为什么基于范围的 for 语句会通过 auto&& 获取范围? [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中定义为等价于:

<代码>{汽车&&__range = 范围初始化；for ( auto __begin = begin-expr,__end = 结束表达式；__开始！= __结束；++__开始) {for-range-declaration = *__begin;陈述}}

其中 range-init 是为基于范围的 for 的两种形式定义的:

for ( for-range-declaration : expression ) =>( 表达 )for ( for-range-declaration :braced-init-list ) =>花括号初始化列表

(该子句进一步规定了其他子表达式的含义)

为什么 __range 给定推导类型 auto&&?我对 auto&& 的理解是，通过 std::forward 传递表达式，它对于保留表达式的原始值(左值/右值)很有用.但是，__range 不会通过 std::forward 传递到任何地方.它仅在获取范围迭代器时使用，作为 __range、__range.begin() 或 begin(__range) 之一.

使用通用参考"auto&& 有什么好处?auto& 还不够吗?

注意:据我所知，提案 没有说明 auto&& 的选择.

解决方案

不会自动&够了吗?

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左值表达式 或 右值表达式.因此，您无需将范围粘贴到变量中即可使其工作.

或者，换句话说，这是不可能的:

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

<块引用>

const auto& 还不够吗?

不，不会.const std::vector 只会将 const_iterators 返回到它的内容.如果您想对内容进行非const 遍历，那将无济于事.

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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