Home/ Questions/Q 1095421
In Process

The Archive Base Latest Questions

Editorial Team
  • 0
Asked: 2026-05-17T00:07:17+00:00

How is the conditional operator represented using bitwise operators? It is a homework question

  • 0

How is the conditional operator represented using bitwise operators?

It is a homework question where I have to implement the conditional operator using only bitwise operations. It would be simple if if statements were allowed, however it has to be strictly bitwise operators.

Only the operators !, ~, &, ^, |, +, >>, and << can be used. No if statements or loops can be used.

The function takes three ints and works just like the normal conditional operator. The first argument is evaluated as either zero or non-zero. If the first argument is zero then the second argument is returned. If the first argument is non-zero then the third argument is returned.

I was hoping there would be a simple algorithm for this. Any ideas on where to start would be a great help.

Share

Leave an answer

You must login to add an answer.

Need An Account,

1 Answer

  1. Editorial Team

    Are shifts allowed as bitwise operators? Are arithmetic operators allowed?

    Your edit is not entirely clear, but I assume that you need to implement an equivalent of

    a ? b : c

    where a, b and c are integers. This is in turn equivalent to

    a != 0 ? b : c

    One way to achieve that is to find a way to turn non-zero value of a into an all-ones bit pattern using only bitwise operators. If we figure out how to do that, the rest would be easy. Now, I don’t immediately remember any ingenious tricks that would do that (they do exist I believe), and I don’t know for sure which operators are allowed and which are not, so for now I will just use something like

    a |= a >> 1; a |= a >> 2; a |= a >> 4; a |= a >> 8; a |= a >> 16;
a |= a << 1; a |= a << 2; a |= a << 4; a |= a << 8; a |= a << 16;

    For a 32-bit integer type, if (and only if) there was at least one bit set in the original a, the above should result in all bits of a set to 1. (Let’s assume we are working with unsigned integers, to avoid the issues associated with shifting of signed values). Again, there must be a more clever way to do that, I’m sure. For example: a = !a - 1, but I don’t know if ! and - are allowed.

    Once we’ve done that, the original conditional operator becomes equivalent to

    (a & b) | (~a & c)

    Done.

    • 0