Why does the disjunction assignment operator |= not work with vectors of bools?

Question

If I have a vector<bool> vec_bool then I cannot modify the contents of the vector using the |= assignment operator. That is, the lines

vec_bool[0] |= true;
vec_bool[0] |= vec_bool[1];

give compiler errors, while the lines

bool a = false;
a |= true;
a |= vec_bool[0];
vec_bool[0] = vec_bool[0] | vec_bool[1];
vec_bool[0] = vec_bool[0] || vec_bool[1];

vector<int> vec_int(3);
vec_int[0] |= vec_int[1];

do not. What is the reason for this?

The error given (by gcc) is:

test.cpp:21:17: error: no match for ‘operator|=’ (operand types are ‘std::vector::reference {aka std::_Bit_reference}’ and ‘bool’)

Alvarez

You have to remember that std::vector<bool> is not an ordinary vector. It's not an actual vector of bool, but more like a vector of bits, whose implementation is not specified by the C++ specification. – Varien

Possible duplicate of Why does std::vector<bool> has no .data()? – Josephjosepha

You can take a look at en.cppreference.com/w/cpp/container/vector_bool for more information about it – Fremantle

@GauravSehgal related, but not a duplicate. The same underlying cause, but very different question/goal. – Dudgeon

It is probably worth mentioning that std::array<bool> has not this kind of optimization, so it behaves different ([...]This container is an aggregate type with the same semantics as a struct holding a C-style array T[N] as its only non-static data member.[...]). However you lose the ability to resize it. For example. – Breakout

To paraphrase Scott Meyers "There are two problems with vector<bool>. It is not a vector. It does not store bools" – Septuplicate

Answer 1

The reference returned from operator[] of std::vector<bool> is not an alias for bool&, as it is for the primary specialization of std::vector. It is rather specified by the C++ standard as this:

// bit reference:
class reference {
  friend class vector;
  reference() noexcept;
public:
  ~reference();
  operator bool() const noexcept;
  reference& operator=(const bool x) noexcept;
  reference& operator=(const reference& x) noexcept;
  void flip() noexcept;     // flips the bit
};

And as you can see, there is no operator |= declared. So you can't apply it on the reference returned from vec_bool[0].

The reason that vec_bool[0] = vec_bool[0] | vec_bool[1]; works is that there are overloads above that facilitate it. operator bool() converts the two operands of the built-in | into bool values. And then the assignment operator of reference assigns the result back to vec_bool[0].

As specified by the C++ standard, std::vector<bool> isn't a particularly good abstraction, IMO.

Parfait

Thank you! Could you perhaps tell me if vector<bool> is the only type of vector that deviates from the 'standard' implementation of the vector type? – Alvarez

@MeesdeVries - It's the only one that the C++ standard mandates to deviate from the primary declaration. All others you can assume to be of the primary template. – Parfait

... which is to say, others may be specialized for performance reasons, but you can't detect that in portable code. – Lading