C++核心準則R.4: 原始引用(T&)不包含所有權(quán)

R.4: A raw reference (a?T&) is non-owning

R.4:?原始引用(T&)不包含所有權(quán)

Reason(原因)

There is nothing (in the C++ standard or in most code) to say otherwise and most raw references are non-owning. We want owners identified so that we can reliably and efficiently delete the objects pointed to by owning pointers.

這一點不存在例外(無論是C++標準還是大部分代碼中)，實際上大多數(shù)原始引用就是不包含所有權(quán)的。我希望所有者被明確下來以便我們可以可靠而且高效的刪除所有權(quán)指針指向的對象。

Example(示例)

void f()
{
    int& r = *new int{7};  // bad: raw owning reference
    // ...
    delete &r;             // bad: violated the rule against deleting raw pointers
}

See also:?The raw pointer rule:

https://github.com/isocpp/CppCoreGuidelines/blob/master/CppCoreGuidelines.md#Rr-ptr

參見原始指針原則。

Enforcement(實施建議)

See the raw pointer rule：

https://github.com/isocpp/CppCoreGuidelines/blob/master/CppCoreGidelines.md#Rr-ptr

參見原始指針原則

原文鏈接：

https://github.com/isocpp/CppCoreGuidelines/blob/master/CppCoreGuidelines.md#r3-a-raw-pointer-a-t-is-non-owning

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