Function pointer without arguments types?

Question

I was trying to declare a function pointer that points to any function that returns the same type. I omitted the arguments types in the pointer declaration to see what error will be generated. But the program was compiled successfully and executed without any issue.

Is this a correct declaration? Shouldn't we specify the arguments types?

#include <stdio.h>
#include <stdlib.h>

void add(int a, int b)
{
    printf("a + b = %d", a + b);
}

void (*pointer)() = &add;

int main()
{
    add(5, 5);
    return 0;
}

Output:

a + b = 10

Answer 1

Empty parentheses within a type name means unspecified arguments. Note that this is an obsolescent feature.

C11(ISO/IEC 9899:201x) §6.11.6 Function declarators

The use of function declarators with empty parentheses (not prototype-format parameter type declarators) is an obsolescent feature.

Answer 2

void (*pointer)() explains function pointed have unspecified number of argument. It is not similar to void (*pointer)(void). So later when you used two arguments that fits successfully according to definition.