为什么是vector< bool>不是STL容器？ [英] Why is vector<bool> not a STL container?

问题描述

Scott Meyers的Item 18表示避免向量< bool> ，因为它不是一个STL容器，它并不真正持有bools。

以下代码：

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

无法编译，违反STL容器的要求。

错误：

 无法转换std :: vector< bool> :: reference * {aka std :: _ Bit_reference *}'到初始化中的'bool *'
  

向量< T& ：operator [] 返回类型应该是T&，但为什么它是向量< bool> b
$ b

向量< bool> 是什么？

进一步说：

  deque< bool> v; //是一个STL容器，它真的包含bools 
  

这可以用来替代<$

任何人都可以解释一下吗？

vector 作为一个特殊的标准容器，其中每个bool只使用一个位的空间，而不是一个字节作为一个正常的bool（实现一种动态bitset）。作为这种优化的交换，它不提供正常标准容器的所有功能和接口。

在这种情况下，由于您不能获取一个字节内的位，例如 operator [] 不能返回 bool& ，而是返回一个代理对象其允许操纵所讨论的特定位。由于此代理对象不是 bool& ，因此您不能将其地址分配给 bool * 这种操作符调用在正常容器上的结果。这意味着 bool * pb =& v [0]; 是无效的代码。

另一方面， deque 没有任何这样的特殊化调用所以每个bool需要一个字节，你可以取地址的值从

最后注意，MS标准库实现（可以说是）次优的，因为它使用小块大小的deques，这意味着使用deque作为替代并不总是正确的答案。

Scott Meyers's Item 18 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 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 it's 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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