在C中实现回调函数

Question

I am a newbie to C. I am trying to implement callback function using function pointers.

I am getting an error

:test_callback.c:10: error: expected identifier or ‘(’ before ‘void’

when I try to compile the following program:

#include<stdio.h>

void (*callback) (void);

void callback_proc ()
{
  printf ("Inside callback function\n");
}

void register ((void (*callback) (void)))
{
  printf ("Inside registration \n");
  callback (); /* Calling an initial callback with function pointer */
}

int main ()
{
  callback = callback_proc;/* Assigning function to the function pointer */
  register (callback);/* Passing the function pointer */
  return 0;
}

---

What is this error?Can anyone help?

Answer 1

1. register is a C keyword: Use another name for the function.

2. You have extra parantheses around the callback parameter. It should be:

    void funcName(void (*callback) (void)) 

Answer 2

I would recommend to use a typedef

#include<stdio.h>

typedef void (*callback_t) (void);
callback_t callback;

void callback_proc(void)
{
    printf ("Inside callback function\n");
}

void reg( callback_t _callback )
{
    printf ("Inside registration \n");
    _callback();
}

int main ()
{
    callback = callback_proc;
    reg(callback);

    return 0;
}

EDIT: removed the register issue

Answer 3

您不能将'register'用作函数名，因为它是C关键字。

Answer 4

2 problems:

• you can't use the name register as it's a keyword (not used often anymore, but it's still there)

• change the definition of the function from

   void wasRegister((void (*callback) (void))) 

  to:

   void wasRegister(void (*callback) (void)) 

  (get rid of the parens around the parameter's declaration.

Also you might get a warning about callback_proc() not having a matching delaration to the callback variable (depending on how you compile the program - as C or C++), so you might want to change its declaration to:

void callback_proc (void)

to make it explicit that it takes no parameters.

Answer 5

Have a look at type safe callbacks from ccan . Its one thing to expose a typed function pointer for the world to use, its another to ensure sane casting.

Answer 6

#include<stdio.h>

typedef void (*callback_func) (void);

static callback_func the_callback = 0;

void process (void)
{
  printf ("Inside process function\n");
}

void callback_register (callback_func cb)
{
  the_callback = cb;
  printf ("Inside registration \n");
}

void callback(void)
{
    the_callback();
}

int main (void)
{
  callback_register(process); /* Passing the function pointer */
  callback();
  return 0;
}

Declaring the_callback static would make more sense if this code was modularized and then you would be forced to call callback_register in order to set it, and callback in order to call it - the_callback would not be accessible outside of the implementation (.c) only the function declarations would be in the header (.h).