nothing bad with recycling a fitting old term per se, but one has to somehow make it clear which meaning is used each time. Not doing that is exactly what keeps causing confusion.
It's a way how to pass arguments to functions. Passing by reference means the called functions' parameter will be the same as the callers' passed argument (not the value, but the identity - the variable itself). Pass by value means the called functions' parameter will be a copy of the callers' passed argument. The value will be the same, but the identity - the variable - is different. Thus changes to a parameter done by the called function in one case changes the argument passed and in the other case just changes the value of the parameter in the called function (which is only a copy). In a quick hurry:
Java only supports pass by value. Always copies arguments, even though when copying a reference to an object, the parameter in the called function will point to the same object and changes to that object will be see in the caller. Since this can be confusing, here is what Jon Skeet has to say about this.
C# supports pass by value and pass by reference (keyword ref used at caller and called function). Jon Skeet also has a nice explanation of this here.
C++ supports pass by value and pass by reference (reference parameter type used at called function). You will find an explanation of this below.
Codes
Since my language is C++, i will use that here
// passes a pointer (called reference in java) to an integer
void call_by_value(int *p) { // :1
p = NULL;
}
// passes an integer
void call_by_value(int p) { // :2
p = 42;
}
// passes an integer by reference
void call_by_reference(int & p) { // :3
p = 42;
}
// this is the java style of passing references. NULL is called "null" there.
void call_by_value_special(int *p) { // :4
*p = 10; // changes what p points to ("what p references" in java)
// only changes the value of the parameter, but *not* of
// the argument passed by the caller. thus it's pass-by-value:
p = NULL;
}
int main() {
int value = 10;
int * pointer = &value;
call_by_value(pointer); // :1
assert(pointer == &value); // pointer was copied
call_by_value(value); // :2
assert(value == 10); // value was copied
call_by_reference(value); // :3
assert(value == 42); // value was passed by reference
call_by_value_special(pointer); // :4
// pointer was copied but what pointer references was changed.
assert(value == 10 && pointer == &value);
}
And an example in Java won't hurt:
class Example {
int value = 0;
// similar to :4 case in the c++ example
static void accept_reference(Example e) { // :1
e.value++; // will change the referenced object
e = null; // will only change the parameter
}
// similar to the :2 case in the c++ example
static void accept_primitive(int v) { // :2
v++; // will only change the parameter
}
public static void main(String... args) {
int value = 0;
Example ref = new Example(); // reference
// note what we pass is the reference, not the object. we can't
// pass objects. The reference is copied (pass-by-value).
accept_reference(ref); // :1
assert ref != null && ref.value == 1;
// the primitive int variable is copied
accept_primitive(value); // :2
assert value == 0;
}
}
Wikipedia
http://en.wikipedia.org/wiki/Pass_by_reference#Call_by_value
http://en.wikipedia.org/wiki/Pass_by_reference#Call_by_reference
This guy pretty much nails it:
http://javadude.com/articles/passbyvalue.htm
Many answers here (and in particular the most highly upvoted answer) are factually incorrect, since they misunderstand what "call by reference" really means. Here's my attempt to set matters straight.
TL;DR
In simplest terms:
call by value means that you pass values as function arguments
call by reference means that you pass variables as function arguments
In metaphoric terms:
Call by value is where I write down something on a piece of paper and hand it to you. Maybe it's a URL, maybe it's a complete copy of War and Peace. No matter what it is, it's on a piece of paper which I've given to you, and so now it is effectively your piece of paper. You are now free to scribble on that piece of paper, or use that piece of paper to find something somewhere else and fiddle with it, whatever.
Call by reference is when I give you my notebook which has something written down in it. You may scribble in my notebook (maybe I want you to, maybe I don't), and afterwards I keep my notebook, with whatever scribbles you've put there. Also, if what either you or I wrote there is information about how to find something somewhere else, either you or I can go there and fiddle with that information.
What "call by value" and "call by reference" don't mean
Note that both of these concepts are completely independent and orthogonal from the concept of reference types (which in Java is all types that are subtypes of Object, and in C# all class types), or the concept of pointer types like in C (which are semantically equivalent to Java's "reference types", simply with different syntax).
The notion of reference type corresponds to a URL: it is both itself a piece of information, and it is a reference (a pointer, if you will) to other information. You can have many copies of a URL in different places, and they don't change what website they all link to; if the website is updated then every URL copy will still lead to the updated information. Conversely, changing the URL in any one place won't affect any other written copy of the URL.
Note that C++ has a notion of "references" (e.g. int&) that is not like Java and C#'s "reference types", but is like "call by reference". Java and C#'s "reference types", and all types in Python, are like what C and C++ call "pointer types" (e.g. int*).
OK, here's the longer and more formal explanation.
Terminology
To start with, I want to highlight some important bits of terminology, to help clarify my answer and to ensure we're all referring to the same ideas when we are using words. (In practice, I believe the vast majority of confusion about topics such as these stems from using words in ways that to not fully communicate the meaning that was intended.)
To start, here's an example in some C-like language of a function declaration:
void foo(int param) { // line 1
param += 1;
}
And here's an example of calling this function:
void bar() {
int arg = 1; // line 2
foo(arg); // line 3
}
Using this example, I want to define some important bits of terminology:
foo is a function declared on line 1 (Java insists on making all functions methods, but the concept is the same without loss of generality; C and C++ make a distinction between declaration and definition which I won't go into here)
param is a formal parameter to foo, also declared on line 1
arg is a variable, specifically a local variable of the function bar, declared and initialized on line 2
arg is also an argument to a specific invocation of foo on line 3
There are two very important sets of concepts to distinguish here. The first is value versus variable:
A value is the result of evaluating an expression in the language. For example, in the bar function above, after the line int arg = 1;, the expression arg has the value 1.
A variable is a container for values. A variable can be mutable (this is the default in most C-like languages), read-only (e.g. declared using Java's final or C#'s readonly) or deeply immutable (e.g. using C++'s const).
The other important pair of concepts to distinguish is parameter versus argument:
A parameter (also called a formal parameter) is a variable which must be supplied by the caller when calling a function.
An argument is a value that is supplied by the caller of a function to satisfy a specific formal parameter of that function
Call by value
In call by value, the function's formal parameters are variables that are newly created for the function invocation, and which are initialized with the values of their arguments.
This works exactly the same way that any other kinds of variables are initialized with values. For example:
int arg = 1;
int another_variable = arg;
Here arg and another_variable are completely independent variables -- their values can change independently of each other. However, at the point where another_variable is declared, it is initialized to hold the same value that arg holds -- which is 1.
Since they are independent variables, changes to another_variable do not affect arg:
int arg = 1;
int another_variable = arg;
another_variable = 2;
assert arg == 1; // true
assert another_variable == 2; // true
This is exactly the same as the relationship between arg and param in our example above, which I'll repeat here for symmetry:
void foo(int param) {
param += 1;
}
void bar() {
int arg = 1;
foo(arg);
}
It is exactly as if we had written the code this way:
// entering function "bar" here
int arg = 1;
// entering function "foo" here
int param = arg;
param += 1;
// exiting function "foo" here
// exiting function "bar" here
That is, the defining characteristic of what call by value means is that the callee (foo in this case) receives values as arguments, but has its own separate variables for those values from the variables of the caller (bar in this case).
Going back to my metaphor above, if I'm bar and you're foo, when I call you, I hand you a piece of paper with a value written on it. You call that piece of paper param. That value is a copy of the value I have written in my notebook (my local variables), in a variable I call arg.
(As an aside: depending on hardware and operating system, there are various calling conventions about how you call one function from another. The calling convention is like us deciding whether I write the value on a piece of my paper and then hand it to you, or if you have a piece of paper that I write it on, or if I write it on the wall in front of both of us. This is an interesting subject as well, but far beyond the scope of this already long answer.)
Call by reference
In call by reference, the function's formal parameters are simply new names for the same variables that the caller supplies as arguments.
Going back to our example above, it's equivalent to:
// entering function "bar" here
int arg = 1;
// entering function "foo" here
// aha! I note that "param" is just another name for "arg"
arg /* param */ += 1;
// exiting function "foo" here
// exiting function "bar" here
Since param is just another name for arg -- that is, they are the same variable, changes to param are reflected in arg. This is the fundamental way in which call by reference differs from call by value.
Very few languages support call by reference, but C++ can do it like this:
void foo(int& param) {
param += 1;
}
void bar() {
int arg = 1;
foo(arg);
}
In this case, param doesn't just have the same value as arg, it actually is arg (just by a different name) and so bar can observe that arg has been incremented.
Note that this is not how any of Java, JavaScript, C, Objective-C, Python, or nearly any other popular language today works. This means that those languages are not call by reference, they are call by value.
Addendum: call by object sharing
If what you have is call by value, but the actual value is a reference type or pointer type, then the "value" itself isn't very interesting (e.g. in C it's just an integer of a platform-specific size) -- what's interesting is what that value points to.
If what that reference type (that is, pointer) points to is mutable then an interesting effect is possible: you can modify the pointed-to value, and the caller can observe changes to the pointed-to value, even though the caller cannot observe changes to the pointer itself.
To borrow the analogy of the URL again, the fact that I gave you a copy of the URL to a website is not particularly interesting if the thing we both care about is the website, not the URL. The fact that you scribbling over your copy of the URL doesn't affect my copy of the URL isn't a thing we care about (and in fact, in languages like Java and Python the "URL", or reference type value, can't be modified at all, only the thing pointed to by it can).
Barbara Liskov, when she invented the CLU programming language (which had these semantics), realized that the existing terms "call by value" and "call by reference" weren't particularly useful for describing the semantics of this new language. So she invented a new term: call by object sharing.
When discussing languages that are technically call by value, but where common types in use are reference or pointer types (that is: nearly every modern imperative, object-oriented, or multi-paradigm programming language), I find it's a lot less confusing to simply avoid talking about call by value or call by reference. Stick to call by object sharing (or simply call by object) and nobody will be confused. :-)
Before understanding the two terms, you must understand the following. Every object has two things that can make it be distinguished.
Its value.
Its address.
So if you say employee.name = "John", know that there are two things about name. Its value which is "John" and also its location in the memory which is some hexadecimal number maybe like this: 0x7fd5d258dd00.
Depending on the language's architecture or the type (class, struct, etc.) of your object, you would be either transferring "John" or 0x7fd5d258dd00
Passing "John" is known as passing by value.
Passing 0x7fd5d258dd00 is known as passing by reference. Anyone who is pointing to this memory location will have access to the value of "John".
For more on this, I recommend you to read about dereferencing a pointer and also why choose struct (value type) over class (reference type).
Here is an example:
#include <iostream>
void by_val(int arg) { arg += 2; }
void by_ref(int&arg) { arg += 2; }
int main()
{
int x = 0;
by_val(x); std::cout << x << std::endl; // prints 0
by_ref(x); std::cout << x << std::endl; // prints 2
int y = 0;
by_ref(y); std::cout << y << std::endl; // prints 2
by_val(y); std::cout << y << std::endl; // prints 2
}
The simplest way to get this is on an Excel file. Let’s say for example that you have two numbers, 5 and 2 in cells A1 and B1 accordingly, and you want to find their sum in a third cell, let's say A2.
You can do this in two ways.
Either by passing their values to cell A2 by typing = 5 + 2 into this cell. In this case, if the values of the cells A1 or B1 change, the sum in A2 remains the same.
Or by passing the “references” of the cells A1 and B1 to cell A2 by typing = A1 + B1. In this case, if the values of the cells A1 or B1 change, the sum in A2 changes too.
When passing by reference you are basically passing a pointer to the variable. Pass by value you are passing a copy of the variable.
In basic usage this normally means pass by reference, changes to the variable will seen be in the calling method and in pass by value they won’t.
Pass by value sends a copy of the data stored in the variable you specify, and pass by reference sends a direct link to the variable itself.
So if you pass a variable by reference and then change the variable inside the block you passed it into, the original variable will be changed. If you simply pass by value, the original variable will not be able to be changed by the block you passed it into, but you will get a copy of whatever it contained at the time of the call.
Take a look at this photo:
In the first case (pass by reference), when the variable is set or changed inside the function, the external variable also changes.
But in the second case (pass by value), changing the variable inside the function doesn't have any effect on the external variable.
For reading the article, see this link.
Pass by value - The function copies the variable and works with a copy (so it doesn't change anything in the original variable)
Pass by reference - The function uses the original variable. If you change the variable in the other function, it changes in the original variable too.
Example (copy and use/try this yourself and see):
#include <iostream>
using namespace std;
void funct1(int a) // Pass-by-value
{
a = 6; // Now "a" is 6 only in funct1, but not in main or anywhere else
}
void funct2(int &a) // Pass-by-reference
{
a = 7; // Now "a" is 7 both in funct2, main and everywhere else it'll be used
}
int main()
{
int a = 5;
funct1(a);
cout << endl << "A is currently " << a << endl << endl; // Will output 5
funct2(a);
cout << endl << "A is currently " << a << endl << endl; // Will output 7
return 0;
}
Keep it simple, peeps. Walls of text can be a bad habit.
A major difference between them is that value-type variables store values, so specifying a value-type variable in a method call passes a copy of that variable's value to the method. Reference-type variables store references to objects, so specifying a reference-type variable as an argument passes the method a copy of the actual reference that refers to the object. Even though the reference itself is passed by value, the method can still use the reference it receives to interact with—and possibly modify—the original object. Similarly, when returning information from a method via a return statement, the method returns a copy of the value stored in a value-type variable or a copy of the reference stored in a reference-type variable. When a reference is returned, the calling method can use that reference to interact with the referenced object. So, in effect, objects are always passed by reference.
In c#, to pass a variable by reference so the called method can modify the variable's, C# provides keywords ref and out. Applying the ref keyword to a parameter declaration allows you to pass a variable to a method by reference—the called method will be able to modify the original variable in the caller. The ref keyword is used for variables that already have been initialized in the calling method. Normally, when a method call contains an uninitialized variable as an argument, the compiler generates an error. Preceding a parameter with keyword out creates an output parameter. This indicates to the compiler that the argument will be passed into the called method by reference and that the called method will assign a value to the original variable in the caller. If the method does not assign a value to the output parameter in every possible path of execution, the compiler generates an error. This also prevents the compiler from generating an error message for an uninitialized variable that is passed as an argument to a method. A method can return only one value to its caller via a return statement, but can return many values by specifying multiple output (ref and/or out) parameters.
see c# discussion and examples here link text
Examples:
class Dog
{
public:
barkAt( const std::string& pOtherDog ); // const reference
barkAt( std::string pOtherDog ); // value
};
const & is generally best. You don't incur the construction and destruction penalty. If the reference isn't const your interface is suggesting that it will change the passed in data.
If you don't want to change the value of the original variable after passing it into a function, the function should be constructed with a "pass by value" parameter.
Then the function will have only the value, but not the address of the passed in variable. Without the variable's address, the code inside the function cannot change the variable value as seen from the outside of the function.
But if you want to give the function the ability to change the value of the variable as seen from the outside, you need to use pass by reference. As both the value and the address (reference) are passed in and are available inside the function.
In short, Passed by value is WHAT it is and passed by reference is WHERE it is.
If your value is VAR1 = "Happy Guy!", you will only see "Happy Guy!". If VAR1 changes to "Happy Gal!", you won't know that. If it's passed by reference, and VAR1 changes, you will.
Pass by value means how to pass a value to a function by making use of arguments. In pass by value, we copy the data stored in the variable we specify, and it is slower than pass by reference because the data is copied.
Or we make changes in the copied data. The original data is not affected. And in pass by reference or pass by address, we send a direct link to the variable itself. Or passing a pointer to a variable. It is faster because less time is consumed.
Here is an example that demonstrates the differences between pass by value - pointer value - reference:
void swap_by_value(int a, int b){
int temp;
temp = a;
a = b;
b = temp;
}
void swap_by_pointer(int *a, int *b){
int temp;
temp = *a;
*a = *b;
*b = temp;
}
void swap_by_reference(int &a, int &b){
int temp;
temp = a;
a = b;
b = temp;
}
int main(void){
int arg1 = 1, arg2 = 2;
swap_by_value(arg1, arg2);
cout << arg1 << " " << arg2 << endl; //prints 1 2
swap_by_pointer(&arg1, &arg2);
cout << arg1 << " " << arg2 << endl; //prints 2 1
arg1 = 1; //reset values
arg2 = 2;
swap_by_reference(arg1, arg2);
cout << arg1 << " " << arg2 << endl; //prints 2 1
}
The “passing by reference” method has an important limitation. If a parameter is declared as passed by reference (so it is preceded by the & sign) its corresponding actual parameter must be a variable.
An actual parameter referring to “passed by value” formal parameter may be an expression in general, so it is allowed to use not only a variable but also a literal or even a function invocation's result.
The function is not able to place a value in something other than a variable. It cannot assign a new value to a literal or force an expression to change its result.
PS: You can also check Dylan Beattie answer in the current thread that explains it in plain words.
1. Pass By Value / Call By Value
void printvalue(int x)
{
x = x + 1 ;
cout << x ; // 6
}
int x = 5;
printvalue(x);
cout << x; // 5
In call by value, when you pass a value to printvalue(x) i.e. the argument which is 5, it is copied to void printvalue(int x). Now, we have two different values 5 and the copied value 5 and these two values are stored in different memory locations. So if you make any change inside void printvalue(int x) it won't reflect back to the argument.
2. Pass By Reference/ Call By Reference
void printvalue(int &x)
{
x = x + 1 ;
cout << x ; // 6
}
int x = 5;
printvalue(x);
cout << x; // 6
In call by reference, there's only one difference. We use & i.e. the address operator. By doing
void printvalue(int &x) we are referring to the address of x which tells us that it both refers to the same location. Hence, any changes made inside the function will reflect outside.
Now that you're here, you should also know about ...
3. Pass By Pointer/ Call By Address
void printvalue(int* x)
{
*x = *x + 1 ;
cout << *x ; // 6
}
int x = 5;
printvalue(&x);
cout << x; // 6
In pass by address, the pointer int* x holds the address passed to it printvalue(&x). Hence, any changes done inside the function will reflect outside.
The question is "vs".
And nobody has pointed to an important point. In passing with values, additional memory is occupied to store the passed variable values.
While in passing with a reference, no additional memory is occupied for the values (memory efficient in circumstances).
New to Julia and hence, likely a basic question.
x = 1
function someFn()
print(x)
x = 3
end
This throws an error at print(x) because global x inside is not seen inside the function. This makes sense.
x = [1,2]
function someFn()
print(x)
x[1] = 4
end
print(x)
Here print(x) is able to see the global array x and x[1]=4 changes value of global x globally. A similar behavior is observed when I make x a dictionary.
Three questions
Is this behavior of variable / array and its scope inside a function consistent how Julia should work?
When x is an array, it is visible inside the function even without passing a reference as an input to function. Is this correct?
Also, changing a value of an entry in the array is reflected globally. Is this so because Julia treats array x as a reference everywhere?
Let me comment on my understanding of the issue.
First, variable names point to values. The most basic way to make a binding between the variable name and the value is assignment of the form:
variable_name = value
Note that it is crucial that on the left hand side of = the only thing that is present is variable name and nothing else.
(side note: apart from = there are other ways to make a binding, e.g. += etc. or function definitions; but let us focus here on the core of the issue)
Now if you see = this does not mean that it is an assignment that creates a new binding. In particular:
variable_name[index] = value
is not an assignment operation (that creates a new binding) but setindex! operation. Writing variable_name[index] = value is the same as writing setindex!(variable_name, value, index.
(side note: this is true except for some corner cases like e.g. handling of begin or end, handling by #view macro etc., but again this is not crucial in the discussion)
Now to directly answer your questions given these comments:
Is this behavior of variable / array and its scope inside a function consistent how Julia should work?
Yes. But note that you get an error in:
x = 1
function someFn()
print(x)
x = 3
end
because Julia knows that x = 3 is present in the body of the function which makes Julia aware that x is a local variable even before it is bound to the value (a variable is either local or global).
Here is a more extreme example of the situation:
julia> x = 10
10
julia> function f()
if true
println(x)
else
x = 5
end
end
f (generic function with 1 method)
julia> f()
ERROR: UndefVarError: x not defined
although we know that the branch after else will never executed still it makes variable x to be local.
When x is an array, it is visible inside the function even without passing a reference as an input to function. Is this correct?
It is visible beause x[1] = 4 does not create a local variable named x but is simply a call to setindex! function. So there is no x variable defined locally, therefore a global variable x is used.
Also, changing a value of an entry in the array is reflected globally. Is this so because Julia treats array x as a reference everywhere?
Here be sure to remember that x is not an array. It is a variable that is bound to point to an array. Then, again, x[1] = 4 is the same as if you have written setindex!(x, 4, 1) so you simply call a function on a value (an array in this case) that a variable x is bound to. And this works, because scoping rules in Julia say that if there is no local variable of a given name in the scope Julia searches global scope for this name.
Finally let me comment that using global variables in Julia is discouraged as this is slow. So while this discussion is highly relevant to understand how Julia works, in practice you almost never need to know this (you will use local variables 99.99% of the time, and if you use global variables better make them const).
Weird title, anyway, in C function parameters are as follows:
void func_name(int a, int b) {
}
However in Rust:
fn func_name(a: int, b: int) {
}
Is this just a preference in syntax and was appealing to the creators of Rust, or is this for a specific purpose that I don't know about? For example, Go has "optional semi-colons", but they are actually to show when an expression ends. Please bare in mind that I'm a complete novice at Rust, so if you try to provide some fancy examples in Rust I probably wont understand :(
The declaration of a function argument is just a special case of variable declarations in Rust, therefore the answer to your question lies in variable declaration in general.
Let us start with C:
a b = 1;
a = 2;
From a grammar point of view, C is not quite regular:
in a b = 1;, a is the type and b is the name of a new variable being declared (and initialized)
in a = 1;, a is the name of a variable that was declared previously and is now either initialized or assigned a new value (overwriting the previous one).
Therefore, in C, knowing whether a is a type or a variable name requires looking ahead (ie, if followed by another variable then it's a type, otherwise it's a variable).
Now, in Rust:
let a = 1;
a = 2;
The syntax for introducing a new variable requires using the let keyword, there is no ambiguity and no need to look ahead to disambiguate. This is all the more important because of shadowing in Rust (let a = ...; let a = a.foo;).
The question was about types though, so let's extend the example:
let a: b = 1;
a = 2;
In this case, again, there is no need to look ahead. Immediately after let comes the variable name, and only after parsing a : comes the variable type.
Therefore, the syntax of Rust is simply meant to avoid look ahead (Rust aims at having a LL(1) syntax) and the syntax of function arguments simply follows the regular syntax.
Oh, and by the way, not all arguments have a type:
impl Foo {
fn doit(&self);
}
In normal Rust code a variable is declared this way:
let x : i32 = 0;
The C style is not possible because the type is optional, so the former is equivalent to this one:
let x = 0i32;
You need the let keyword to declare the intention of declaring a name.
In a function declaration the type is mandatory, the initialization is not allowed and the let keyword makes no sense. Other than that the syntax is the same:
fn foo(x : i32)
It would be weird to have a different syntax for declaring local variables and function arguments, don't you think?
What is the difference between
a parameter passed by reference
a parameter passed by value?
Could you give me some examples, please?
First and foremost, the "pass by value vs. pass by reference" distinction as defined in the CS theory is now obsolete because the technique originally defined as "pass by reference" has since fallen out of favor and is seldom used now.1
Newer languages2 tend to use a different (but similar) pair of techniques to achieve the same effects (see below) which is the primary source of confusion.
A secondary source of confusion is the fact that in "pass by reference", "reference" has a narrower meaning than the general term "reference" (because the phrase predates it).
Now, the authentic definition is:
When a parameter is passed by reference, the caller and the callee use the same variable for the parameter. If the callee modifies the parameter variable, the effect is visible to the caller's variable.
When a parameter is passed by value, the caller and callee have two independent variables with the same value. If the callee modifies the parameter variable, the effect is not visible to the caller.
Things to note in this definition are:
"Variable" here means the caller's (local or global) variable itself -- i.e. if I pass a local variable by reference and assign to it, I'll change the caller's variable itself, not e.g. whatever it is pointing to if it's a pointer.
This is now considered bad practice (as an implicit dependency). As such, virtually all newer languages are exclusively, or almost exclusively pass-by-value. Pass-by-reference is now chiefly used in the form of "output/inout arguments" in languages where a function cannot return more than one value.
The meaning of "reference" in "pass by reference". The difference with the general "reference" term is that this "reference" is temporary and implicit. What the callee basically gets is a "variable" that is somehow "the same" as the original one. How specifically this effect is achieved is irrelevant (e.g. the language may also expose some implementation details -- addresses, pointers, dereferencing -- this is all irrelevant; if the net effect is this, it's pass-by-reference).
Now, in modern languages, variables tend to be of "reference types" (another concept invented later than "pass by reference" and inspired by it), i.e. the actual object data is stored separately somewhere (usually, on the heap), and only "references" to it are ever held in variables and passed as parameters.3
Passing such a reference falls under pass-by-value because a variable's value is technically the reference itself, not the referred object. However, the net effect on the program can be the same as either pass-by-value or pass-by-reference:
If a reference is just taken from a caller's variable and passed as an argument, this has the same effect as pass-by-reference: if the referred object is mutated in the callee, the caller will see the change.
However, if a variable holding this reference is reassigned, it will stop pointing to that object, so any further operations on this variable will instead affect whatever it is pointing to now.
To have the same effect as pass-by-value, a copy of the object is made at some point. Options include:
The caller can just make a private copy before the call and give the callee a reference to that instead.
In some languages, some object types are "immutable": any operation on them that seems to alter the value actually creates a completely new object without affecting the original one. So, passing an object of such a type as an argument always has the effect of pass-by-value: a copy for the callee will be made automatically if and when it needs a change, and the caller's object will never be affected.
In functional languages, all objects are immutable.
As you may see, this pair of techniques is almost the same as those in the definition, only with a level of indirection: just replace "variable" with "referenced object".
There's no agreed-upon name for them, which leads to contorted explanations like "call by value where the value is a reference". In 1975, Barbara Liskov suggested the term "call-by-object-sharing" (or sometimes just "call-by-sharing") though it never quite caught on. Moreover, neither of these phrases draws a parallel with the original pair. No wonder the old terms ended up being reused in the absence of anything better, leading to confusion.4
(I would use the terms "new" or "indirect" pass-by-value/pass-by-reference for the new techniques.)
NOTE: For a long time, this answer used to say:
Say I want to share a web page with you. If I tell you the URL, I'm
passing by reference. You can use that URL to see the same web page I
can see. If that page is changed, we both see the changes. If you
delete the URL, all you're doing is destroying your reference to that
page - you're not deleting the actual page itself.
If I print out the page and give you the printout, I'm passing by
value. Your page is a disconnected copy of the original. You won't see
any subsequent changes, and any changes that you make (e.g. scribbling
on your printout) will not show up on the original page. If you
destroy the printout, you have actually destroyed your copy of the
object - but the original web page remains intact.
This is mostly correct except the narrower meaning of "reference" -- it being both temporary and implicit (it doesn't have to, but being explicit and/or persistent are additional features, not a part of the pass-by-reference semantic, as explained above). A closer analogy would be giving you a copy of a document vs inviting you to work on the original.
1Unless you are programming in Fortran or Visual Basic, it's not the default behavior, and in most languages in modern use, true call-by-reference is not even possible.
2A fair amount of older ones support it, too
3In several modern languages, all types are reference types. This approach was pioneered by the language CLU in 1975 and has since been adopted by many other languages, including Python and Ruby. And many more languages use a hybrid approach, where some types are "value types" and others are "reference types" -- among them are C#, Java, and JavaScript.
4There's nothing bad with recycling a fitting old term per se, but one has to somehow make it clear which meaning is used each time. Not doing that is exactly what keeps causing confusion.
It's a way how to pass arguments to functions. Passing by reference means the called functions' parameter will be the same as the callers' passed argument (not the value, but the identity - the variable itself). Pass by value means the called functions' parameter will be a copy of the callers' passed argument. The value will be the same, but the identity - the variable - is different. Thus changes to a parameter done by the called function in one case changes the argument passed and in the other case just changes the value of the parameter in the called function (which is only a copy). In a quick hurry:
Java only supports pass by value. Always copies arguments, even though when copying a reference to an object, the parameter in the called function will point to the same object and changes to that object will be see in the caller. Since this can be confusing, here is what Jon Skeet has to say about this.
C# supports pass by value and pass by reference (keyword ref used at caller and called function). Jon Skeet also has a nice explanation of this here.
C++ supports pass by value and pass by reference (reference parameter type used at called function). You will find an explanation of this below.
Codes
Since my language is C++, i will use that here
// passes a pointer (called reference in java) to an integer
void call_by_value(int *p) { // :1
p = NULL;
}
// passes an integer
void call_by_value(int p) { // :2
p = 42;
}
// passes an integer by reference
void call_by_reference(int & p) { // :3
p = 42;
}
// this is the java style of passing references. NULL is called "null" there.
void call_by_value_special(int *p) { // :4
*p = 10; // changes what p points to ("what p references" in java)
// only changes the value of the parameter, but *not* of
// the argument passed by the caller. thus it's pass-by-value:
p = NULL;
}
int main() {
int value = 10;
int * pointer = &value;
call_by_value(pointer); // :1
assert(pointer == &value); // pointer was copied
call_by_value(value); // :2
assert(value == 10); // value was copied
call_by_reference(value); // :3
assert(value == 42); // value was passed by reference
call_by_value_special(pointer); // :4
// pointer was copied but what pointer references was changed.
assert(value == 10 && pointer == &value);
}
And an example in Java won't hurt:
class Example {
int value = 0;
// similar to :4 case in the c++ example
static void accept_reference(Example e) { // :1
e.value++; // will change the referenced object
e = null; // will only change the parameter
}
// similar to the :2 case in the c++ example
static void accept_primitive(int v) { // :2
v++; // will only change the parameter
}
public static void main(String... args) {
int value = 0;
Example ref = new Example(); // reference
// note what we pass is the reference, not the object. we can't
// pass objects. The reference is copied (pass-by-value).
accept_reference(ref); // :1
assert ref != null && ref.value == 1;
// the primitive int variable is copied
accept_primitive(value); // :2
assert value == 0;
}
}
Wikipedia
http://en.wikipedia.org/wiki/Pass_by_reference#Call_by_value
http://en.wikipedia.org/wiki/Pass_by_reference#Call_by_reference
This guy pretty much nails it:
http://javadude.com/articles/passbyvalue.htm
Many answers here (and in particular the most highly upvoted answer) are factually incorrect, since they misunderstand what "call by reference" really means. Here's my attempt to set matters straight.
TL;DR
In simplest terms:
call by value means that you pass values as function arguments
call by reference means that you pass variables as function arguments
In metaphoric terms:
Call by value is where I write down something on a piece of paper and hand it to you. Maybe it's a URL, maybe it's a complete copy of War and Peace. No matter what it is, it's on a piece of paper which I've given to you, and so now it is effectively your piece of paper. You are now free to scribble on that piece of paper, or use that piece of paper to find something somewhere else and fiddle with it, whatever.
Call by reference is when I give you my notebook which has something written down in it. You may scribble in my notebook (maybe I want you to, maybe I don't), and afterwards I keep my notebook, with whatever scribbles you've put there. Also, if what either you or I wrote there is information about how to find something somewhere else, either you or I can go there and fiddle with that information.
What "call by value" and "call by reference" don't mean
Note that both of these concepts are completely independent and orthogonal from the concept of reference types (which in Java is all types that are subtypes of Object, and in C# all class types), or the concept of pointer types like in C (which are semantically equivalent to Java's "reference types", simply with different syntax).
The notion of reference type corresponds to a URL: it is both itself a piece of information, and it is a reference (a pointer, if you will) to other information. You can have many copies of a URL in different places, and they don't change what website they all link to; if the website is updated then every URL copy will still lead to the updated information. Conversely, changing the URL in any one place won't affect any other written copy of the URL.
Note that C++ has a notion of "references" (e.g. int&) that is not like Java and C#'s "reference types", but is like "call by reference". Java and C#'s "reference types", and all types in Python, are like what C and C++ call "pointer types" (e.g. int*).
OK, here's the longer and more formal explanation.
Terminology
To start with, I want to highlight some important bits of terminology, to help clarify my answer and to ensure we're all referring to the same ideas when we are using words. (In practice, I believe the vast majority of confusion about topics such as these stems from using words in ways that to not fully communicate the meaning that was intended.)
To start, here's an example in some C-like language of a function declaration:
void foo(int param) { // line 1
param += 1;
}
And here's an example of calling this function:
void bar() {
int arg = 1; // line 2
foo(arg); // line 3
}
Using this example, I want to define some important bits of terminology:
foo is a function declared on line 1 (Java insists on making all functions methods, but the concept is the same without loss of generality; C and C++ make a distinction between declaration and definition which I won't go into here)
param is a formal parameter to foo, also declared on line 1
arg is a variable, specifically a local variable of the function bar, declared and initialized on line 2
arg is also an argument to a specific invocation of foo on line 3
There are two very important sets of concepts to distinguish here. The first is value versus variable:
A value is the result of evaluating an expression in the language. For example, in the bar function above, after the line int arg = 1;, the expression arg has the value 1.
A variable is a container for values. A variable can be mutable (this is the default in most C-like languages), read-only (e.g. declared using Java's final or C#'s readonly) or deeply immutable (e.g. using C++'s const).
The other important pair of concepts to distinguish is parameter versus argument:
A parameter (also called a formal parameter) is a variable which must be supplied by the caller when calling a function.
An argument is a value that is supplied by the caller of a function to satisfy a specific formal parameter of that function
Call by value
In call by value, the function's formal parameters are variables that are newly created for the function invocation, and which are initialized with the values of their arguments.
This works exactly the same way that any other kinds of variables are initialized with values. For example:
int arg = 1;
int another_variable = arg;
Here arg and another_variable are completely independent variables -- their values can change independently of each other. However, at the point where another_variable is declared, it is initialized to hold the same value that arg holds -- which is 1.
Since they are independent variables, changes to another_variable do not affect arg:
int arg = 1;
int another_variable = arg;
another_variable = 2;
assert arg == 1; // true
assert another_variable == 2; // true
This is exactly the same as the relationship between arg and param in our example above, which I'll repeat here for symmetry:
void foo(int param) {
param += 1;
}
void bar() {
int arg = 1;
foo(arg);
}
It is exactly as if we had written the code this way:
// entering function "bar" here
int arg = 1;
// entering function "foo" here
int param = arg;
param += 1;
// exiting function "foo" here
// exiting function "bar" here
That is, the defining characteristic of what call by value means is that the callee (foo in this case) receives values as arguments, but has its own separate variables for those values from the variables of the caller (bar in this case).
Going back to my metaphor above, if I'm bar and you're foo, when I call you, I hand you a piece of paper with a value written on it. You call that piece of paper param. That value is a copy of the value I have written in my notebook (my local variables), in a variable I call arg.
(As an aside: depending on hardware and operating system, there are various calling conventions about how you call one function from another. The calling convention is like us deciding whether I write the value on a piece of my paper and then hand it to you, or if you have a piece of paper that I write it on, or if I write it on the wall in front of both of us. This is an interesting subject as well, but far beyond the scope of this already long answer.)
Call by reference
In call by reference, the function's formal parameters are simply new names for the same variables that the caller supplies as arguments.
Going back to our example above, it's equivalent to:
// entering function "bar" here
int arg = 1;
// entering function "foo" here
// aha! I note that "param" is just another name for "arg"
arg /* param */ += 1;
// exiting function "foo" here
// exiting function "bar" here
Since param is just another name for arg -- that is, they are the same variable, changes to param are reflected in arg. This is the fundamental way in which call by reference differs from call by value.
Very few languages support call by reference, but C++ can do it like this:
void foo(int& param) {
param += 1;
}
void bar() {
int arg = 1;
foo(arg);
}
In this case, param doesn't just have the same value as arg, it actually is arg (just by a different name) and so bar can observe that arg has been incremented.
Note that this is not how any of Java, JavaScript, C, Objective-C, Python, or nearly any other popular language today works. This means that those languages are not call by reference, they are call by value.
Addendum: call by object sharing
If what you have is call by value, but the actual value is a reference type or pointer type, then the "value" itself isn't very interesting (e.g. in C it's just an integer of a platform-specific size) -- what's interesting is what that value points to.
If what that reference type (that is, pointer) points to is mutable then an interesting effect is possible: you can modify the pointed-to value, and the caller can observe changes to the pointed-to value, even though the caller cannot observe changes to the pointer itself.
To borrow the analogy of the URL again, the fact that I gave you a copy of the URL to a website is not particularly interesting if the thing we both care about is the website, not the URL. The fact that you scribbling over your copy of the URL doesn't affect my copy of the URL isn't a thing we care about (and in fact, in languages like Java and Python the "URL", or reference type value, can't be modified at all, only the thing pointed to by it can).
Barbara Liskov, when she invented the CLU programming language (which had these semantics), realized that the existing terms "call by value" and "call by reference" weren't particularly useful for describing the semantics of this new language. So she invented a new term: call by object sharing.
When discussing languages that are technically call by value, but where common types in use are reference or pointer types (that is: nearly every modern imperative, object-oriented, or multi-paradigm programming language), I find it's a lot less confusing to simply avoid talking about call by value or call by reference. Stick to call by object sharing (or simply call by object) and nobody will be confused. :-)
Before understanding the two terms, you must understand the following. Every object has two things that can make it be distinguished.
Its value.
Its address.
So if you say employee.name = "John", know that there are two things about name. Its value which is "John" and also its location in the memory which is some hexadecimal number maybe like this: 0x7fd5d258dd00.
Depending on the language's architecture or the type (class, struct, etc.) of your object, you would be either transferring "John" or 0x7fd5d258dd00
Passing "John" is known as passing by value.
Passing 0x7fd5d258dd00 is known as passing by reference. Anyone who is pointing to this memory location will have access to the value of "John".
For more on this, I recommend you to read about dereferencing a pointer and also why choose struct (value type) over class (reference type).
Here is an example:
#include <iostream>
void by_val(int arg) { arg += 2; }
void by_ref(int&arg) { arg += 2; }
int main()
{
int x = 0;
by_val(x); std::cout << x << std::endl; // prints 0
by_ref(x); std::cout << x << std::endl; // prints 2
int y = 0;
by_ref(y); std::cout << y << std::endl; // prints 2
by_val(y); std::cout << y << std::endl; // prints 2
}
The simplest way to get this is on an Excel file. Let’s say for example that you have two numbers, 5 and 2 in cells A1 and B1 accordingly, and you want to find their sum in a third cell, let's say A2.
You can do this in two ways.
Either by passing their values to cell A2 by typing = 5 + 2 into this cell. In this case, if the values of the cells A1 or B1 change, the sum in A2 remains the same.
Or by passing the “references” of the cells A1 and B1 to cell A2 by typing = A1 + B1. In this case, if the values of the cells A1 or B1 change, the sum in A2 changes too.
When passing by reference you are basically passing a pointer to the variable. Pass by value you are passing a copy of the variable.
In basic usage this normally means pass by reference, changes to the variable will seen be in the calling method and in pass by value they won’t.
Pass by value sends a copy of the data stored in the variable you specify, and pass by reference sends a direct link to the variable itself.
So if you pass a variable by reference and then change the variable inside the block you passed it into, the original variable will be changed. If you simply pass by value, the original variable will not be able to be changed by the block you passed it into, but you will get a copy of whatever it contained at the time of the call.
Take a look at this photo:
In the first case (pass by reference), when the variable is set or changed inside the function, the external variable also changes.
But in the second case (pass by value), changing the variable inside the function doesn't have any effect on the external variable.
For reading the article, see this link.
Pass by value - The function copies the variable and works with a copy (so it doesn't change anything in the original variable)
Pass by reference - The function uses the original variable. If you change the variable in the other function, it changes in the original variable too.
Example (copy and use/try this yourself and see):
#include <iostream>
using namespace std;
void funct1(int a) // Pass-by-value
{
a = 6; // Now "a" is 6 only in funct1, but not in main or anywhere else
}
void funct2(int &a) // Pass-by-reference
{
a = 7; // Now "a" is 7 both in funct2, main and everywhere else it'll be used
}
int main()
{
int a = 5;
funct1(a);
cout << endl << "A is currently " << a << endl << endl; // Will output 5
funct2(a);
cout << endl << "A is currently " << a << endl << endl; // Will output 7
return 0;
}
Keep it simple, peeps. Walls of text can be a bad habit.
A major difference between them is that value-type variables store values, so specifying a value-type variable in a method call passes a copy of that variable's value to the method. Reference-type variables store references to objects, so specifying a reference-type variable as an argument passes the method a copy of the actual reference that refers to the object. Even though the reference itself is passed by value, the method can still use the reference it receives to interact with—and possibly modify—the original object. Similarly, when returning information from a method via a return statement, the method returns a copy of the value stored in a value-type variable or a copy of the reference stored in a reference-type variable. When a reference is returned, the calling method can use that reference to interact with the referenced object. So, in effect, objects are always passed by reference.
In c#, to pass a variable by reference so the called method can modify the variable's, C# provides keywords ref and out. Applying the ref keyword to a parameter declaration allows you to pass a variable to a method by reference—the called method will be able to modify the original variable in the caller. The ref keyword is used for variables that already have been initialized in the calling method. Normally, when a method call contains an uninitialized variable as an argument, the compiler generates an error. Preceding a parameter with keyword out creates an output parameter. This indicates to the compiler that the argument will be passed into the called method by reference and that the called method will assign a value to the original variable in the caller. If the method does not assign a value to the output parameter in every possible path of execution, the compiler generates an error. This also prevents the compiler from generating an error message for an uninitialized variable that is passed as an argument to a method. A method can return only one value to its caller via a return statement, but can return many values by specifying multiple output (ref and/or out) parameters.
see c# discussion and examples here link text
Examples:
class Dog
{
public:
barkAt( const std::string& pOtherDog ); // const reference
barkAt( std::string pOtherDog ); // value
};
const & is generally best. You don't incur the construction and destruction penalty. If the reference isn't const your interface is suggesting that it will change the passed in data.
If you don't want to change the value of the original variable after passing it into a function, the function should be constructed with a "pass by value" parameter.
Then the function will have only the value, but not the address of the passed in variable. Without the variable's address, the code inside the function cannot change the variable value as seen from the outside of the function.
But if you want to give the function the ability to change the value of the variable as seen from the outside, you need to use pass by reference. As both the value and the address (reference) are passed in and are available inside the function.
In short, Passed by value is WHAT it is and passed by reference is WHERE it is.
If your value is VAR1 = "Happy Guy!", you will only see "Happy Guy!". If VAR1 changes to "Happy Gal!", you won't know that. If it's passed by reference, and VAR1 changes, you will.
Pass by value means how to pass a value to a function by making use of arguments. In pass by value, we copy the data stored in the variable we specify, and it is slower than pass by reference because the data is copied.
Or we make changes in the copied data. The original data is not affected. And in pass by reference or pass by address, we send a direct link to the variable itself. Or passing a pointer to a variable. It is faster because less time is consumed.
Here is an example that demonstrates the differences between pass by value - pointer value - reference:
void swap_by_value(int a, int b){
int temp;
temp = a;
a = b;
b = temp;
}
void swap_by_pointer(int *a, int *b){
int temp;
temp = *a;
*a = *b;
*b = temp;
}
void swap_by_reference(int &a, int &b){
int temp;
temp = a;
a = b;
b = temp;
}
int main(void){
int arg1 = 1, arg2 = 2;
swap_by_value(arg1, arg2);
cout << arg1 << " " << arg2 << endl; //prints 1 2
swap_by_pointer(&arg1, &arg2);
cout << arg1 << " " << arg2 << endl; //prints 2 1
arg1 = 1; //reset values
arg2 = 2;
swap_by_reference(arg1, arg2);
cout << arg1 << " " << arg2 << endl; //prints 2 1
}
The “passing by reference” method has an important limitation. If a parameter is declared as passed by reference (so it is preceded by the & sign) its corresponding actual parameter must be a variable.
An actual parameter referring to “passed by value” formal parameter may be an expression in general, so it is allowed to use not only a variable but also a literal or even a function invocation's result.
The function is not able to place a value in something other than a variable. It cannot assign a new value to a literal or force an expression to change its result.
PS: You can also check Dylan Beattie answer in the current thread that explains it in plain words.
1. Pass By Value / Call By Value
void printvalue(int x)
{
x = x + 1 ;
cout << x ; // 6
}
int x = 5;
printvalue(x);
cout << x; // 5
In call by value, when you pass a value to printvalue(x) i.e. the argument which is 5, it is copied to void printvalue(int x). Now, we have two different values 5 and the copied value 5 and these two values are stored in different memory locations. So if you make any change inside void printvalue(int x) it won't reflect back to the argument.
2. Pass By Reference/ Call By Reference
void printvalue(int &x)
{
x = x + 1 ;
cout << x ; // 6
}
int x = 5;
printvalue(x);
cout << x; // 6
In call by reference, there's only one difference. We use & i.e. the address operator. By doing
void printvalue(int &x) we are referring to the address of x which tells us that it both refers to the same location. Hence, any changes made inside the function will reflect outside.
Now that you're here, you should also know about ...
3. Pass By Pointer/ Call By Address
void printvalue(int* x)
{
*x = *x + 1 ;
cout << *x ; // 6
}
int x = 5;
printvalue(&x);
cout << x; // 6
In pass by address, the pointer int* x holds the address passed to it printvalue(&x). Hence, any changes done inside the function will reflect outside.
The question is "vs".
And nobody has pointed to an important point. In passing with values, additional memory is occupied to store the passed variable values.
While in passing with a reference, no additional memory is occupied for the values (memory efficient in circumstances).