c++ - C++运算符的继承

Question

我在该网站上阅读过许多类似的问题，但没有任何内容能回答我正在尝试做的事情。

public class base {
public:
    base(){
        //Default Constructor
    }

    base( int num ){
        // use num to create base
    }

    base& operator=( base&& _data ){
        // do move assignment stuff
    }
};

public class derived : public base {
public:
    derived() : base() {
        int num1;
        //Do some stuff

        // Now I want to assign the base of this class with a new base
        base::operator=( Base( num1 ) );
    }
};

我想在派生类的构造过程中对基类调用移动赋值(或只是常规赋值)。这样，派生类可以在创建其基类之前解析一些信息。这似乎行不通。有人有什么想法吗？

Answer 1

请记住，基类在进入构造函数体之前被初始化。因此，使用您的方法，您首先初始化 base ，然后通过赋值覆盖它。那可不好。

使用委托(delegate)构造函数和私有(private)辅助函数:

class derived : public base {
private:
    static int help () { /* Do some stuff */ }

    // private constructor to be called with the helper values
    derived (int i) : base (i) { }
public:
    derived() : derived (help ()) { }
};

当然，在这种情况下，您可以通过以下方式定义您的构造函数

derived () : base (help ()) { }

然而，如果您必须为基类构造函数计算多个参数值，则上述委托(delegate)构造函数方法会变得很有用:

class base {
public:
    base (int i, double d);
};

class derived : public base {
private:
    struct BaseParams { int i; double d; };
    static BaseParams help () {
        BaseParams p;
        /* Do some stuff and set p.i and p.d */;
        return p;
    }

    // private constructor to be called with the helper values
    derived (BaseParams const & p) : base (p.i, p.d) { }
public:
    derived() : derived (help ()) { }
};

如果你真的想构造一个 base 对象并将其移动进去，请使用 base 的移动构造函数:

class base {
public:
    base (int i, double d);
    base (base &&);
};

class derived : public base {
private:
    static base help () {
        /* Do some stuff and finally construct a base object */;
        return base { /* calculated parameters for constructor of base */ };
    }

    // private constructor to be called with a base object to move from
    derived (base && b) : base (std::move (b)) { }
public:
    derived() : derived (help ()) { }
    // or simply derived () : base (help ()) { }
};