Question

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Why vector<bool>::reference doesn’t return reference to bool?

I used to think that with std::vector::operator[] we get deep copies of the accessed item, but it seems that it is not always true. At least, with vector<bool> the following test code gives a different result:

#include <iostream>
#include <vector>
using namespace std;

template <typename T>
void Test(const T& oldValue, const T& newValue, const char* message)
{
    cout << message << '\n';

    vector<T> v;
    v.push_back(oldValue);
    cout << " before:  v[0] = " << v[0] << '\n';

    // Should be a deep-copy (?)       
    auto x = v[0];   
    x = newValue;

    cout << " after:   v[0] = " << v[0] << '\n';
    cout << "-------------------------------\n";
}

int main()
{
    Test<int>(10, 20, "Testing vector<int>");
    Test<double>(3.14, 6.28, "Testing vector<double>");
    Test<bool>(true, false, "Testing vector<bool>");
}

Output (source code compiled with VC10/VS2010 SP1):

Testing vector<int>
 before:  v[0] = 10
 after:   v[0] = 10
-------------------------------
Testing vector<double>
 before:  v[0] = 3.14
 after:   v[0] = 3.14
-------------------------------
Testing vector<bool>
 before:  v[0] = 1
 after:   v[0] = 0
-------------------------------

I would have expected that v[0] after the x = newValue assignment would still be equal to its previous value, but this seems not true.
Why is that?
Why is vector<bool> special?

Answer 1

vector<bool>::operator[] neither yields a bool nor a reference to a bool. It just returns a little proxy object that acts like a reference. This is because there are no references to single bits and vector<bool> actually stores the bools in a compressed way. So by using auto you just created a copy of that reference-like object. The problem is that C++ does not know that this object acts as a reference. You have to force the “decay to a value” here by replacing auto with T.