How to reverse bits in a bitset?

3

For example, I have the integer

a = 10;

and it's binary representation (for a 32 bit integer) is

00000000000000000000000000001010

and reversed, it becomes

01010000000000000000000000000000

Now I've seen this code, from this topcoder article that can accomplish this

x = ((x & 0xaaaaaaaa) >> 1) | ((x & 0x55555555) << 1);
x = ((x & 0xcccccccc) >> 2) | ((x & 0x33333333) << 2);
x = ((x & 0xf0f0f0f0) >> 4) | ((x & 0x0f0f0f0f) << 4);
x = ((x & 0xff00ff00) >> 8) | ((x & 0x00ff00ff) << 8);
x = ((x & 0xffff0000) >> 16) | ((x & 0x0000ffff) << 16);

Now is there some straightforward way to achieve the same effect. Perhaps by converting our bitset into a string, and then reversing that? The constructors and method for converting bitset to a string of a bitset are so complicated I can't seem to figure out how to do this.

Here's what I tried so far

#include <bitset>
#include <iostream>
#include <algorithm>
#include <cmath>
#include <vector>
#include <cstdlib>
#include <ctime>
#include <typeinfo>
#include <string>

using namespace std;

int main() {

    const unsigned int k = 32;


    int x = 10;
    bitset<k> nf(x);

    cout << nf << endl;

    string str =
        nf.to_string<char,string::traits_type,string::allocator_type>();

    reverse(str.begin(), str.end() + str.size());

    cout << str << endl;

    return 0;
}

But I'm getting this as the output:

00000000000000000000000000001010
G;ÿJG¥±žGsÿkìöUàä˜\éä˜\é
c++
c++11
bits
bitset
asked on Stack Overflow Feb 1, 2018 by Rockstar5645 • edited Feb 1, 2018 by Rockstar5645

4 Answers

5

Less code will win you some time in TopCoder SRMs. Following is what I would use in TopCoder SRMs (see it live here):

#include <algorithm>
#include <bitset>
#include <iostream>
#include <string>

int main() {
  auto x = std::bitset<32>(10);
  std::cout << x << std::endl;

  auto str = x.to_string();
  std::reverse(str.begin(), str.end());
  auto y = std::bitset<32>(str);
  std::cout << y << std::endl;

  return 0;
}
answered on Stack Overflow Feb 1, 2018 by Lingxi
5

This is the trivial inplace approach straight on a bitset:

template<std::size_t N>
void reverse(std::bitset<N> &b) {
    for(std::size_t i = 0; i < N/2; ++i) {
        bool t = b[i];
        b[i] = b[N-i-1];
        b[N-i-1] = t;
    }
}
answered on Stack Overflow Feb 1, 2018 by Matteo Italia
1

Without using any standard library functions (aside from printing the result):

#include<iostream>
#include<bitset>

const int size = sizeof(int)*CHAR_BIT;

int main()
{
    int x = 10;
    int r = 0;
    for(int i = 0; i < size; i++)
    {
        r = r << 1 | (x & 1);
        x >>= 1;
    }
    std::bitset<size> bits(r);
    std::cout << "Reverse " << bits << std::endl;

}

answered on Stack Overflow Feb 1, 2018 by Mats Petersson • edited Oct 25, 2019 by Matteo Italia
0

Please check this one method :)

#include <iostream>
#include <bitset>

template<std::size_t N>
std::bitset<N> reverse(const std::bitset<N> &bit_set) {
  std::bitset<N> reversed;
  for (int i = 0, j = N - 1; i < N; i++, j--) {
    reversed[j] = bit_set[i];
  }
  return reversed;
}

int main() {
  std::bitset<32> b1(10);
  std::bitset<32> reversed_b1 = reverse(b1);
  std::cout << b1;
  std::cout << "\nand reversed\n" << reversed_b1 << std::endl;
  return 0;
}
answered on Stack Overflow Feb 1, 2018 by BartekPL

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