我有一个C库函数,它需要一个函数指针用于回调,我想传入一个C成员函数. C函数修改了一个成员变量,所以我不能使用静态自由函数(如几个类似的帖子所示).我的尝试(如下所示)因编译器错误而失败.
这篇文章最接近我的要求：
Using a C++ class member function as a C callback function
如果没有静态函数,我怎么能这样做？谢谢！
test.h
#ifndef TEST_H_
#define TEST_H_
#ifdef __cplusplus
extern "C" {
#endif
typedef void (*handler_t)(int foo,void *bar);
void set_handler(handler_t h);
#ifdef __cplusplus
}
#endif
#endif
test.c的
#include "test.h"
#include <stdlib.h>
static handler_t handler_ = NULL;
void set_handler(handler_t h) {
handler_ = h;
}
void handle_event(int foo,void *bar) {
if (handler_ != NULL) handler_(foo,bar);
}
TEST.CPP
#include "test.h"
#include <iostream>
using namespace std;
class Foo {
public:
Foo() : ctr_(0) {};
// handler needs to access non-static variable,so it can't be static
void handler(int foo,void *bar) { ++ctr_; }
private:
int ctr_;
};
int main(int argc,char **argv) {
// error: can't convert to "void (*)(int,void*)"
set_handler(&Foo::handler);
cout << "done" << endl;
return 0;
}
GCC barf
$gcc test.cpp test.c test.cpp: In function ‘int main(int,char**)’: test.cpp:18: error: cannot convert ‘void (Foo::*)(int,void*)’ to ‘void (*)(int,void*)’ for argument ‘1’ to ‘void set_handler(void (*)(int,void*))’
解决方法
至少使用handler_t签名是不可能的.
虽然您可以在.cpp上创建一个自由函数来包装成员调用,但您需要一个指向Foo实例的指针：
void my_wrap(int foo,void* bar) {
Foo* some_foo_instance = ...;
some_foo_instance->handler(foo,bar);
}
int main(int argc,char **argv) {
set_handler(&my_wrap);
}
您需要一些void *来将Foo实例作为处理程序属性传递：
// Header
typedef void (*handler_t)(int foo,void *bar,void* arg1);
void set_handler(handler_t h,void* arg1);
// Impl.
void set_handler(handler_t h,void* arg1) {
handler_ = h;
handler_arg1_ = arg1;
}
// cpp
void my_wrap(int foo,void* bar,void* arg1) {
Foo* some_foo_instance = static_cast<Foo*>(arg1);
some_foo_instance->handler(foo,bar);
}
// main
int main(int argc,char **argv) {
Foo some_concrete_instance;
set_handler(&my_wrap,static_cast<void*>(&some_concrete_instance));
}
