功能指针作为参数 [英] Function pointer as parameter

问题描述

我尝试调用一个不带任何参数的函数指针，但是我无法使其正常工作.

I try to call a function which passed as function pointer with no argument, but I can't make it work.

void *disconnectFunc;

void D::setDisconnectFunc(void (*func)){
    disconnectFunc = func;
}

void D::disconnected(){
    *disconnectFunc;
    connected = false;
}

推荐答案

执行此操作的正确方法是:

The correct way to do this is:

typedef void (*callback_function)(void); // type for conciseness

callback_function disconnectFunc; // variable to store function pointer type

void D::setDisconnectFunc(callback_function pFunc)
{
    disconnectFunc = pFunc; // store
}

void D::disconnected()
{
    disconnectFunc(); // call
    connected = false;
}

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