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’)

You have to remember that [`std::vector`](http://en.cppreference.com/w/cpp/container/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. — Some programmer dude, Oct 03 '17 at 09:49
Possible duplicate of [Why does std::vector has no .data()?](https://stackoverflow.com/questions/46115669/why-does-stdvectorbool-has-no-data) — Gaurav Sehgal, Oct 03 '17 at 09:52
You can take a look at http://en.cppreference.com/w/cpp/container/vector_bool for more information about it — cbuchart, Oct 03 '17 at 09:53
@GauravSehgal related, but not a duplicate. The same underlying cause, but very different question/goal. — sehe, Oct 03 '17 at 10:01
It is probably worth mentioning that [std::array](http://en.cppreference.com/w/cpp/container/array) 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](http://coliru.stacked-crooked.com/a/acc6abad754046cc). — user1810087, Oct 03 '17 at 10:36
To paraphrase Scott Meyers *"There are two problems with `vector`. It is not a vector. It does not store `bool`s"* — Happy Green Kid Naps, Oct 03 '17 at 13:32

score 18 · Accepted Answer · answered Oct 03 '17 at 09:58

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.

Thank you! Could you perhaps tell me if `vector` is the only type of vector that deviates from the 'standard' implementation of the vector type? — Mees de Vries, Oct 03 '17 at 10:05
@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. — StoryTeller - Unslander Monica, Oct 03 '17 at 10:07
... which is to say, others may be specialized for performance reasons, but you can't detect that in portable code. — MSalters, Oct 03 '17 at 13:17

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

1 Answers1