Create sum type in C implementation of OCaml function - c

Let's say you had a type declaration:
type foo = Bar | Baz of int
How would you implement a C function to create a Baz? Let's say I declare it like this:
external create_baz : int -> foo = "create_baz"
Then I would need to fill out this code:
CAMLprim value create_baz(value bar) {
// what do I do here?
}
I understand that this is a rather silly thing to do, but it's just and example of what I'm trying to do.

This is described in Chapter 19 of the OCaml manual.
Basically, constructors are numbered in order, in two separate sequences. Nullary constructors (those taking no values, like Bar) are numbered in one sequence, and constructors that take a value (like Baz) are numbered in a second sequence. So both of your constructors are numbered 0.
Nullary constructors are represented by immediate values (a simple bit pattern representing a value like an int). Constructors taking values are represented by pointers to blocks, which have fields that can store the contained values.
So, basically your function wants to make a block of size 1 with tag 0. bar is saved in the block's 0th field (the only field).
It looks something like this:
value create_baz(value bar) {
// Caller guarantees that bar is an int.
//
CAMLparam1(bar);
CAMLlocal1(result);
result = caml_alloc(1, 0);
Store_field(result, 0, bar);
CAMLreturn(result);
}

Related

What is the difference between passing by value and passing by reference? [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).

C enum-like preprocessor define with dot access to "fields"?

I guess, what I want is something similar to this posting for C++ Accessing to enum values by '::' in C++ :
I want to: ... access Color values as Color::Red.
In current C++ (i.e. C++11 and beyond), you can already access enum values like that:
enum Color { Red };
Color c = Color::Red;
So, I would like to know - can I do a similar level of access in C, but through a define (so I don't have to create an extra variable)? For example, say I have this code:
#include <stdio.h>
int model_type = 1;
int model_variant = 4;
int main(void) {
printf("Hello World %d\n", model_variant);
return 0;
}
I would like to write something like this instead:
int model_type = DEFAULT.MODEL_TYPE;
int model_variant = DEFAULT.MODEL_VARIANT;
I'm aware that macro names (via Can you use a.b notation in a #define macro name?):
They cannot contain dot.
... but I'd still like to know - is there maybe some sort of a trick, so I can I achieve something like this (being able to access a numeric value via say DEFAULT.MODEL_TYPE) on the preprocessor level in C? I'd like the preprocessor, since it will just insert raw numbers where needed - otherwise I have to define a struct, then a variable of that struct type, then populate that variable ...
If not, are there other options for similar typing style - say accessing enum fields directly by name? (I've tried enum DEFAULTS { MODEL_TYPE =1, MODEL_VARIANT = 4 };, but DEFAULTS then is "undeclared identifier", and as such, DEFAULT.MODEL_TYPE won't work either - so in that sense, I cannot use it for my purpose) ...
object . foo in C means (approximately) "add an offset (associated with foo) to (char*)&object cast the result to a pointer to the type associated with foo and then dereference".
Unlike in object-oriented programming languages, C's ./-> is not an operator for accessing things somehow associated with an object.
If you don't want to be adding offsets to addresses of lvalues, you should be grouping things together differently, e.g., via a common prefix:
enum DEFAULTS { DEFAULTS__MODEL_TYPE =1, DEFAULTS__MODEL_VARIANT = 4 };

D straight array indexed by an enum

In my D program, I have a read-only array of fixed length and I wish to index the array by an enumerated type.
If I do something like
static const my_struct_t aray[ my_enum_t ] = ... whatever ...;
my_enum_t index;
result = aray[ index ];
then the code produced by GDC is huge, full of calls to the runtime when the array is indexed. So it looks as if either the array is being treated as variable-length or as an associative array (hash table) or something, anyway far from a lightweight C-style array of fixed length with straightforward indexing. Since enums have a fixed cardinality and can't grow, and I have a modest sparse range of values (I'm not misusing the keyword enum just to defined a load of random constants) then I don't know why this happened.
I fixed the issue by changing the line to
static const my_struct_t aray[ my_enum_t.max + 1 ]
and as I understand it that will mean the value in the square brackets is just a known constant of integral type. Since the index is now not an enum at all, I now have an array indexed by an integer, so I have lost type checking, I could index it with any random integer typed variable rather than ensuring that only the correct (strong) type is used.
What should I be doing?
In the more general case, (silly example)
static const value_t aray[ bool ] = blah
for example, where I have an index type that is perfectly sensible semantically, but not just a typeless size_t/int/uint I presume I would get the same problem.
I wouldn't want to say that this is a compiler design problem. It's certainly a case of sub-optimal behaviour. But to be fair to the compiler what exactly is telling it whether the array is fixed-length or variable, and sparse or dense? I want two things; type checking of the index and non-variable length. Actually, in this particular case the array is const (I could have put immutable just as well) so it clearly can't be variable-length any way. But with an array that has modifiable content but is of fixed length you need to be able to declare that it is fixed-length.
V[K] name is the syntax for an associative array which does indeed do runtime calls and such, even when the type is limited to a small number of values like bool or an enum. The compiler probably could optimize that, making it act to the program like an AA while implementing it as a simple fixed-length array, but it doesn't; it treats all key types the same.
I would suggest going with what you started: T[enum.max + 1], but then doing a wrapper if you want to force type safety. You can make the index overloads static if you only want one instance of it:
enum Foo {
one,
two
}
struct struct_t {}
struct array {
static private struct_t[Foo.max + 1] content;
static struct_t opIndex(Foo idx) { return content[cast(int) idx]; }
}
void main() {
struct_t a = array[Foo.one];
}
Then, you can just genericize that if you want simpler reuse.
struct enum_array(Key, Value) {
static private struct_t[Key.max + 1] content;
static Value opIndex(Key idx) { return content[cast(int) idx]; }
}
alias array = enum_array!(Foo, struct_t);
Or, of course, you don't need to make it static, you could do a regular instance too, and initialize the contents inside and such.
In D, both static and dynamic arrays are indexed by size_t, just like they would be in C and C++. And you can't change the type of the index in D any more than you can in C or C++. So, in D, if you put a type between the brackets in the array declaration, you're defining an associative array and not a static array. If you want a static array, you must provide an integer literal or compile-time constant, and there is no way to require that a naked, static array be indexed by an enum type that has a base type of size_t or a type that implicitly converts to size_t.
If you want to require that your static array be indexed by a type other than size_t, then you need to wrap it in a struct or class and control the access to the static array via the member functions. You could overload opIndex to take your enum type and treat your struct type as if it were a static array. So, the effect should then be effectively what you were trying to do with putting the enum type in the static array declaration, but it would be the member function that took the enum value and called the static array with it rather than doing anything to the static array itself.

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).

Return One Value of different kinds in a function in c

Well I have a library for simulate a dynamic array in c. So I have different functions that create a list and adds nodes to it. In the main function it is supposed that the people that it is using the library will not have an idea of what list is.
main()
{
list dinarr = new_list(3,'d');
In this case, it will create a a list of three nodes. Inside the struct list I just have a data type char data[60]. However, the user well think that it is an array type of double.
double x = 2.5;
dinarr = put_list(dinnar,1,&x);
Here I have a function that, given a position, will put a value in that node. The array was declared as double so it is logical that the user enters a double array. However since I don't know what type of array will be the other that the user created, I have in that function a void pointer, that given the type of the array (double in this case) convert the void pointer to a double pointer and that to a char, so I can put that data in the list. Here I don't have any problems.
double y = 0;
y = get_list(dinnar,1);
Here is the thing I have a double var call y, because I create a double array, then in get_list I send the position and the list, so the function will bring me the data that the user need. In that function I have a switch that in case of the datatype of the array is d (double), c(char), s(string),i(int) or f(float). It converts the char data of the struct list for the kind of type that the array is in this case double. But I do not know how return different types of data with just one argument. I used a struct but then I have to declare that struct in the main, and the main purpose of this, is that the user doesn't notice.
}
So what kind of type, should I declare my function?
I used void pointer and struct, but it does not work for me because the user shouldn't know what it is happening inside my library.
You cannot implement that, C doesn't support overloading or having multiple functions with the same name but different return types.
You can solve it by making the list getter take a pointer to where the caller wants the result:
double y;
get_list(dinnar, &y);
Under the hood your list module copies the value since it knows (from the 'd' format specifier) the size to expect. You can of course generalize that part to just take a size_t itemSize instead of a format letter, to make the list capable of handling any-size items.
It's not as convenient to use though but it's hard to improve on.

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