C++核心准则C.168: 将重载的运算符定义在操作对象的命名空间内-技术圈

C.168: Define overloaded operators in the namespace of their operands

C.168: 将重载的运算符定义在操作对象的命名空间内

Reason(原因)

Readability. Ability for find operators using ADL. Avoiding inconsistent definition in different namespaces

可读性。提供使用ADL发现操作符的能力。避免不同命名空间中的不一致。

ADL,Argument-dependent lookup.详细请参照以下链接：

https://en.cppreference.com/w/cpp/language/adl

--译者注

Example(示例)

struct S { };
bool operator==(S, S);   // OK: in the same namespace as S, and even next to S
S s;

bool x = (s == s);

This is what a default == would do, if we had such defaults.

这正是默认相等比较运算符做的事情，如果存在这么一个默认的话。

Example(示例)

namespace N {
    struct S { };
    bool operator==(S, S);   // OK: in the same namespace as S, and even next to S
}

N::S s;

bool x = (s == s);  // finds N::operator==() by ADL

Example, bad(反面示例)

struct S { };
S s;

namespace N {
    S::operator!(S a) { return true; }
    S not_s = !s;
}

namespace M {
    S::operator!(S a) { return false; }
    S not_s = !s;
}

Here, the meaning of !s differs in N and M. This can be most confusing. Remove the definition of namespace M and the confusion is replaced by an opportunity to make the mistake.

代码中N和M两个命名空间中!s的含义不一样。这会非常混乱。如果去掉命名空间M的定义又会增加出错的可能。

Note(注意)

If a binary operator is defined for two types that are defined in different namespaces, you cannot follow this rule. For example:

如果为不同命名空间内的两个不同的类型定义二目运算符，你无法遵守本准则。例如：

Vec::Vector operator*(const Vec::Vector&, const Mat::Matrix&);

This may be something best avoided.

这可能是最好状态了。

See also(参照)

This is a special case of the rule that helper functions should be defined in the same namespace as their class.

这可以说是【帮助函数应该和它帮助的类定义在一个命名空间内】规则的特例。

Enforcement(实施建议)

Flag operator definitions that are not it the namespace of their operands
标记没有和操作对象定义在同一个命名空间中的运算符。

原文链接：

https://github.com/isocpp/CppCoreGuidelines/blob/master/CppCoreGuidelines.md#c168-define-overloaded-operators-in-the-namespace-of-their-operands

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