Why does this code work
std::vector<int> intVector(10);
for(auto& i : intVector)
std::cout << i;
And this doesn't?
std::vector<bool> boolVector(10);
for(auto& i : boolVector)
std::cout << i;
In the latter case, I get an error
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)
Because std::vector<bool> is not a container !
std::vector<T>'s iterators usually dereference to a T&, which you can bind to your own 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&.
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.
std::vector<bool> does not obey the standard container rules.
In particular, its operator[] does not return bool&.
The loop in the invalid code
#include <vector>
#include <iostream>
int main() {
std::vector<bool> boolVector(10);
for (auto& i: boolVector)
std::cout << i;
}
can be rewritten in any of three ways to iterate through the values:
(read-only)
for (auto i: boolVector)
std::cout << i;
(read-only, possibly inefficient)
for (auto const& i: boolVector)
std::cout << i;
(read/write)
for (auto&& i: boolVector)
std::cout << i;
The choice between the first and last is down to whether you need to modify the values in the vector, or just to read them.
vector<bool> is (usually) specialized explicitly to store each bool in a single bit, reducing the storage costs from one byte per value to one byte per eight values. No processor I know of offhand is bit addressable, so it's impossible to store a reference to the values in the vector<bool>. You need to use plain auto, not auto& for the iteration value i.
