在类中声明的朋友操作符的左手参数的隐式转换

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问题描述

我使用CRTP为类提供模板参数相关的函数添加，在这种情况下添加 operator + 和 operator + = ，使用模板类 ImplAdd 。对于前者，应该对两个参数执行隐式转换，这意味着我必须使用类中的friend运算符，如下所示：

I am using CRTP to provide template-argument dependent addition of functions to a class, in this case the addition of operator + and operator +=, using the template class ImplAdd. For the former, implicit conversions should be performed on both arguments, which means I have to use an in-class friend operator like this:

template<class Type, bool active>
struct ImplAdd{
    virtual int get_val_() const = 0;
    virtual void set_val_(int) = 0;
};

//if activated is true, the operators + and += will be defined
template<class Type>
class ImplAdd < Type, true > {
    virtual int get_val_() const = 0;
    virtual void set_val_(int) = 0;
    Type* this_(){ return (Type*)this; }
public:
    Type& operator +=(const Type& x){
        set_val_(get_val_() + x.get_val_());
        return *this_();
    }

    //This should enable conversions on the lefthand argument
    friend Type& operator+(const Type& lhs, const Type& rhs){
        Type ret = lhs;
        return ret += rhs;
    }
};

这是必需的，因为类实际继承自 ImplAdd 为这些常量定义常量值和唯一值类型，很像一个范围化的枚举。

This is needed because classes that actually inherit from ImplAdd define constant values and a unique value type for those constants, much like a scoped enum.

//by using true as the template argument, the operators + and += will be defined
class MyEnum : public ImplAdd<MyEnum, true>{
    int get_val_() const override{
        return (int)value;
    }
    void set_val_(int v) override{
        value = (ValueT)v;
    }
public:
    enum class ValueT{
        zero, one, two
    };
private:
    typedef  int UnderlyingT;
    ValueT value;
public:
    static const ValueT zero = ValueT::zero;
    static const ValueT one = ValueT::one;
    static const ValueT two = ValueT::two;
    MyEnum(ValueT x) : value(x){}
    MyEnum(const MyEnum& other) : value(other.value){}
};

从我的角度来看，下面的代码现在很容易编译，但它不能。 >

From my perspective, the following code should now easily compile, but it doesn't.

int main(int argc, char* argv[])
{
    MyEnum my = MyEnum::zero;                //works
    my += MyEnum::one;                       //works
    my = MyEnum(MyEnum::zero) + MyEnum::two; //works
    my = MyEnum::zero + MyEnum(MyEnum::two); //ERROR C2676
    my = MyEnum::zero + MyEnum::two;         //ERROR C2676
    MyEnum my2 = my + my;                    //works
    return 0;
}

对于标有C2676的行，将打印以下错误消息： p>

For both lines marked with C2676, the following error message is printed:

error C2676: binary '+' : 'const MyEnum::ValueT' does not define this operator or a conversion to a type acceptable to the predefined operator

我做错了什么？不是使用定义一个操作符作为类的朋友通常的方式来启用隐式转换两个参数？

What am I doing wrong? Isn't the use of defining an operator as an in-class friend the common way to enable implicit conversion on both arguments? If not, how can I do it in this case?

myenum

在类中声明的朋友操作符的左手参数的隐式转换

问题描述

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