C.62: Make copy assignment safe for self-assignment
C.62:保證拷貝賦值對(duì)自我賦值安全

Reason(原因)

If x = x changes the value of x, people will be surprised and bad errors will occur (often including leaks).

如果x=x改變了x的值，人們會(huì)覺(jué)得很奇怪，同時(shí)也會(huì)發(fā)生很不好的錯(cuò)誤。(通常會(huì)包含泄露)

Example(示例)

The standard-library containers handle self-assignment elegantly and efficiently:

標(biāo)準(zhǔn)庫(kù)容器處理自我賦值的方式優(yōu)雅且高效：

std::vector v = {3, 1, 4, 1, 5, 9};v = v;// the value of v is still {3, 1, 4, 1, 5, 9}

Note(注意)

The default assignment generated from members that handle self-assignment correctly handles self-assignment.

產(chǎn)生于正確處理了自我賦值的成員的默認(rèn)的賦值操作會(huì)處理自我賦值問(wèn)題。

struct Bar { ? ?vector> v; ? ?map m; ? ?string s;};Bar b;// ...b = b; ? // correct and efficient

Note(注意)

You can handle self-assignment by explicitly testing for self-assignment, but often it is faster and more elegant to cope without such a test (e.g., using swap).

你可以通過(guò)明確地對(duì)自我賦值進(jìn)行檢查的方式處理自我賦值，但是通常不使用上述檢查的處理方式(例如使用swap)都會(huì)更快，更優(yōu)雅。

class Foo { ? ?string s; ? ?int i;public: ? ?Foo& operator=(const Foo& a); ? ?// ...};Foo& Foo::operator=(const Foo& a) ? // OK, but there is a cost{ ? ?if (this == &a) return *this; ? ?s = a.s; ? ?i = a.i; ? ?return *this;}

This is obviously safe and apparently efficient. However, what if we do one self-assignment per million assignments? That's about a million redundant tests (but since the answer is essentially always the same, the computer's branch predictor will guess right essentially every time). Consider:

這種做法顯然安全并且看起來(lái)高效。但是如果在一百萬(wàn)次賦值中發(fā)生一次自我賦值的情況下會(huì)怎么樣呢？大概有一百萬(wàn)次多余的檢查(但是由于本質(zhì)上結(jié)果總是一樣的，計(jì)算機(jī)的分支預(yù)測(cè)會(huì)每次都猜對(duì))。考慮下面的代碼：

Foo& Foo::operator=(const Foo& a) ? // simpler, and probably much better{ ? ?s = a.s; ? ?i = a.i; ? ?return *this;}

std::string is safe for self-assignment and so are int. All the cost is carried by the (rare) case of self-assignment.

std::string對(duì)自我賦值安全，int也是。所有的代價(jià)都來(lái)自(極少)發(fā)生的自我賦值。

Enforcement(實(shí)施建議)

(Simple) Assignment operators should not contain the pattern if (this == &a) return *this; ???

(簡(jiǎn)單)賦值運(yùn)算符不應(yīng)該包含以下的檢查：if (this == &a) return *this;

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總結(jié)

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