Pointers and References

Are reference and pointers same?

No.

I have seen this confusion crumbling up among the student from the first day. So better clear out this confusion at thevery beginning.

Pointers and reference both hold the address of other variables. Up to this they look similar, but their syntax and further consequences are totally different. Just consider the following pieces of code

Code 1                                         Code 2

int i;                                             int i;

int *p = &i;                                 int &r = i ;

Here in code 1 we have declared and defined one integer pointer p which points to variable i, that is now , p holds the address of i.

In code 2, we have declared and defined one integer reference r which points to variable i, that is now, r holds the address of i. This is completely same as that of p.

So where is the difference?

The first difference can be found just by looking at the code. Their syntaxes!

Secondly the difference will come up when they would be used differently to assign a value (suppose 10) to i.

If you are using a pointer, you can do it like *p = 10; but if you are using a reference, you can do it like r = 10.  Just be careful to understand that when you are using pointers, the address must be dereferenced using the *, whereas, when you are using references, the address is dereferenced without using any operators at all.

This notion leaves a huge effect as consequences. As the address of the variable is dereferenced by * operator, while using a pointer, you are free to do any arithmetic operations on it. That is you can increment the pointer p to point to the next address just by doing p++. But, this is not possible using references.  So a pointer can point to many different elements during its lifetime; where as a reference can refer to only one element during its life time.

Does C language support references?

No. The concept of reference has been added to C++, not in C. So if you run the following code, C compiler will object then and there.

#include<stdio.h>

#include<conio.h>

int main(void)

{

    int i;

    int &r = i;

    r = 10;

    printf("\n Value of i assigned with reference r = %d",i);

    getch();

    return 0;

}

But if you are using any C++ compiler, this code will work fine as expected.

If there is no concept of reference in C language, then how come there exists C function call by reference?

Strictly speaking, there is no concept of function call by reference in C language. C only supports function call by value. Though in some books ( I will not name any one) it is written that C supports function call by reference or the simulation of  function call by reference can be achieved through pointers, I will strongly say that C language neither directly supports function call by reference, nor provides any other mechanism to simulate the same effect.

I know you are at your toes to argue that what about calling a C function with address of a variable and receiving it with a pointer? The change made to that variable within the function has a global effect. How this cannot be treated as an example of function call by reference?

You probably argue with a code like following

#include<stdio.h>

#include<conio.h>

void foo(int* p)

{

     *p = 5;

      printf("\n Inside foo() the value of the variable: %d",*p);

}

int main(void)

{

    int i = 10;

    printf("\n before calling  foo() the value of the variable: %d",i);

    foo(&i);

    printf("\n after calling  foo() the value of the variable: %d",i);

    getch();

    return 0;

}      

  Your code will show the result as 

Your points are well taken. But the thing is what you are showing is not at all calling a function by reference. It just the function call by value! Here you are essentially copying the value of address of your variable i and calling the function foo with that copy. Now eventually in this case, the value that is being passed contains the address of another variable. Within the function, you are accepting this value with a pointer and changing the 

value of the content addressed by that pointer. So it is nothing but a function call by value only.

Please note that to change the value of the content addressed by a pointer, you are to use *, no way could it be thought of as a reference.

Now let me give you one example of true function call by reference

#include<stdio.h>

#include<conio.h>

void foo (int& r1)

{

     r1 = 5;

     printf("\n Inside foo() the value of the variable: %d", r1);

}

int main(void)

{

    int i = 10;

    int &r = i;

    printf("\n before calling  foo() the value of the variable: %d",i);

    foo(r);

    printf("\n after calling  foo() the value of the variable: %d",i);

    getch();

    return 0;

}

Will this run with your C compiler? No.

Note:: I have used DevC++ as the coding platform