type of function pointer

Question

I have a short program with which I am going to test the “infinite recursion”. However, I meet another tough question associated to function pointer...

the program is as follow:

#include <stdio.h>

int repeat(char (*func)(int, int), int a) {
  func(a, a+1);
  return repeat(func, a+1);
}

char f (int a , int b){
  printf("%d\n", a);
  //return (double)(a + b);
  return 'c';
}

int main(int argc, char const *argv[])
{
  repeat(&f, 1);
  return 0;
}

As for me, I think the 'f' in “repeat(&f, 1)” has the type of “char *” and the 'func' in “int repeat(char (*func)(int, int), int a)” has the type of “char *”,too.

However, it seems strange. If I change “(char (*func)(int, int), int a) ” to “(char func(int, int), int a) ”, the compiler(testing in gcc and vc6) do not give any warnings and the running result is the same.

If I change “(char (*func)(int, int), int a) ” to “(double func(int, int), int a) ”, the compiler throws:

warning: incompatible pointer types passing 'char ()(int, int)' to parameter of type 'double ()(int, int)' [-Wincompatible-pointer-types] repeat(&f, 1);

It seems that the compiler treat “double (int, int)” and “double (*)(int, int)” the same stuff. In other words, If I change “(char (*func)(int, int), int a)” to “double (*func)(int, int)”, the compiler will throw the same message (I have tested it):

warning: incompatible pointer types passing 'char ()(int, int)' to parameter of type 'double ()(int, int)' [-Wincompatible-pointer-types] repeat(&f, 1);

God... Well, it is not the end. After that, I change the “repeat(&f, 1)” to “repeat(f, 1)” with other parts unchanged. That is, the whole program is:

#include <stdio.h>

int repeat(char (*func)(int, int), int a) {
  func(a, a+1);
  return repeat(func, a+1);
}

char f (int a , int b){
  printf("%d\n", a);
  //return (double)(a + b);
  return 'c';
}

int main(int argc, char const *argv[])
{
  repeat(f, 1);
  return 0;
}

the compiler do not give any warnings and the running results are the same.

So I am confused that all of the following paris yields no warnings and correct results.

    1   repeat(&f, 1)   int repeat(char (*func)(int, int), int a)
    2   repeat(&f, 1)   int repeat(char func(int, int), int a) 
    3   repeat(f, 1)    int repeat(char (*func)(int, int), int a) 
    4   repeat(f, 1)    int repeat(char func(int, int), int a) 

Of course the 'f' in “repeat(f, 1)” has the type of “char (int, int)”, and the 'func' in "int repeat(double (func)(int, int), int a)" has the type of “char ()(int, int)”.

You can infer this by changing “int repeat(char (*func)(int, int), int a) ” to “int repeat(double (*func)(int, int), int a) ” and check the warning message:

warning: incompatible pointer types passing 'char (int, int)' to parameter of type 'double ()(int, int)' [-Wincompatible-pointer-types] repeat(f, 1);*

Who can give some comments? I totally think that only the first of the four is correct regarding the type.

Answer 1

When used as a function parameter, both of the syntax

foo(char bar(int, int), int a)  

and

foo(char (*bar)(int, int), int a)

are identical from the compiler's standpoint.

Also note that when you call function foo, the name of the function passed to it is treated as pointer to that function. These both calls are valid

foo(&f, 1)
foo(f, 1)

As for me, I think the f in repeat(&f, 1) has the type of char * and the func in int repeat(char (*func)(int, int), int a) has the type of char *,too.

No . f and func both are of type char (*)(int, int), i.e. they are pointer to a function that takes two int as argument and returns char

Answer 2

Type of the parameter:

The order of precendence imposed by () in the syntax of :

    int repeat(char (*func)(int, int), int a)

means that (*func)(int,int) is a function taking two int arguments and returning a char. The * means that func a pointer to a function returning char.

The following more modern syntax would have worked as well :

    int repeat(char func(int, int), int a)

ATTENTION, the following looks close to the first one, but would be totally different, the function argument would be a pointer to a function returning a char *:

    int repeat(char *func(int, int), int a)

And yes, just for completeness, int repeat((char *)func(int, int), int a) would be a syntax error, as (char*) would be understood as a cast operator whereas a type is expected.

Use of the f and &f syntax:

The C11 standard section 6.5.3.2 states that "The unary & operator yields the address of its operand. If the operand has type ‘‘type’’, the result has type ‘‘pointer to type’’":

• f being a function taking two int as argument and returning a char
• So &f is a pointer to a function taking two int as argument and returning a char.

But the standard in section 6.3.2.1/4 also states that "A function designator is an expression that has function type. (...) a function designator with type ‘‘function returning type’’ is converted to an expression that has type ‘‘pointer to function returning type’’":

• f being a function taking two int as argument and returning a char
• if provided in an expression (of course not in the syntax of a function call), f is also understood as pointer to a function taking two int as argument and returning a char.

Nowadays, you'll find both syntaxes.

Historically speaking, in former times the use of * in the (*func)() syntax was mandatory to call the function pointed to (see original K&R).