Function pointer as parameter

问题

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;
}

回答1:

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;
}

回答2:

Replace void *disconnectFunc; with void (*disconnectFunc)(); to declare function pointer type variable. Or even better use a typedef:

typedef void (*func_t)(); // pointer to function with no args and void return
...
func_t fptr; // variable of pointer to function
...
void D::setDisconnectFunc( func_t func )
{
    fptr = func;
}

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

回答3:

You need to declare disconnectFunc as a function pointer, not a void pointer. You also need to call it as a function (with parentheses), and no "*" is needed.

来源：https://stackoverflow.com/questions/2582161/function-pointer-as-parameter