Problems about function pointers in C/C++ [duplicate]

#include <stdio.h> typedef int Myfunc(int); 

Myfunc is the name of a type; it is a function taking an int argument and returning an int.

typedef int (*point_to_myfunc)(int); 

point_to_myfunc is a pointer to a function taking an int argument and returning an int. You could also have: typedef Myfunc *ptr_to_myfunc; if you wished (another name for the same type).

static Myfunc example; 

This says 'there exists a function called example of type Myfunc'.

static int example(int a) { printf("example a=%d\n", a); return 1; } 

This is a possible implementation of example. You can't use a typedef name to like Myfunc in the definition of a function of that type.

static void example2(Myfunc *f) { printf("example2\n"); f(2); } 

This is a function that takes a pointer to a Myfunc. The line f(2); invokes the function pointed at with the argument 2 and ignores the returned value.

static void example3(int (*)(int)); 

This declares example3 as a function taking a pointer to a function that takes an int argument and returns an int result. It could have been written as static void example3(point_to_myfunc); or static void example3(ptr_to_myfunc); or static void example3(Myfunc *);.

static void example3(int (*point_to_Myfunc)(int)) { printf("example3\n"); point_to_Myfunc(3); } 

This is an implementation of example3.

int main(void) { point_to_myfunc f = &example; example2(f); example3(f); return 0; } 

This program has a variable f that's a pointer to a function. Interestingly, you could have:

 point_to_myfunc f2 = example; point_to_myfunc f3 = *example; 

Etc. And they all mean the same thing.

You could also invoke them using:

 (*f2)(101); (**f3)(103); 

The standard notation for the initialization would use neither the & nor the *. If you're an old school C programmer, you may well invoke the function pointer using the (*f2)(101) notation; before the C89 standard, that was the only way to invoke function pointers. Modern style tends to be f2(101); instead.