meaning of &variable (passed to function) - c

int main()
{
int a = 3;
func(&a)
}
(func being some arbitrary user defined function)
What is the meaning of this &? i know that its the reference variable, but so far I have only seen it being used in a function definition.
(Also, the variable a in the actual code is probably global or extern or something, I'm not sure :-/ ).

func being some arbitrary user defined function
It couldn't be "arbitrary" - it must take a pointer to int or a void* in order for the call to be legal.
This ampersand is the "take address" operator. It passes func the address of a, so that the func could, for example, modify it:
void func(int *pa) {
*pa = 4; // Note the asterisk - it "undoes" the effect of the ampersand
}
If your main prints a after the call to func, it prints 4 instead of 3.
Note that if you pass a instead of a pointer to a to a function that takes an int, not an int*, then modifications done to that int inside the function will have no effect on the parameter that you pass, because in C parameters are passed by value.
the variable a in the actual code is probably global or extern or something
It is probably not global, because there is no point in passing globals around: by virtue of being global, they are already accessible from everywhere.

the & means, that you are passing the address of that variable to the function.
That is needed when the function takes a pointer argument and you only have an atomic variable.

& is used in function call to pass the address of the variable (following &) to the function.
Since in C, there is no call by reference, you need to pass the address of the variable to the function if you are interested to reflect the changes to the variable in the caller.

The expression &<variable> yields the address of <variable>.
In your example it would allow func() to directly modify the value of main()'s local variable a, by dereferencing (e.g. *a = b).
In your example, the result might more simply and safely be returned from the function normally: a = func() for example), but the technique is more generally useful for:
Obtaining more than one result from a function through multiple pointer-arguments in addition to or instead of the return value.
Efficiently passing very large objects to functions (where the argument might be const qualified if it is for input only).

Related

I need to pass a file.txt to a string buffer [duplicate]

I am currently reading on the ways of passing arguments to C functions. And while reading I came to know that in C there are two methods to pass arguments as pass by value and pass by reference.
Then again I read that we can pass a pointer to the variable as a parameter too.
The beginnersbook website this method is mentioned as pass by reference.
In this example, we are passing a pointer to a function. When we pass a pointer as an argument instead of a variable then the address of the variable is passed instead of the value. So any change made by the function using the pointer is permanently made at the address of passed variable. This technique is known as call by reference in C.
I know that we can pass a pointer variable or the address of the variable with & operator when we need to change or access the original variable. I want to know whether this pass by pointer can also be called pass by reference.
Is pass by pointer also a method of pass by reference?
Technically, all arguments to functions in C are pass by value. The language doesn't have support for true references unlike C++.
Passing by reference can be emulated in C by passing (by value) the address of the variable you want to modify, then subsequently dereferencing that address to modify the pointed-to value. This is not true pass by reference, however.
Pointers in C are references. C 2018 6.2.5 20 says:
… A pointer type describes an object whose value provides a reference to an entity of the referenced type…
When you pass a pointer to a function, you are passing the function a reference to whatever object the pointer points to. The pointer itself is passed by value, but, because the pointer is a reference and you are passing it, the description that you are passing an object by reference is accurate.
Prior to the development of C++, there was no dispute about this; people said they were passing an object by reference to describe passing a pointer to the object. C++ adopted the word “reference” as a name for a new feature in its language, so, in C++ terminology, “reference” generally refers to that feature unless otherwise stated or made apparent by context. However, in C, we are not obliged by C++ terminology, and the original meaning of pass by reference remains accurate.
In general, all arguments to functions in 'c' are passed 'by values'. It means that If an argument is modified in a function, this modification will not be visible by the caller:
void foo(int val) {
...
val = newval;
...
}
void bar() {
foo(4);
}
The value val in the above example can be modified and used in 'foo' but 'bar' will not see the update.
In order to make an update visible in the caller, pointers are used.
void foo(int *arg) {
...
*arg = 5;
...
}
void bar () {
int x = 4;
foo(&x);
// x will be 5 here
}
In the above case the address of 'x' will be passed to the function foo and it will be used inside the function *arg to modify the value of 'x'.
One can modify the value of the pointer inside the foo, since it itself is passed by value
void foo(int *arg) {
arg = newAddress;
}
but function 'bar' will not see this modification as well.
So, there are no real 'references' in 'c', however some people use it in reference to the passing of pointers.

Can Pass by Reference also be called as Pass by Pointer?

I am currently reading on the ways of passing arguments to C functions. And while reading I came to know that in C there are two methods to pass arguments as pass by value and pass by reference.
Then again I read that we can pass a pointer to the variable as a parameter too.
The beginnersbook website this method is mentioned as pass by reference.
In this example, we are passing a pointer to a function. When we pass a pointer as an argument instead of a variable then the address of the variable is passed instead of the value. So any change made by the function using the pointer is permanently made at the address of passed variable. This technique is known as call by reference in C.
I know that we can pass a pointer variable or the address of the variable with & operator when we need to change or access the original variable. I want to know whether this pass by pointer can also be called pass by reference.
Is pass by pointer also a method of pass by reference?
Technically, all arguments to functions in C are pass by value. The language doesn't have support for true references unlike C++.
Passing by reference can be emulated in C by passing (by value) the address of the variable you want to modify, then subsequently dereferencing that address to modify the pointed-to value. This is not true pass by reference, however.
Pointers in C are references. C 2018 6.2.5 20 says:
… A pointer type describes an object whose value provides a reference to an entity of the referenced type…
When you pass a pointer to a function, you are passing the function a reference to whatever object the pointer points to. The pointer itself is passed by value, but, because the pointer is a reference and you are passing it, the description that you are passing an object by reference is accurate.
Prior to the development of C++, there was no dispute about this; people said they were passing an object by reference to describe passing a pointer to the object. C++ adopted the word “reference” as a name for a new feature in its language, so, in C++ terminology, “reference” generally refers to that feature unless otherwise stated or made apparent by context. However, in C, we are not obliged by C++ terminology, and the original meaning of pass by reference remains accurate.
In general, all arguments to functions in 'c' are passed 'by values'. It means that If an argument is modified in a function, this modification will not be visible by the caller:
void foo(int val) {
...
val = newval;
...
}
void bar() {
foo(4);
}
The value val in the above example can be modified and used in 'foo' but 'bar' will not see the update.
In order to make an update visible in the caller, pointers are used.
void foo(int *arg) {
...
*arg = 5;
...
}
void bar () {
int x = 4;
foo(&x);
// x will be 5 here
}
In the above case the address of 'x' will be passed to the function foo and it will be used inside the function *arg to modify the value of 'x'.
One can modify the value of the pointer inside the foo, since it itself is passed by value
void foo(int *arg) {
arg = newAddress;
}
but function 'bar' will not see this modification as well.
So, there are no real 'references' in 'c', however some people use it in reference to the passing of pointers.

Function parameters in my program [duplicate]

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).

What is meaning of a pointer to a constant function?

Pointers can be declared as pointing to mutable (non-const) data or pointer to constant data.
Pointers can be defined to point to a function.
My coworkers and I were discussing the use of "const" with pointers and the question came up regarding the use of const with function pointers.
Here are some questions:
What is the meaning of a pointer to a constant function versus a
pointer to a non-constant function?
Can a function be const?
Can a function be non-const (mutable)?
What is the proper (safe) syntax for passing a function pointer?
Edit 1: Function pointer syntax
typedef void (*Function_Pointer)(void); // Pointer to void function returning void.
void function_a(Function_Pointer p_func); // Example 1.
void function_b(const Function_Pointer p_func); // Example 2.
void function_c(Function_Pointer const p_func); // Example 3.
void function_d(const Function_Pointer const p_func); // Example 4.
The above declarations are examples of treating a function pointer like a pointer to an intrinsic type.
A data, variable or memory pointer allows for the above combinations.
So the questions are: can a function pointer have the same combinations and what is meant by a pointer to a const function (such as Example 2)?
In C, there's no such thing as a function being const or otherwise, so a pointer to a const function is meaningless (shouldn't compile, though I haven't checked with any particular compiler).
Note that although it's different, you can have a const pointer to a function, a pointer to function returning const, etc. Essentially everything but the function itself can be const. Consider a few examples:
// normal pointer to function
int (*func)(int);
// pointer to const function -- not allowed
int (const *func)(int);
// const pointer to function. Allowed, must be initialized.
int (*const func)(int) = some_func;
// Bonus: pointer to function returning pointer to const
void const *(*func)(int);
// triple bonus: const pointer to function returning pointer to const.
void const *(*const func)(int) = func.
As far as passing a pointer to a function as a parameter goes, it's pretty simple. You normally want to just pass a pointer to the correct type. However, a pointer to any type of function can be converted to a pointer to some other type of function, then back to its original type, and retain the original value.
According to the C spec (C99, section 6.7.3):
The properties associated with qualified types are meaningful only for
expressions that are lvalues.
When the spec says "qualified types", it means things defined with the const, restrict, orvolatile keyword. Snice functions are not lvalues, the const keyword on a function isn't meaningful. You may be looking at some sort of compiler-specific extension. Some compilers will throw an error if you try to declare a function as const.
Are you sure that you're looking at a pointer to a constant function and not a constant pointer to a function (that is, it's the pointer that's const, not the function)?
Regarding #4: see this guide for a helpful overview of creating, passing, and using function pointers.
Under C, there is no such thing as a const function. const is a type-qualifier, and so can only be used to qualify a type, not a function. Maybe you mean a const pointer to a function or a non-const pointer to a function?
In C++, methods can be const. If a method is const, it means that after you call that method, the object containing the method will be in the same state as before you called the method (none of the instance variables[1] have been modified). Thus, you can point to a const method and a non-const method, and those methods are different.
You can accept a function pointer in an argument list as retType (*variableName)(arguments).
[1] Unless they are mutable.
In C, functions can be const if you are in the world of GCC! Functions can be declared const through the use of attributes attached to declarations of functions and other symbols. It is basically used to provide information to the compiler on what the function does, even though its body is not available so that the compiler can do some kind of optimizations with it.
A constant function is generally defined in terms of a pure function.
A pure function is a function with basically no side effect. This
means that pure functions return a value that is calculated based on
given parameters and global memory, but cannot affect the value of any
other global variable. Pure functions cannot reasonably lack a return
type (i.e. have a void return type).
And now we can define what is a const function,
A pure function that does not access global memory, but only its
parameters, is called a constant function. This is because the
function, being unrelated to the state of global memory, will always
return the same value when given the same parameters. The return value
is thus derived directly and exclusively from the values of the
parameters given.
Here const doesn't imply anything about function mutability. But it is a function which doesn't touch global memory. You can assign normal pointers to such functions. Anyways the code region will generally ( forgetting self modifying code for a while ) be RO and you cannot modify it through the normal pointer.
Read the full insightful article here.
So when it comes to GCC Constant functions we are talking about optimizations and not function mutability.
1.What is the meaning of a pointer to a constant function versus a pointer to a non-constant function?
There is no difference between const and non-const: the function itself is not modifiable
Note: In C++ if the function is a member function of a class, const means that the state of the object within this function cannot be changed (member variables assigned to, non-const memeber functions called). In this case the const keyword is part of the member function's signature and therefore makes a difference in terms of the pointer.
2.Can a function be const?
See above.
3.Can a function be non-const (mutable)?
See above
4.What is the proper (safe) syntax for passing a function pointer?
All pointers to free functions can be cast to any other pointer to free function (i.e. their size is the same). So you could define a type for a (hypothetical) function: void f(); and convert all function pointers to this type to store. Note that you should not call the function via this common type: you need to cast it bact to its original pointer-to-function type, otherwise you get undefined behavior (and most likely crash)
For C++: pointers to member functions are not guaranteed to be convertible to pointers to free functions
1.
There is syntactically nowhere to place 'const' to make a function's contents constant.
You will run into 'function is not an l-value' error regardless if you have const or not.
typedef void (* FUNC)(void);
FUNC pFunc;
pFunc = 0; // OK
*pFunc = 0; // error: Cannot assign to a function (a function is not an l-value)
typedef void (* const FUNC)(void);
FUNC pFunc;
pFunc = 0; // error (const)
*pFunc = 0; // error: Cannot assign to a function (a function is not an l-value)
typedef void (const * FUNC)(void); // error: <cv-qualifier> (lol?)
2 & 3.
Function pointers- yes.. Function contents, doesn't look like.
4.
I don't think there is any way to make passing a function pointer safer. With all the consts in the world, the only thing you can protect is that 'SetCallback' can't change it's own local copy of the parameter.
typedef void (* const FUNC)(void);
void SetCallback(const FUNC const pCallback)
{
FUNC pTemp = pCallback; // OK (even though pTemp is not const!!)
pCallback = 0; // error (const)
}

Pointer to function in ROM

I have microcontroler that I am working with. When debugging it is necessary to call a function from that is hard coded in ROM. Technical Reference shows how to do this:
# define Device_cal (void(*)(void))0x3D7C80
and calling procedure looks like this:
(*Device_cal)()
I can't understand what actually happens here, so my question is:
How does it work?
void (*) (void) is a type. It's a pointer to a function that takes no parameter and returns void.
(void(*)(void)) 0x3D7C80 casts the 0x3D7C80 integer to this function pointer.
(*Device_cal)() calls the function.
(Device_cal)() would do the exactly the same.
The parentheses around *Device_cal and Device_cal are required because otherwise the cast to the integer would not have the higher precedence.
The #define causes (*Device_cal)() to be expanded into this immediately before compiling:
(*(void(*)(void))0x3D7C80)()
The void(*)(void) is a declaration for a function pointer that takes void and returns void types. The (*()) represents a cast for the next token in the expression (0x3D7C80). Thus this asks to treat the data at location 0x3D7C80 as a function. The final () calls the function with no arguments.
well, you "define" a pointer to function, and call it.
void(*)(void) mean a pointer to function, that gets no arguments, and return void.
If you cast 0x3D7C80 to that type, and call it, you basically call the function that its address is 0x3D7C80.
This is not an answer (that has already been done satisfactorily), but some advice:
I would suggest the following method instead:
typedef void (*tVOID_ROMFUNCTION_VOID)( void ) ;
tVOID_ROMFUNCTION_VOID Device_cal = (tVOID_ROMFUNCTION_VOID)0x3D7C80 ;
Device_cal() ;
That way you can create any number of global function pointers on initialisation while the calls look like normal statically linked functions. And you avoid confusing pre-processor macros voodoo at the same time.
By creating different function-pointer types with different signatures, the compiler will be able to perform some parameter type checking for you too.
The symbol is pasted in which creates a temporary (un named ) pointer to a function at a fixed memory location and then calls it via dereferencing.

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