为什么vector< bool>一个STL容器？

问题描述

Scott Meyers的书中的第18条有效的STL：改进对标准模板库的使用的50种特定方法说避免使用 vector< bool> ，因为它不是STL容器，并且实际上并不容纳 bool s。

以下代码：

  vector< bool> v; 
 bool * pb =&v [0]; 
  

不会编译，这违反了STL容器的要求。

错误：

 无法转换'std :: vector< bool> :: reference * {aka std :: _Bit_reference *}'初始化为
  

vector< T> :: operator [] 返回类型应该是 T& ，但是为什么它是 vector<的特殊情况？ bool> ？

vector< bool> 到底由什么组成？

该项目还说：

  deque< bool> v; //是一个STL容器，它实际上包含布尔
  

是否可以用作<$的替代c $ c> vector< bool> ？

有人可以解释吗？

解决方案

出于空间优化的原因，C ++标准（最早可追溯至C ++ 98）明确调用 vector< bool> 作为特殊的标准容器，其中每个布尔仅使用一个空间位，而不是普通布尔所使用的一个字节（实现一种动态位集）。作为这种优化的交换，它没有提供普通标准容器的所有功能和接口。

在这种情况下，因为您不能使用a的地址在一个字节内，例如 operator [] 之类的内容不能返回 bool& ，而是返回一个代理对象这样就可以操纵相关的特定位。由于此代理对象不是 bool& ，因此无法像其可能的那样将其地址分配给 bool * 这样的结果就是操作员在普通容器上调用。反过来，这意味着 bool * pb =& v [0]; 无效。

另一方面， deque 没有任何这样的特殊化，因此每个布尔值都占用一个字节，您可以从中获取值返回的地址。最后请注意，MS标准库实现（可以说）不是最优的，因为它使用较小的块大小进行双端队列，这意味着使用双端队列代替并不总是正确的答案。

Item 18 of Scott Meyers's book Effective STL: 50 Specific Ways to Improve Your Use of the Standard Template Library says to avoid vector <bool> as it's not an STL container and it doesn't really hold bools.

The following code:

vector <bool> v;
bool *pb =&v[0];

will not compile, violating a requirement of STL containers.

Error:

cannot convert 'std::vector<bool>::reference* {aka std::_Bit_reference*}' to 'bool*' in initialization

vector<T>::operator [] return type is supposed to be T&, but why is it a special case for vector<bool>?

What does vector<bool> really consist of?

The Item further says:

deque<bool> v; // is a STL container and it really contains bools

Can this be used as an alternative to vector<bool>?

Can anyone please explain this?

解决方案

For space-optimization reasons, the C++ standard (as far back as C++98) explicitly calls out vector<bool> as a special standard container where each bool uses only one bit of space rather than one byte as a normal bool would (implementing a kind of "dynamic bitset"). In exchange for this optimization it doesn't offer all the capabilities and interface of a normal standard container.

In this case, since you can't take the address of a bit within a byte, things such as operator[] can't return a bool& but instead return a proxy object that allows to manipulate the particular bit in question. Since this proxy object is not a bool&, you can't assign its address to a bool* like you could with the result of such an operator call on a "normal" container. In turn this means that bool *pb =&v[0]; isn't valid code.

On the other hand deque doesn't have any such specialization called out so each bool takes a byte and you can take the address of the value return from operator[].

Finally note that the MS standard library implementation is (arguably) suboptimal in that it uses a small chunk size for deques, which means that using deque as a substitute isn't always the right answer.

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