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>' toconst 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?
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;
}
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