假设我在使用基于范围的循环编程时的当前规则说
Assuming my current rule when programming with range-based loops says
尽可能使用 for(auto const &e :...)
或 for(auto &e:...)
而不是 for(自动 a: ...)
.
Use
for(auto const &e :...)
orfor(auto &e:...)
when possible overfor(auto a: ...)
.
我以我自己的经验和这个问题为例.
I base this on my own experience and this question for example.
但是在阅读了新的 简洁的循环 我想知道,我不应该用 &&
替换我的规则中的 &
吗?正如此处所写,这看起来像Meyers 的通用参考文献.
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,我问自己,我的新规则是否应该是
So, I ask myself, should my new rule either be
尽可能使用 for(auto const &&e :...)
或 for(auto &&e:...)
...
Use
for(auto const &&e :...)
orfor(auto &&e:...)
when possible ...
或者这并不总是有效,因此应该是相当复杂的
or does that not always work and therefore should rather be the quite complicated one
检查 for(auto const &&e:...)
或 for(auto &&e:...)
是否可行,然后考虑 for(auto const &e :...)
或 for(auto &e:...)
,并且仅在需要时才使用引用.
Check if
for(auto const &&e :...)
orfor(auto &&e:...)
is possible, then considerfor(auto const &e :...)
orfor(auto &e:...)
, and only when needed do not use references.
Howard Hinnant 此处.
When and if you should use auto&&
in for loops has been explained very nicely by Howard Hinnant here.
这就留下了x
在
auto &&x = ...expr...
实际上是.并且像有函数模板定义一样处理
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
的类型是按照与u
[§7.1.6.4.(7)]相同的规则推导出来的.
and the type of x
is deduced by the same rules as u
[§7.1.6.4.(7)].
这意味着它不作为 RValue 参考处理,而是作为通用/转发参考"处理.--参考折叠规则"申请.
This means it is not handled as a RValue Reference, but as a "Universal/Forwarding Reference" -- the "Reference Collapsing Rules" apply.
这也适用于
const auto &&x = ...expr...
作为 §7.1.6.4.(7) 中的例子,至少对于 const auto &x
.
as the example in §7.1.6.4.(7) states, at least for const auto &x
.
但是,正如 PiotrS 在问题评论中所说,任何限定词都会使 URef-ness 无效:
But, as PiotrS says in the questions comments, any qualifiers nullifies the URef-ness:
不,因为template
是转发引用,T
都没有.void f(const T&&)const auto&&
也不是.T&&
出现在参数声明中的事实并不意味着它是转发引用.只有没有像 const
或 volatile
这样的限定符的纯 T&&
是转发引用,这意味着它必须是 template
或 auto&&
,从不const T&&
或 const auto&&
代码>
no, because neither
T
intemplate<class T> void f(const T&&)
is a forwarding reference, norconst auto&&
is. The fact thatT&&
occurs in parameter declaration does not imply it is forwarding reference. Only pureT&&
with no qualifiers likeconst
orvolatile
is forwarding reference, meaning it has to betemplate<class T> void f(T&&)
orauto&&
, and neverconst T&&
orconst auto&&
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