Printing Function Address

Question

I've been trying a way to figure out how to print the address of the function, This is what I came up with

#include<stdio.h>
int test(int a, int b)
{
     return a+b;
}
int main(void)
{
     int (*ptr)(int,int);
     ptr=&test;
     printf("The address of the function is =%p\n",ptr);
     printf("The address of the function pointer is =%p\n",&ptr);
     return 0;
}

It o/p something like this without any warning and errors

address of function is =0x4006fa
address of function pointer  is =0x7fffae4f73d8

My question whether using %p format specifier is the correct way to print the address of the function or is there any other way to do so?

The name of the function is the address of it. You can achieve the same result by using: `printf ("Address of function: %p\n", test);` And yes: %p is the correct format specifier to print pointer values (memory addresses) because the implementation knows how your pointers look like. — mcleod_ideafix, Mar 23 '15 at 05:04
@mcleod_ideafix: That's is incorrect. `%p` can be used with `void *` pointers and (after conversion) with other object pointers. `%p` cannot be used with function pointers. — AnT stands with Russia, Mar 23 '15 at 05:11
I've just tested it on Linux and gcc. Compiling with -Wall doesn't trigger any warnings, and it prints the actual address of main — mcleod_ideafix, Mar 23 '15 at 12:13

score 9 · Accepted Answer · answered Mar 23 '15 at 05:09

9

This is not correct. %p is only for object pointer types (in fact, void * specifically). There is no format specifier for a function pointer.

answered Mar 23 '15 at 05:09

M.M

138,810
21
208
365

score 1 · Answer 2 · answered Mar 23 '15 at 05:08

1

Without using the separate pointer for print the address of the function, You can use the name of the function in the printf.

 printf("The address of the function is =%p\n",test);

For printing the address in the hexa-decimal format you can use the %p.

answered Mar 23 '15 at 05:08

sharon

734
6
15

1

This fails when compiling with `-pedantic-errors`, with the message `format specifies type 'void *' but the argument has type 'void (*)()'` – Rafael Eyng Jan 30 '20 at 17:15

Printing Function Address

2 Answers2