Thread: Referencing and De-referencing Question

Code:

int *a = NULL;

a is memory location which should contain another memory location. With this statement, you've set the contained memory location to 0, which is not a valid location in memory.

Code:

int b = 1;

b is a memory location containing an integer.

Code:

a = &b;

Because a contains a memory location, and the & operator returns a memory location, you've assigned a memory location -- a -- to point to b's memory location. Perfectly good.

Code:

int c = 2;
*a = c;

a currently has a valid memory location, so you're going to that memory location -- which contains the value in b -- and storing the value in memory location c there. Print b after and you'll see what I mean.

If you had not previously assigned the address in a to the address of b, it would in fact be illegal.

Originally Posted by audinue

But,... What's happen here?

Code:

int c = 2;
*a = b;

I think this is illegal ( undefined behaviour, maybe? ) since "a" is pointing to nothing
and there is no memory allocated for its value.

Or 0 (NULL) has its own value?

[/code]Thanks in advance.

Code:

int main()
{
         int  *a,b=2,c;
         *a = b;
         printf("*a = %d\n",*a);
         return 0;
}

Output:
*a = 2

Hope this clears your doubt?

Code:

#include "stdio.h"

int main( int argc, char *argv[] ) {

  int *a = NULL;
  printf( "%d\n", a );    

  int b = 1;
  a = &b;
  printf( "%d\n", *a ); // >> 1

  int c = 2;
  *a = c;
  printf( "%d\n", *a ); // >> 2 ??
}

The first printf is an unfortunate use of printf, but it would work in most systems. "%d" assumes a signed integer, and if the size of an integer pointer is the same as the size of an integer, you'll see a zero printed - interpreting the address (set to null) AS an integer. I say unfortunate because if the size of an integer were different than the size of the pointer (which is possible is some systems), this could result in odd behavior.

The second printf does what you expect. Dereferencing "a" results in the value stored in b.

So, too, with the 3rd printf - in the context where a has been given the address of b (and so storage is available, a is no longer null), *a = c will also mean that b == 2.

As to your question "what's happening here" - I just answered it, but let me break it down to be clear.

"b" is a local variable, it is stored on the stack. The address of b will be a location on the stack, which is of minor importance limited to the scope of our discussion, but has other implications.

a = &b;

In this case, "a" is a pionter TO an integer, and after this statement it points to the location on the stack where b is stored (&b is taking the address of b). At this point "what's stored at 'a'" - or *a - is in fact what is stored in b - both a and b are operating on the same location in memory.

such that *a = 3 will mean that what is stored in b will become 3, as if it were stated b = 3;

It may also be of note that the code b = 3 generates assembler that is slightly smaller than *a = 3 in most cases. b = 3 is interpreted such that the compiler need only utilize the address of b and move the constant 3 into that location.

On the other hand, *a = 3 cause the compiler to first must 'load' the address of a, a pointer, and then in a second step load what is stored there (the content of a), which is a pointer to an integer. It is this second value which is used to move the constant 3 into position, which in this discussion happens to be the same location where b is stored.