C++核心準(zhǔn)則C.166:? 重載的單目運算符&只能用于智能指針和引用

C.166: Overload unary?&?only as part of a system of smart pointers and references

C.166: 重載的單目運算符&只能用于智能指針和引用

Reason(原因)

The?&?operator is fundamental in C++. Many parts of the C++ semantics assumes its default meaning.

取地址運算符&是C++的基本要素，C++語義的很多地方為它設(shè)定了默認(rèn)含義。

Example(示例)

class Ptr { // a somewhat smart pointer
    Ptr(X* pp) :p(pp) { /* check */ }
    X* operator->() { /* check */ return p; }
    X operator[](int i);
    X operator*();
private:
    T* p;
};

class X {
    Ptr operator&() { return Ptr{this}; }
    // ...
};

Note(注意)

If you "mess with" operator?&?be sure that its definition has matching meanings for?->,?[],?*, and?.?on the result type. Note that operator?.?currently cannot be overloaded so a perfect system is impossible. We hope to remedy that:?http://www.open-std.org/jtc1/sc22/wg21/docs/papers/2015/n4477.pdf. Note that?std::addressof()?always yields a built-in pointer.

如果你要"招惹"&運算符，一定要確保它的結(jié)果類型和->，[]，*和 . 相匹配。注意 . 運算符現(xiàn)在不能被重載，因此不可能實現(xiàn)完美系統(tǒng)。我們可以希望可以改進這一點：

http://www.open-std.org/jtc1/sc22/wg21/docs/papers/2015/n4477.pdf。

注意std::addressof()總是返回一個內(nèi)置類型的指針。

Enforcement(實施建議)

Tricky. Warn if?&?is user-defined without also defining?->?for the result type.

不好辦。如果定制了&運算符卻沒有為結(jié)果定義->運算符，報警。

原文鏈接：

https://github.com/isocpp/CppCoreGuidelines/blob/master/CppCoreGuidelines.md#c166-overload-unary--only-as-part-of-a-system-of-smart-pointers-and-references

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