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why typedef void (*f_ptr)(int); and typedef void (*f_ptr)(); work the same in C?

I accidentally forgot to add (int) when defining a function pointer, and my program still worked. I wold like to know if there is any case where it would not work. my code:

#include <stdio.h>

void f1(int var)
{
    printf("this is f1 and var is: %d\n", var);
}

void f2(int var)
{
    printf("this is f2 and var is: %d\n", var);
}

void f3(int var)
{
    printf("this is f3 and var is: %d\n", var);
}
typedef void (*f_ptr)(int);
// pq eu poderia escrever: typedef void (*f_ptr)(); e o programa funcionaria normalmente?
typedef int n_casa;

int main()
{
    f_ptr ptr[] = {f1, f2, f3};

    int c = 0;
    while (c < 3)
    {   
        ptr[c](c);
        ++c;
    }

    return 0;
}

both typedef void (*f_ptr)(int); and typedef void (*f_ptr)(); worked in my program.

Solution

  • They are different.

    typedef void (*f_ptr)(int) declares a funciton pointer that takes only one int argument, and returns nothing.

    While for typedef void (*f_ptr)(), the function pointer takes an UNSPECIFIED number of arguments, and returns nothing.

    According to the SEI CERT C Coding Standard, it is recommended to explicitly specify void when a function accepts no arguments.