C++ 定义到基类的转换 [英] C++ Define Conversion to Base Class

问题描述

我在定义和使用到基类的转换运算符时遇到了困难.考虑以下几点:

I'm having difficulty defining and using a conversion operator to a base class. Consider the following:

class base 
{
public:
    base(const base&);
    base& operator=(const base&);
    //some stuff
};
class child : public base
{
public:
    operator base() const;
    //some more stuff
};

int main()
{
    child c;
    base b=c;
    b=c;
}

如果我尝试将子项转换为基数，则永远不会调用 operator base()(即使我使转换显式).而是直接调用基类中的复制构造函数或赋值运算符，无需转换.

If I attempt to convert a child to a base, operator base() is never called (even if I make the conversion explicit). Instead, the copy constructor or assignment operator from the base class is called directly, without a conversion.

当将 child 分配(或复制构造)到 base 时，如何调用 operator base()?

How can I cause operator base() to be called when a child is assigned (or copy-constructed) to a base?

推荐答案

到基类的转换函数在语法上是合法的，但永远不会被调用.来自标准草案 n3337:

A conversion function to a base class is syntactically legal, but it will never be called. From standard draft n3337:

12.3.2 转换函数 [class.conv.fct] §1

[...] 从来没有使用转换函数来转换一个(可能cv 限定)对象到(可能有 cv 限定)相同的对象类型(或对它的引用)，到它的(可能是 cv 限定的)基类类型(或对它的引用)，或(可能是 cv 限定的)void.[...]

[...] A conversion function is never used to convert a (possibly cv-qualified) object to the (possibly cv-qualified) same object type (or a reference to it), to a (possibly cv-qualified) base class of that type (or a reference to it), or to (possibly cv-qualified) void. [...]

该语言有不同的从派生到基类的转换机制.阅读对象切片.简而言之 - 它会自动为您完成.

The language has got different mechanism for conversions from derived to base. Read about object slicing. To put it simply - it's done automatically for you.

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