CODEWITHSUNDEEP
cfunction-pointers
Destructor
Destructor

Reputation: 14438

c function pointer & user defined function identifier collision

i tested this simple program of function pointer.

#include <stdio.h>
#include <conio.h>
void ptr();
void fun()
{
    printf("fun() is called\n");
}
void ptr()
{
    printf("ptr() is called\n");
}
int main()
{
  void(*ptr)();
  ptr=fun;
  ptr();        // Why it works fine?
  //(ptr)();
  void(*a)()=fun;
  (*a)();
  _getch();
  return 0; 
}

output:

fun() is called

fun() is called

I have a question that why the function call statement ptr(); won't call user defined function ptr(), but calls fun() function? Why parentheses around ptr isn't necessary? What changes my program will require so my program displays output as following.

ptr() is called

fun() is called

Upvotes: 0

Views: 171

Answers (2)

Jonathan Leffler
Jonathan Leffler

Reputation: 754700

It is an application of the usual scoping rules; local variables hide global variables of the same name.

Or, more accurately, variables defined in an inner scope hide variables of the same name in any outer scopes.

And scopes are basically controlled by sets of braces around blocks of statements (simplifying a bit).

You can get GCC to warn you about such problems with -Wshadow.

The other part of the issue is that you can call functions via a pointer to function in (at least) two ways:

(*ptr_to_function)();    // Pre-standard notation
ptr_to_function();       // Notation allowed by standard C

So the call ptr(); is identical to (*ptr)(); in your program.

What changes does my program require to display:

ptr() is called
fun() is called

The simplest change is to remove the line void (*ptr)() = fun;.

You could also do something like:

#include <stdio.h>

void fun(void)
{
    printf("fun() is called\n");
}
void ptr(void)
{
    printf("ptr() is called\n");
}
int main(void)
{
  void (*xyz)(void) = ptr;
  void (*ptr)(void) = fun;
  xyz();
  ptr();
  return 0; 
}

Upvotes: 2

Yu Hao
Yu Hao

Reputation: 122483

When you define a local variable ptr inside main, it shadows the same name function ptr which is global. So when you call ptr(), it's the local function pointer ptr that is called.

To make the function ptr being called, you need to make sure the local variable ptr is out of scope, as this artificial example:

int main()
{
    {
      void(*ptr)();
      ptr=fun;
    }
    ptr();
}

Upvotes: 4

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