How do I flip part of number/bitset in C++ efficiently?

Question

Consider I have a number 100101 of length 6. I wish to flip bits starting from position 2 to 5 so my answer will be 111011. I know that I can flip individual bits in a loop. But is there an efficient way of doing this without a for loop?

Presumably you aren't always trying to flip all unset bits? Please clarify your input/output a little bit. It also seems like you're assuming indices to start at `1`? And is it big-endian or little endian? — AndyG, Jan 03 '21 at 20:17
@AsteroidsWithWings I am trying to solve a competitive coding problem. Using a loop increases gets me a Time limit Exceeded output. — Leena, Jan 03 '21 at 20:28
@AsteroidsWithWings Leetcode 995. There may be a different solution to the problem, but when i looked for flipping part of bitset, i couldnt find anything — Leena, Jan 03 '21 at 20:36

Bo R · Answer 1 · 2021-01-06T13:19:58.110

2

If I understand you correctly, try

namespace {
unsigned flip(unsigned a)
{
   return a ^ 0b011110u;
}
}  // namespace

Just adjust the constant to the actual bits you want to flip by having them as 1's in the constant.

On the otherhand, if you need just to update an individual variable you also use

unsigned value = 0b100101u;
value ^= 0b011110u;
assert(value == 0b111011u);

EDIT And here is the same using std::bitset<6u> and C++98:

#include <bitset>
#include <cassert>

int main()
{
   std::bitset<6u> const kFlipBit2to6WithXor(0x1Eu);  // aka 0b011110u
   std::bitset<6u> number(0x25u);                     // aka 0b100101u
   number ^= kFlipBit2to6WithXor;
   assert(number.to_ulong() == 0x3Bu);                // aka 0b111011u
   return 0;
}

edited Jan 06 '21 at 13:19

answered Jan 03 '21 at 20:25

Bo R

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XOR is more likely to give the result the OP was looking for than OR – Alan Birtles Jan 03 '21 at 20:27
Now do the other one — flip the bits, so that 0s become 1s and 1s become 0s. Hint: you only have to change one character in your code. – Pete Becker Jan 03 '21 at 20:27

score 1 · Answer 2 · answered Jan 03 '21 at 20:25

0b1111 is 0b10000-1.

constexpr unsigned low_mask(unsigned x){
  return (1u<<x)-1;
}

0b1100 is 0b1111-0b11.

constexpr unsigned mask(unsigned width, unsigned offset){
  return low_mask(width+offset)-low_mask(offset);
}

then use xor to flip bits.

unsigned result = 0b100001 ^ mask(4,2);

How do I flip part of number/bitset in C++ efficiently?

2 Answers2