c - right way to define pointer to the function -

everyone, have piece of code:

void foo(int var, int var1) {  printf("%d\n", var);  printf("%d\n", var1); }  void foo_for_foo( void (*some_function)(int, int)) {  int x = 5;  some_function(x, x); }  int main() {  void (*ptr_foo);  // <- here  ptr_foo = &foo;  foo_for_foo(ptr_foo);   return 0; } 

does matter how define pointer function:

1) void (*ptr_foo); 2) void (*ptr_foo)(int, int); 

my compiler receives both versions

thanks in advance explanations of difference

the 2 forms not equivalent.

void (*ptr_foo) not function pointer @ all. it's normal, non-function void pointer. parentheses superfluous , misleading. it's if had written void* ptr_foo.

void (*ptr_foo)(int, int) proper way declare function pointer function taking 2 ints , returning void.

the reason works because in c, void pointers implicitly convertible other type of pointer. is, can assign other pointer void*, , can assign void* other pointer.

but fact works in example accident of syntax. cannot in general replace void (*foo)(int, int) void (*foo).

if try doing some_function in argument list foo_for_foo, compiler complain when try invoke some_function because not function pointer.

similarly, if foo function happened return int instead of void, notice problem right away. declaring ptr_foo int (*ptr_foo) have resulted in error on statement ptr_foo = &foo because unlike void pointers, int pointers not implicitly convertible other pointer types.

in short, always use second form. 1 correct in general, despite fluky case.