转换因参数类型而异的函数指针 [英] Cast function pointers that differs by argument type

问题描述

为什么这不合法：

class Base
{
public:
    Base(){};
    virtual ~Base(){};
};

class Derived : public Base{};

void takeDerived(Derived * c){};

// main
void(*ptr)(Base*) = static_cast<void(*)(Base*)>(&takeDerived); // doesn't work

// but this work ok, as well as reinterpret_cast
// void(*ptr)(Base*) = (void(*)(Base*))(&takeDerived);

派生是基本。为什么不能将其强制转换为函数参数？例如，即使不进行强制转换，我也可以轻松做到这一点：

Derived is a Base. Why can't it be casted in function parameter? For example, I can do this easily even without casting:

void takeBase(Base* c){};
takeBase(new Derived{});

推荐答案

它就是这样设计的。 基本不是派生的。类派生的 is-a 关系不可逆。

It's just designed to be that way. A Base isn't a Derived. The is-a relationship for class derivation can't be reversed.

函数参数的类型表示接受，而对象的类型表示适合。强制转换函数的类型会更改其接受的内容。允许函数接受最初不接受的内容是危险的。

The type of a function parameter means "accept", while the type of an object means "fit". Casting the type of a function changes what it accepts. It's dangerous to allow a function to accept whatever isn't what it originally accepts.

考虑以下代码：

class Base {};
class Derived : public Base {
    public:
    int t;
    void useDerived() {}
};

void useDerived(Derived *d){
    d->useDerived();
}

如果 Base 对象已传递？

Base b;
((void(*)(Base*))useDerived) (&b);

更糟糕的是，如果另外导出 Base

Even worse, what if another derivation of Base is passed?

class AnotherDerived : public Base {
    public:
    double t;
    void useDerived() {}
};
AnotherDerived ad;
((void(*)(Base*))useDerived) (&ad);

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