为什么基于范围的for语句通过auto&&？ [英] Why does a range-based for statement take the range by auto&&?

问题描述

基于范围的 for 语句在§6.5.4中定义为等同于：

  {
 auto&&& __range = range-init; 
 for（auto __begin = begin-expr，
 __end = end-expr; 
 __begin！= __end; 
 ++ __ begin）{
 for-range-declaration = * __ begin; 
语句
} 
} 
  

其中 c>

$

pre> for（for-range-declaration：expression）=> （expression）
for（for-range-declaration：braced-init-list）=> braced-init-list


（该子句进一步指定其他子表达式的含义） p>

为什么是 __ range 给定推导类型 auto&& ？我对 auto&&& 的理解是，通过传递表达式的原始价值（lvalue / rvalue）通过 std ::转发。但是， __ range 不会通过 std :: forward 传递到任何地方。它只在获取范围迭代器时使用，例如 __ range ， __ range.begin（）或 begin（__ range）。

使用通用参考的好处是 auto&& ; ？不会 auto& 足够了？

注意：据我所知，提案没有说明选择 auto& ;& 。

解决方案

足够了？

不，不会。它不允许使用计算范围的 r值表达式。使用 auto&&& 是因为它可以绑定到l值表达式或 r值表达式。因此，您不需要将范围粘贴到变量中以使其工作。

或者，换句话说，这是不可能的： p>

  for（const auto& v：std :: vector< int> {1，43，5，2，4}）
 {
} 
  




c $ c> const auto& 够用吗？

不，一个 const std :: vector 只会返回 const_iterator s到它的内容。如果你想做一个非 - const 遍历内容，这将无济于事。

A range-based for statement is defined in §6.5.4 to be equivalent to:

{
  auto && __range = range-init;
  for ( auto __begin = begin-expr,
             __end = end-expr;
        __begin != __end;
        ++__begin ) {
    for-range-declaration = *__begin;
    statement
  }
}

where range-init is defined for the two forms of range-based for as:

for ( for-range-declaration : expression )         =>   ( expression )
for ( for-range-declaration : braced-init-list )   =>   braced-init-list

(the clause further specifies the meaning of the other sub-expressions)

Why is __range given the deduced type auto&&? My understanding of auto&& is that it's useful for preserving the original valueness (lvalue/rvalue) of an expression by passing it through std::forward. However, __range isn't passed anywhere through std::forward. It's only used when getting the range iterators, as one of __range, __range.begin(), or begin(__range).

What's the benefit here of using the "universal reference" auto&&? Wouldn't auto& suffice?

Note: As far as I can tell, the proposal doesn't say anything about the choice of auto&&.

解决方案

Wouldn't auto& suffice?

No, it wouldn't. It wouldn't allow the use of an r-value expression that computes a range. auto&& is used because it can bind to an l-value expression or an r-value expression. So you don't need to stick the range into a variable to make it work.

Or, to put it another way, this wouldn't be possible:

for(const auto &v : std::vector<int>{1, 43, 5, 2, 4})
{
}

Wouldn't const auto& suffice?

No, it wouldn't. A const std::vector will only ever return const_iterators to its contents. If you want to do a non-const traversal over the contents, that won't help.

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