Question

请考虑这个例子，我们如何在函数中强制隐式转换，其第二个参数是指向成员函数的指针。在函数的参数列表中明确地进行转换并不是我现在想要的。相反，我希望那个编译器以某种方式像FIRST参数那样做...

struct Base
{
    virtual ~Base() = 0 {}
};

struct Derived : public Base
{
    void f(){}
};

typedef void(Base::*polymorph)();

// how do I force IMPLICIT conversion here: EDIT: (polymorph type work only for polymorph pointer type no conversion)
void func(Base* arg1, polymorph arg2) // void* arg2, (void*) arg2 etc...  dosn't work
{
    polymorph temp = reinterpret_cast<polymorph>(arg2); // to achive this
}

int main()
{
    Derived* test = new Derived;
    // first parameter work but another gives an error
    func(test, &Derived::f); // BY NOT CHANGING THIS!
    delete test;
    return 0;
}

Answer 1

尽可能干净。代码如下。但是当我真正调用“temp”时，我不知道是谁会引用“this”指针。

typedef void(Base::*polymorph)(); 

void func(Base* arg1, polymorph arg2)
{ 
    polymorph temp = arg2;
} 


int main() 
{ 
    Derived* test = new Derived; 
    // first parameter work but another gives an error 
    func(test, static_cast<polymorph>(&Derived::f)); 
    delete test; 
    return 0; 
}

如何在多态中强制隐式转换？

1 个答案: