CODEWITHSUNDEEP
c++booleanbit-shiftinitializer-listbitset
Jonathan Mee
Jonathan Mee

Reputation: 38949

Using an initializer_list with bitset

Is there a way to use an initializer_list to construct a bitset?

For example I'd like to do:

const auto msb = false;
const auto b = true;
const auto lsb = false;
const bitset<3> foo = {msb, b, lsb};

But when I try this I get:

error: could not convert {msb, b, lsb} from '<brace-enclosed initializer list>' to const std::bitset<3u>

Do I have to use shifts to construct an unsigned long to initialize foo, or is there a way to do this that I don't know about?

Upvotes: 4

Views: 1659

Answers (2)

Oleksandr Yefremov
Oleksandr Yefremov

Reputation: 98

Short example using string representation of bitset. In some cases this implementation may not be optimal.

template <typename ... args>
auto make_bitset(args... bools) -> std::bitset<sizeof...(bools)>
{
    char data[] = {bools ? '1':'0'...};
    return std::bitset<sizeof...(bools)>(data, sizeof...(bools));
}

int main()
{
  std::cout << make_bitset(1,1,0,0,1,0, 1) << "\n";
  std::cout << make_bitset(1,1,1) << "\n";
  std::cout << make_bitset(0,0,0,0) << "\n";
}

Another implementation using "almost recursive" calls:

template <size_t N, int pos>
void init_bitset(std::bitset<N>&){}

template <size_t N, int pos = 0, typename ...Bits>
void init_bitset(std::bitset<N>& bitset, bool head, Bits ...tail)
{   
   bitset.set(pos, head);
   init_bitset<N, pos + 1>(bitset, tail...);
}

template <typename ... args>
auto make_bitset(args... bools) -> std::bitset<sizeof...(bools)>
{
    std::bitset<sizeof...(bools)> bitset;
    init_bitset(bitset, bools...);
    return bitset;
}

int main()
{
  std::cout << make_bitset(1,1,0,0,1,0, 1) << "\n";
  std::cout << make_bitset(1,1,1) << "\n";
  std::cout << make_bitset(0,0,0,0) << "\n";
}

See compiler explorer for details https://godbolt.org/g/9z2HZa

bonus version using C++17

template <typename ... args>
auto make_bitset(args... bools)
{
    unsigned n = 0;
    static_assert( sizeof...(bools) <= sizeof(n) * 8, "overflow" );
    ((n = n * 2 + static_cast<bool>(bools)), ...);
    return std::bitset<sizeof...(bools)>(n);
}

Upvotes: 3

Richard Hodges
Richard Hodges

Reputation: 69912

There is no constructor to directly construct a bitset from an initialiser list. You'll need a function:

#include <bitset>
#include <initializer_list>
#include <iostream>

auto to_bitset(std::initializer_list<bool> il)
{
    using ul = unsigned long;
    auto bits = ul(0);
    if (il.size())
    {
        auto mask = ul(1) << (il.size() - 1);

        for (auto b : il) {
            if (b) {
                bits |= mask;
            }
            mask >>= 1;
        }
    }
    return std::bitset<3> { bits };

}

int main()
{
    auto bs = to_bitset({true, false, true});

    std::cout << bs << std::endl;
}

expected results:

101

As mentioned in comments, a variadic version is also possible.

#include <bitset>
#include <iostream>
#include <utility>

namespace detail {
    template<std::size_t...Is, class Tuple>
    auto to_bitset(std::index_sequence<Is...>, Tuple&& tuple)
    {
        static constexpr auto size = sizeof...(Is);
        using expand = int[];
        unsigned long bits = 0;
        void(expand {
            0,
            ((bits |= std::get<Is>(tuple) ? 1ul << (size - Is - 1) : 0),0)...
        });
        return std::bitset<size>(bits);
    }
}

template<class...Bools>
auto to_bitset(Bools&&...bools)
{
    return detail::to_bitset(std::make_index_sequence<sizeof...(Bools)>(),
                             std::make_tuple(bool(bools)...));
}

int main()
{
    auto bs = to_bitset(true, false, true);

    std::cout << bs << std::endl;
}

Upvotes: 5

Related Questions