C++ 函数指针作为参数
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Function pointer as parameter
提问by Roland Soós
I try to call a function which passed as function pointer with no argument, but I can't make it work.
我尝试调用一个不带参数作为函数指针传递的函数，但我无法让它工作。
void *disconnectFunc;
void D::setDisconnectFunc(void (*func)){
disconnectFunc = func;
}
void D::disconnected(){
*disconnectFunc;
connected = false;
}
回答by GManNickG
The correct way to do this is:
正确的做法是：
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by Nikolai Fetissov
Replace void *disconnectFunc;with void (*disconnectFunc)();to declare function pointer type variable. Or even better use a typedef:
替换void *disconnectFunc;与void (*disconnectFunc)();要声明的函数指针类型的变量。或者甚至更好地使用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by WhirlWind
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.
您需要将 disconnectFunc 声明为函数指针，而不是 void 指针。您还需要将其作为函数调用（带括号），并且不需要“*”。
