Range-for-loops 和 std::vector<bool>;
为什么这段代码有效
std::vector<int> intVector(10);
for(auto& i : intVector)
std::cout << i;
这不是吗?
std::vector<bool> boolVector(10);
for(auto& i : boolVector)
std::cout << i;
在后一种情况下，我得到一个错误
In the latter case, I get an error
错误:从std::_Bit_iterator::reference {aka std::_Bit_reference}"类型的右值对std::_Bit_reference&"类型的非常量引用进行无效初始化
error: invalid initialization of non-const reference of type ‘std::_Bit_reference&’ from an rvalue of type ‘std::_Bit_iterator::reference {aka std::_Bit_reference}’
for(auto& i : boolVector)
推荐答案
因为 std::vector<bool> 不是容器 ！
Because std::vector<bool> is not a container !
std::vector<T> 的迭代器通常取消对 T& 的引用，您可以将其绑定到自己的 auto&.
std::vector<T>'s iterators usually dereference to a T&, which you can bind to your own auto&.
std::vector<bool> 将其 bool 打包在整数中，因此您需要代理在访问它们时进行位掩码.因此，它的迭代器返回一个 Proxy.
并且由于返回的 Proxy 是一个纯右值(一个临时的)，它不能绑定到一个左值引用，例如 auto&.
std::vector<bool>, however, packs its bools together inside integers, so you need a proxy to do the bit-masking when accessing them. Thus, its iterators return a Proxy.
And since the returned Proxy is an prvalue (a temporary), it cannot bind to an lvalue reference such as auto&.
解决方案:使用 auto&&，如果给定一个左值引用，它将正确折叠成一个左值引用，或者如果给定一个代理，则绑定并保持临时活动.
The solution : use auto&&, which will correctly collapse into an lvalue reference if given one, or bind and maintain the temporary alive if it's given a proxy.
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