指向 std::vector<bool> 的元素时出现编译器错误? [英] compiler error when pointing to an element of std::vector&lt;bool&gt;?

问题描述

尝试使用 std::vector 我有一个编译器错误，这让我感到非常惊讶.

Trying to use std::vector<bool> I have a compiler error that is very surprising to me.

简而言之，获取 std::vector 元素的地址并将其分配给 unsigned char 指针:

In short taking the address of an element of a std::vector<unsigned char> and assigning it to a unsigned char pointer:

std::vector<unsigned char> test(10);
unsigned char *pb = &test[0];

工作得很好，同时尝试用 std::vector 做同样的事情会导致编译器错误:

works perfectly well, while trying to do the same thing with a std::vector<bool> results in a compiler error:

int main() {

    std::vector<bool> test(10);
    bool *pb = &test[0];    // line 4, compile error

    return 0;
}

在 Visual Studio 上，它说的是:

On Visual Studio, it says something like:

std::vector bool cannot convert std::_Vb_reference<_Alloc> * to bool *

而键盘(参见 http://codepad.org/vaiN3iEq 上的示例)说:

while codepad (see example at http://codepad.org/vaiN3iEq) says:

cc1plus: warnings being treated as errors
In function 'int main()':
Line 4: warning: taking address of temporary
Line 4: error: cannot convert '__gnu_norm::_Bit_reference*' to 'bool*' in initialization
compilation terminated due to -Wfatal-errors.

我认为 bool 和 unsigned char 在内部是相同的(只是一个 1 字节类型，有一些编译器的东西来强制 bool只允许真/假值).但没想到会出现这样的问题！知道为什么吗?！

I thought both bool and unsigned char were internally the same (just a 1 byte type, with some compiler stuffs to enforce bool to allow only true/false values). But I was not expecting such problem! Any idea why?!

请注意，我知道 bitsets 并且对在此处使用它们不感兴趣.

Note that I know of bitsets and am not interested in using them here.

推荐答案

是的，bool 和 unsigned char 通常各自占用相同数量的内存，但是不会使 vector 和 vector 相同！

Yes, bool and unsigned char typically take the same amount of memory on their own, but that does not make vector<bool> and vector<unsigned char> the same!

vector 被标准给予了非常、非常 的特殊处理，以便尽可能接近地打包元素(1990 年代有人认为这很聪明，因为 bool 只有两种状态之一)，结果就是你所看到的:它的元素是不可寻址的.

vector<bool> is given very, very special treatment by the standard in order to pack elements as close as possible (which someone in the 1990s thought would be clever, since a bool has one of only two states), and the result is what you've seen: its elements are non-addressable.

避免！

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