is it possible to copy the value of bitset<8> to a string of arrays without converting them?

Question

I tried char a[9]=bitset<8>(f[1]) but as many of you know that did not work. i want to copy the binary value to an array so basically if f[1] will be 'a' the char a will be 01100001

And also , this is not a duplication the question asked before was not what I searched.

Can you explain what you mean by a string of arrays, and how the bits in the bitfield would map to entries in that "string"? — juanchopanza, Jan 18 '15 at 18:33
What is `f`'s type? Maybe [`reinterpret_cast`](http://en.cppreference.com/w/cpp/language/reinterpret_cast)? — IllusiveBrian, Jan 18 '15 at 18:37

score 1 · Answer 1 · answered Jan 18 '15 at 18:57

You can't direct-initialize an array from an object. You'll need to define the array an then initialize it. Even then, sadly, std::bitset<...> doesn't provide iterators to allow use of std::transform() to fill in the characters. However, a with a simple auxiliary function the initialization can be done:

#include <iostream>
#include <bitset>
template <std::size_t Size>
void initialize(char (&array)[Size], std::bitset<Size - 1> bits) {
    for (std::size_t idx(0); idx != Size - 1; ++idx) {
         array[idx] = '0' + bits[idx];
    }
    array[Size-1] = 0;
}

int main(int, char* av[]) {
    char a[9];
    initialize(a, av[0][0]);
    std::cout << "array=" << a << '\n';
}

is it possible to copy the value of bitset<8> to a string of arrays without converting them?

1 Answers1