右值绑定到非const引用

2014-01-13 06:28:00 304 阅读 AI 写同款 · GEO 优化AI 选题数据监测一键分发 ›

[错误分析][Error]no match for ‘operator＜＜’：当右值遇上非const引用本文深入分析了C++编译错误`[Error]no match for 'operator右值与非const引用的绑定问题。通过典型代码示例和详细解释，展示了如何正确重载运算符`operator右值概念的直观理解。文章还总结了运算符重载的最佳实践和现代C++中的相关特性，帮助开发者编写更健壮的代码。阅读详情

c++ 2003

13.3.1.5

During overload resolution, the implied object argument is indistinguishable from other arguments. The
implicit object parameter, however, retains its identity since conversions on the corresponding argument
shall obey these additional rules:
— no temporary object can be introduced to hold the argument for the implicit object parameter;
— no user-defined conversions can be applied to achieve a type match with it; and
— even if the implicit object parameter is not const-qualified, an rvalue temporary can be bound to the
parameter as long as in all other respects the temporary can be converted to the type of the implicit
object parameter.

最后一点说的是右值可以绑定到非const引用是什么情况?
求举例说明.

...全文

C++11新特性之 rvalue Reference(右值引用) 右值引用可以使我们区分表达式的左值和右值。C++11引入了右值引用的概念，使得我们把引用与右值进行绑定。使用两个“取地址符号”：int&& rvalue_ref = 99;需要注意的是，只有左值可以付给引用，如：int& ref = 9; 我们会得到这样的错误： “invalid initialization of non-const reference of type int& from an 阅读详情

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vipcxj 2014-01-13

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我擦，杯具，已经结贴了~

vipcxj 2014-01-13

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引用 4 楼 vipcxj 的回复:


struct S {
    void f() const & { std::cout << typeid(this).name() << "-const-lv" << std::endl; }
    void f() & { std::cout << typeid(this).name() << "-lv" << std::endl; }
    void f() &&{ std::cout << typeid(this).name() << "-rv" << std::endl; }
    void f() const &&{ std::cout << typeid(this).name() << "-const-rv" << std::endl; }
};

const S fun()
{
    const S s;
    return s;
}
 
int main(){
    S s;
    s.f();            // prints "lvalue"
    std::move(s).f(); // prints "rvalue"
    S().f();          // prints "rvalue"
    const S cs;
    cs.f();
    fun().f(); 
/*
注意这里，注释了这个才能编译过去，之所以编译不过去，是因为这条语句让编译器难以绝定改重载哪个，可能是void f() const &，又可能是void f() const &&。换句话说常右值引用对编译器来说和常左值引用没区别
*/
}

如果注释了fun().f();得到的结果是 P1S-lv P1S-rv P1S-rv PK1S-const-lv

其实上面的例子中fun()返回的显然是右值，而且还是常的，照理应该重载void f() const && 但事实上它就是编译不过去，编译器认为void f() const && 和 void f() const &是一样滴。换句话说常的右值木有啥意义，右值只能绑定到非常版本的成员函数，个人理解，可能不对~

vipcxj 2014-01-13

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struct S {
    void f() const & { std::cout << typeid(this).name() << "-const-lv" << std::endl; }
    void f() & { std::cout << typeid(this).name() << "-lv" << std::endl; }
    void f() &&{ std::cout << typeid(this).name() << "-rv" << std::endl; }
    void f() const &&{ std::cout << typeid(this).name() << "-const-rv" << std::endl; }
};

const S fun()
{
    const S s;
    return s;
}
 
int main(){
    S s;
    s.f();            // prints "lvalue"
    std::move(s).f(); // prints "rvalue"
    S().f();          // prints "rvalue"
    const S cs;
    cs.f();
    fun().f(); 
/*
注意这里，注释了这个才能编译过去，之所以编译不过去，是因为这条语句让编译器难以绝定改重载哪个，可能是void f() const &，又可能是void f() const &&。换句话说常右值引用对编译器来说和常左值引用没区别
*/
}

如果注释了fun().f();得到的结果是 P1S-lv P1S-rv P1S-rv PK1S-const-lv

iamnobody 2014-01-13

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引用 2 楼 supermegaboy 的回复:

[quote=引用楼主 mingliang1212 的回复:] c++ 2003 13.3.1.5 During overload resolution, the implied object argument is indistinguishable from other arguments. The implicit object parameter, however, retains its identity since conversions on the corresponding argument shall obey these additional rules: — no temporary object can be introduced to hold the argument for the implicit object parameter; — no user-defined conversions can be applied to achieve a type match with it; and — even if the implicit object parameter is not const-qualified, an rvalue temporary can be bound to the parameter as long as in all other respects the temporary can be converted to the type of the implicit object parameter. 最后一点说的是右值可以绑定到非const引用是什么情况? 求举例说明.

请注意看前面一段第一句话的说明： For non-static member functions, the type of the implicit object parameter is “reference to cv X” where X is the class of which the function is a member and cv is the cv-qualification on the member function declaration. [Example: for a const member function of class X, the extra parameter is assumed to have type “reference to const X”. ] 隐含形参是否cv-qualified是依据成员函数的const而定的，当一个临时对象调用非const的成员函数时，即使此时隐含形参的类型是reference to X，这个临时对象也必须保持到实参身上的转换完成为止（否则无法保证不发生非确定性结果），例如：


struct A
{
    void Do() const
    {
        std::cout<<"const"<<std::endl;
    }
    void Do()
    {
        std::cout<<"noconst"<<std::endl;
    }
};

struct B : A
{

};

int main( void )
{
    B().Do();
    return 0;
}

[/quote] 又看多了几遍,好像看懂了,之前在纠结的是这句:as long as in all other respects the temporary can be converted to the type of the implicit object parameter 我以为all other respects 会包括这种情况: A &r = A(); (显然是不行的),,所以我不知道它的all other respects 是指什么..

飞天御剑流 2014-01-13

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引用楼主 mingliang1212 的回复:

c++ 2003 13.3.1.5 During overload resolution, the implied object argument is indistinguishable from other arguments. The implicit object parameter, however, retains its identity since conversions on the corresponding argument shall obey these additional rules: — no temporary object can be introduced to hold the argument for the implicit object parameter; — no user-defined conversions can be applied to achieve a type match with it; and — even if the implicit object parameter is not const-qualified, an rvalue temporary can be bound to the parameter as long as in all other respects the temporary can be converted to the type of the implicit object parameter. 最后一点说的是右值可以绑定到非const引用是什么情况? 求举例说明.


struct A
{
    void Do() const
    {
        std::cout<<"const"<<std::endl;
    }
    void Do()
    {
        std::cout<<"noconst"<<std::endl;
    }
};

struct B : A
{

};

int main( void )
{
    B().Do();
    return 0;
}

偏爱风流 2014-01-13

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见过const引用的，但没见过非const引用。不知道这样算不算非const引用 int &&a = (1+2); 某书见过const的右值引用 const int& a = (1+2);

右值绑定到非const引用

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