c++ - 在类内部定义 friend 用户定义的文字运算符

Question

为什么在类 give error 中定义用户定义的文字？

class test
{
    long double x;
    public:    
    friend test operator""_UNIT(long double v)
    {
        test t;
        t.x = v;
        return t;
    }       
};

int main()
{
    test T = 10.0_UNIT;        
    return 0;
}

错误:

unable to find numeric literal operator 'operator""_UNIT'

注意:可以定义any friend function在类里面。

class test
{
    int x;
    public:
    test():x(10) {}
    friend std::ostream& operator<< (std::ostream& o, test t)
    {
        o << t.x ;
        return o;
    }
};

int main() {
    test T;
    std::cout << T;
    return 0;
}

可以定义相同的 friend 用户定义文字outside the class .

class test
{
    long double x;
    public:    
    friend test operator""_UNIT(long double v);
};

test operator""_UNIT(long double v)
{
    test t;
    t.x = v;
    return t;
}

int main()
{
    test T = 10.0_UNIT;        
    return 0;
}

该标准的引用是否有影响？

A friend function defined in a class is in the (lexical) scope of the class in which it is defined. A friend function defined outside the class is not

Answer 1

您遇到的问题是，来自 [namespace.memdef]:

The friend declaration does not by itself make the name visible to unqualified lookup (3.4.1) or qualified lookup (3.4.3).

找到它们的方式是通过参数依赖查找，[basic.lookup.argdep]:

When considering an associated namespace, the lookup is the same as the lookup performed when the associated namespace is used as a qualifier (3.4.3.2) except that: [...] Any namespace-scope friend functions or friend function templates declared in associated classes are visible within their respective namespaces even if they are not visible during an ordinary lookup (11.3).

std::ostream& operator<< (std::ostream& o, test t)由于对第二个参数的参数依赖查找而被发现。

当您定义 _UNIT 时在类的外部，它使函数可见。

但是，当您定义 _UNIT内联 - 它对普通查找不可见，唯一的参数 ( double ) 没有任何关联的 namespace ，因此也无法通过参数相关查找找到它。