suppose i have 3 functions:
a()
b()
and c()
both a() and b() can called c(). if function c() is called how do i know who is the caller??

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You dont', and you shouldn't. c() should not care who called it, but just do its job and return the results (if any) to the caller.

actually i just made a code in which i need to know who's the caller but later i managed the alternate way to solve my problem.
but still i wanna know whos is the caller..???

You can pass the name of the caller into the callee:

void a(const char *caller)
{
    printf("%s called from %s\n", __func__, caller);
}

void b(const char *caller)
{
    printf("%s called from %s\n", __func__, caller);
    a(__func__);
}

int main(void)
{
    b(__func__);
    return 0;
}

The above code uses C99's predefined __func__ for simplicity. In compilers that don't support it, you'll have to come up with your own naming convention and pass string literals.

There are techniques for building a call-tree by judicious use of macros, so in your code you can do something like this:

#ifdef DEBUG
#define MAX_TREE_DEPTH 10000
size_t ct_level = 0;
const char* call_tree[MAX_TREE_DEPTH];

#define CALLTREE_ENTER call_tree[ct_level++] = __func__;
#define CALLTREE_EXIT  call_tree[ct_level--] = 0;
#else
#define CALLTREE_ENTER
#define CALLTREE_EXIT
#endif /* DEBUG */

void a(void)
{
#ifdef DEBUG
  printf("a() called from %s\n", call_tree[ct_level - 1]);
#endif
}


void b(void)
{
#ifdef DEBUG
  printf("b() called from %s\n", call_tree[ct_level - 1]);
#endif

  CALLTREE_ENTER
  a();
  CALLTREE_EXIT
}

int main(void)
{
   CALLTREE_ENTER
   b();
   CALLTREE_EXIT
   return 0;
}

Now, if you haven't compiled your code for debugging, this will have no performance impact upon you, but if you did, you get the call tree output.

I leave it as an exercise to the poster to add the macro needed so you can eliminate the #ifdef DEBUG blocks in the functions, so you could do something like this:

PRINT_CALLER(__func__);

instead if debugging is on.

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