What Exactly Is the Difference Between "Pass by Reference" in C and in C++

What exactly is the difference between pass by reference in C and in C++?

There are questions that already deal with the difference between passing by reference and passing by value. In essence, passing an argument by value to a function means that the function will have its own copy of the argument - its value is copied. Modifying that copy will not modify the original object. However, when passing by reference, the parameter inside the function refers to the same object that was passed in - any changes inside the function will be seen outside.

Unfortunately, there are two ways in which the phrases "pass by value" and "pass by reference" are used which can cause confusion. I believe this is partly why pointers and references can be difficult for new C++ programmers to adopt, especially when they've come from a background in C.

C

In C, everything is passed by value in the technical sense. That is, whatever you give as an argument to a function, it will be copied into that function. For example, calling a function void foo(int) with foo(x) copies the value of x as the parameter of foo. This can be seen in a simple example:

void foo(int param) { param++; }

int main()
{
  int x = 5;
  foo(x);
  printf("%d\n",x); // x == 5
}

The value of x is copied into foo and that copy is incremented. The x in main continues to have its original value.

As I'm sure you're aware, objects can be of pointer type. For example, int* p defines p as a pointer to an int. It is important to note that the following code introduces two objects:

int x = 5;
int* p = &x;

The first is of type int and has the value 5. The second is of type int* and its value is the address of the first object.

When passing a pointer to a function, you are still passing it by value. The address it contains is copied into the function. Modifying that pointer inside the function will not change the pointer outside the function - however, modifying the object it points to will change the object outside the function. But why?

As two pointers that have the same value always point at the same object (they contain the same address), the object that is being pointed to may be accessed and modified through both. This gives the semantics of having passed the pointed to object by reference, although no references ever actually existed - there simply are no references in C. Take a look at the changed example:

void foo(int* param) { (*param)++; }

int main()
{
  int x = 5;
  foo(&x);
  printf("%d\n",x); // x == 6
}

We can say when passing the int* into a function, that the int it points to was "passed by reference" but in truth the int was never actually passed anywhere at all - only the pointer was copied into the function. This gives us the colloquial¹ meaning of "pass by value" and "pass by reference".

The usage of this terminology is backed up by terms within the standard. When you have a pointer type, the type that it is pointing to is known as its referenced type. That is, the referenced type of int* is int.

A pointer type may be derived from a function type, an object type, or an incomplete
type, called the referenced type.

While the unary * operator (as in *p) is known as indirection in the standard, it is commonly also known as dereferencing a pointer. This further promotes the notion of "passing by reference" in C.

C++

C++ adopted many of its original language features from C. Among them are pointers and so this colloquial form of "passing by reference" can still be used - *p is still dereferencing p. However, using the term will be confusing, because C++ introduces a feature that C doesn't have: the ability to truly pass references.

A type followed by an ampersand is a reference type². For example, int& is a reference to an int. when passing an argument to a function that takes reference type, the object is truly passed by reference. There are no pointers involved, no copying of objects, no nothing. The name inside the function actually refers to exactly the same object that was passed in. To contrast with the example above:

void foo(int& param) { param++; }

int main()
{
  int x = 5;
  foo(x);
  std::cout << x << std::endl; // x == 6
}

Now the foo function has a parameter that is a reference to an int. Now when passing x, param refers to precisely the same object. Incrementing param has a visible change on the value of x and now x has the value 6.

In this example, nothing was passed by value. Nothing was copied. Unlike in C, where passing by reference was really just passing a pointer by value, in C++ we can genuinely pass by reference.

Because of this potential ambiguity in the term "pass by reference", it's best to only use it in the context of C++ when you are using a reference type. If you are passing a pointer, you are not passing by reference, you are passing a pointer by value (that is, of course, unless you are passing a reference to a pointer! e.g. int*&). You may, however, come across uses of "pass by reference" when pointers are being used, but now at least you know what is really happening.

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Other languages

Other programming languages further complicate things. In some, such as Java, every variable you have is known as a reference to an object (not the same as a reference in C++, more like a pointer), but those references are passed by value. So even though you appear to be passing to a function by reference, what you're actually doing is copying a reference into the function by value. This subtle difference to passing by reference in C++ is noticed when you assign a new object to the reference passed in:

public void foo(Bar param) {
  param.something();
  param = new Bar();
}

If you were to call this function in Java, passing in some object of type Bar, the call to param.something() would be called on the same object you passed in. This is because you passed in a reference to your object. However, even though a new Bar is assigned to param, the object outside the function is still the same old object. The new one is never seen from the outside. That's because the reference inside foo is being reassigned to a new object. This kind of reassigning references is impossible with C++ references.

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¹ By "colloquial", I don't mean to suggest that the C meaning of "pass by reference" is any less truthful than the C++ meaning, just that C++ really does have reference types and so you are genuinely passing by reference. The C meaning is an abstraction over what is really passing by value.

² Of course, these are lvalue references and we now have rvalue references too in C++11.

Meaning of pass by reference in C and C++?

In colloquial usage, "pass by reference" means that, if the callee modifies its arguments, it affects the caller, because the argument as seen by the callee refers to the value as seen by the caller.

The phrase is used independent of the actual programming language, and how it calls things (pointers, references, whatever).

In C++, call-by-reference can be done with references or pointers. In C, call-by-reference can only be achieved by passing a pointer.

"Call by value":

void foo( int x )
{
    // x is a *copy* of whatever argument foo() was called with
    x = 42;
}

int main()
{
    int a = 0;
    foo( a );
    // at this point, a == 0
}

"Call by reference", C style:

void foo( int * x )
{
    // x is still a *copy* of foo()'s argument, but that copy *refers* to
    // the value as seen by the caller
    *x = 42;
}

int main()
{
    int a = 0;
    foo( &a );
    // at this point, a == 42
}

So, strictly speaking, there is no pass-by-reference in C. You either pass the variable by-value, or you pass a pointer to that variable by-value.

Does C even have pass by reference?

C parameters are always passed by value rather than by reference. However, if you think of the address of an object as being a reference to that object then you can pass that reference by value. For example:

void foo(int *x)
{
    *x = 666;
}

You ask in a comment:

So why do we need pointers in C when we can pass all the parameters by value?

Because in a language that only supports pass-by-value, lack of pointers would be limiting. It would mean that you could not write a function like this:

void swap(int *a, int *b)
{
    int temp = *a;
    *b = *a;
    *a = temp;
}

In Java for example, it is not possible to write that function because it only has pass-by-value and has no pointers.

In C++ you would write the function using references like this:

void swap(int &a, int &b)
{
    int temp = a;
    b = a;
    a = temp;
}

And similarly in C#:

void swap(ref int a, ref int b)
{
    int temp = a;
    b = a;
    a = temp;
}

pass by reference and pass by value in c++ and java

Java is pass by value, but the value you pass may be a reference.

In C++, you choose whether to pass by value or by reference. In C, you always pass by value, and there is no such thing as a reference; you have to pass a pointer to do something similar.

If you pass by value, the function gets a copy of the variable you pass. If you change the copy, the original value remains the same (though, it the value passed was itself a reference, you may change the referred object).

If you pass by reference, the function gets the actual variable passed. If the function changes that variable, you are actually changing the variable that was passed to the function.

Passing by reference in C

Because you're passing the value of the pointer to the method and then dereferencing it to get the integer that is pointed to.

C - pass by reference?

pass them as pointers. If you do not return the value - declare functions void, or return something

void doStuff(int *v)
{
    *v = *v + 10; // main->value = 10
}

void otherStuff(int *v)
{
    *v = *v - 10; // main->value = 0
}

and in the main

        doStuff(&value); // "value" = 10
/*....*/
        otherStuff(&value); // "value" = 0

int *v in the function means that v is the pointer to the int object.
in the function call &value passes the pointer (address) to the value.
Dereferencing the pointer v in the function - all poerations are actuqally done on the value.

Why do programmers say that pass by reference is really passing references by value? Why is that important?

Two main points:

• There is no call by reference in C.
• Pass by value and pass by reference are different. They are not same.

Pass by value: the called function creates a new set of variables in stack and copies the values of the arguments into them.

Pass by reference: instead of passing values to the function being called, references/pointers to the original variables are passed.

Why do programmers say that “pass by reference” is really “passing references by value?”

In passing references/pointers to the original variables, in fact objects/addresses are passed by value. So, you can say pass by reference is passing reference by value but this doesn't imply that pass by reference is pseudo name of pass by value. The difference between the two is well explained in this answer. I am copying the excerpt:

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.

can somebody explain me what does passing by value and Passing by reference mean in C#?

In simple terms...

"Passing by value" means that you pass the actual value of the variable into the function. So, in your example, it would pass the value 9.

"Passing by reference" means that you pass the variable itself into the function (not just the value). So, in your example, it would pass an integer object with the value of 9.

This has various consequences, and each is useful in different situations.

This answer has more thorough information:
What's the difference between passing by reference vs. passing by value?

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