How would I change an array of bit sets to a 1d array of ints with each element holding only 1 digit in C++. for example, i have bitset<8> bitArray[n], and I want to bit into int binArray[8*n], where binArray holds something like [0],[1],[1],[0],[1],[0] and so forth.
2 Answers
You can use an std::bitset::operator[] to access the specifit bit. Keep in mind though, that [0] means the least significant bit, but we want to store them in the most significant -> least significant order, so we have to use the 7 - j instead of simply j:
#include <iostream>
#include <bitset>
int main()
{
constexpr int SIZE = 5;
std::bitset<8> bitArray[SIZE] = {3, 8, 125, 40, 13};
int binArray[8 * SIZE];
for(int i = 0, index = 0; i < SIZE; ++i){
for(int j = 0; j < 8; ++j){
binArray[index++] = bitArray[i][7 - j];
}
}
}
The binArrays content looks like so (newlines added by me to improve readability):
0 0 0 0 0 0 1 1
0 0 0 0 1 0 0 0
0 1 1 1 1 1 0 1
0 0 0 0 1 1 0 1
Comments
Simply construct an array:
std::array<int, 8*n> binArray;
size_t out = 0;
for (const auto& bits : bitArray)
for (size_t ii = 0; ii < n; ++ii)
binArray[out++] = bitArray[ii];
