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?
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.