問題描述

假設我在使用基于范圍的循環編程時的當前規則說

Assuming my current rule when programming with range-based loops says

盡可能使用 for(auto const &e :...) 或 for(auto &e:...) 而不是 for(自動 a: ...).

Use for(auto const &e :...) or for(auto &e:...) when possible over for(auto a: ...).

我以我自己的經驗和這個問題為例.

I base this on my own experience and this question for example.

但是在閱讀了新的 簡潔的循環 我想知道，我不應該用 && 替換我的規則中的 & 嗎?正如此處所寫，這看起來像Meyers 的通用參考文獻.

But after reading about the new terse for loops I wonder, should I not replace my & in my rule with &&? As written here this looks like the Meyers' Universal References.

所以，我問自己，我的新規則是否應該是

So, I ask myself, should my new rule either be

盡可能使用 for(auto const &&e :...) 或 for(auto &&e:...) ...

Use for(auto const &&e :...) or for(auto &&e:...) when possible ...

或者這并不總是有效，因此應該是相當復雜的

or does that not always work and therefore should rather be the quite complicated one

檢查 for(auto const &&e:...) 或 for(auto &&e:...) 是否可行，然后考慮 for(auto const &e :...) 或 for(auto &e:...)，并且僅在需要時才使用引用.

Check if for(auto const &&e :...) or for(auto &&e:...) is possible, then consider for(auto const &e :...) or for(auto &e:...), and only when needed do not use references.

推薦答案

Howard Hinnant 此處.

When and if you should use auto&& in for loops has been explained very nicely by Howard Hinnant here.

這就留下了x在

auto &&x = ...expr...

實際上是.并且像有函數模板定義一樣處理

actually is. And it is handled as if there there were a function template definition

template <class U> void f(const U& u);

并且x的類型是按照與u[§7.1.6.4.(7)]相同的規則推導出來的.

and the type of x is deduced by the same rules as u [§7.1.6.4.(7)].

這意味著它不作為 RValue 參考處理，而是作為通用/轉發參考"處理.--參考折疊規則"申請.

This means it is not handled as a RValue Reference, but as a "Universal/Forwarding Reference" -- the "Reference Collapsing Rules" apply.

這也適用于

const auto &&x = ...expr...

作為 §7.1.6.4.(7) 中的例子，至少對于 const auto &x.

as the example in §7.1.6.4.(7) states, at least for const auto &x.

但是，正如 PiotrS 在問題評論中所說，任何限定詞都會使 URef-ness 無效:

But, as PiotrS says in the questions comments, any qualifiers nullifies the URef-ness:

不，因為template中的T都沒有.void f(const T&&) 是轉發引用，const auto&& 也不是.T&& 出現在參數聲明中的事實并不意味著它是轉發引用.只有沒有像 const 或 volatile 這樣的限定符的純 T&& 是轉發引用，這意味著它必須是 template;void f(T&&) 或 auto&&，從不const T&& 或 const auto&&代碼>

no, because neither T in template<class T> void f(const T&&) is a forwarding reference, nor const auto&& is. The fact that T&& occurs in parameter declaration does not imply it is forwarding reference. Only pure T&& with no qualifiers like const or volatile is forwarding reference, meaning it has to be template<class T> void f(T&&) or auto&&, and never const T&& or const auto&&

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