c++ - move 后右值是否有效？

Question

假设我们有以下类:

class Foo
{
public:
    Foo() { _bar=new Bar };
    Foo(const Foo &right) { _bar=new Bar(right.bar); };
    Foo(Foo &&right) { _bar=right._bar;  right.bar=new Bar(); };

    ~Foo() { delete _bar; }

    Foo &operator=(const Foo &right) { _bar->opertor=(right.bar); return *this;}
    Foo &operator=(Foo &&right) { std::swap(_bar, right._bar); return *this;}

    void func() { _bar->test=1 };

private:
    Bar *_bar;
};

将其更改为以下内容并期望最终用户知道在执行 move 后右值不再有效(因为调用赋值运算符以外的任何东西都可能崩溃)是否合法？

class Foo
{
public:
    Foo() { _bar=new Bar };
    Foo(const Foo &right) { _bar=new Bar(right.bar); };
    Foo(Foo &&right) { _bar=right._bar;  right.bar=nullptr; };

    ~Foo() { if(_bar != nullptr) delete _bar; }

    Foo &operator=(const Foo &right) 
    { 
         if(_bar == nullptr) 
             _bar=new Bar();
         _bar->opertor=(right.bar); 
         return *this;
    }

    Foo &operator=(Foo &&right)
    {
        if(_bar != nullptr)
            delete _bar;  
        _bar=right._bar; 
        right._bar=nullptr; 
        return *this;
    }

    void func() { _bar->test=1 };

private:
    Bar *_bar;
};

我担心的是 func(以及类中的所有其他函数)假设 _bar 存在。

Answer 1

原则上它可能会变得无效，但您可能希望考虑将其保留在可分配状态(您的原始实现，因此编辑，没有这样做)。这将遵循标准库的策略，即:

Unless otherwise specified, all standard library objects that have been moved from are placed in a valid but unspecified state. That is, only the functions without preconditions, such as the assignment operator, can be safely used on the object after it was moved from

我建议重新实现赋值运算符，以便它用新构造的对象交换“this”对象。这通常是避免在执行分配时引入不正确行为的好方法。