为什么不是vector<bool>STL 容器?-C/C++实例代码

Why isn#39;t vectorlt;boolgt; a STL container?(为什么不是vectorlt;boolgt;STL 容器?)

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问题描述

Scott Meyers 的书Effective STL: 50 Specific Ways to Improvement Your Use of the Standard Template Library 的第 18 项说要避免使用 vector <bool>，因为它不是STL 容器，它并没有真正保存 bools.

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.

以下代码:

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

不会编译，违反了 STL 容器的要求.

will not compile, violating a requirement of STL containers.

错误:

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

vector<T>::operator []返回类型应该是T&，但是为什么是vector?


vector<T>::operator [] return type is supposed to be T&, but why is it a special case for vector<bool>?
vector<bool> 真正由什么组成?
物品进一步说:
deque<bool> v; // is a STL container and it really contains bools

这可以用作 vector<bool> 的替代品吗?
Can this be used as an alternative to vector<bool>?
谁能解释一下?
推荐答案
出于空间优化的原因，C++ 标准(早在 C++98)明确调用 vector作为一个特殊的标准容器，其中每个 bool 只使用一位空间，而不是像普通 bool 那样使用一个字节(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.
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.
在这种情况下，由于您无法获取字节内的位地址，因此 operator[] 等内容无法返回 bool& 但而是返回一个允许操作相关特定位的代理对象.由于此代理对象不是 bool&，因此您不能将其地址分配给 bool*，就像在正常"容器.反过来，这意味着 bool *pb =&v[0]; 不是有效代码.
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.
另一方面，deque 没有调用任何此类特殊化，因此每个 bool 都占用一个字节，您可以从 operator[].
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[].
最后请注意，MS 标准库的实现(可以说)是次优的，因为它为 deques 使用了较小的块大小，这意味着使用 deque 作为替代品并不总是正确的答案.
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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