Conditional Statement using Bitwise operators

Question

So I see that this question has already been asked, however the answers were a little vague and unhelpful. Okay, I need to implement a c expression using only "& ^ ~ ! + | >> <<"

The expression needs to resemble: a ? b : c

So, from what I've been able to tell, the expression needs to look something like:

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

This works when a = 0, because anding it with b will give zero, and then the or expression will return the right side, (~a & c) which works because ~0 gives all ones, and anding c with all ones returns c.

However, this doesn't work when a > 0. Can someone try to explain why this is, or how to fix it?

Answer 1

I would convert a to a boolean using !!a, to get 0 or 1. x = !!a.

Then I'd negate that in two's complement. Since you don't have unary minus available, you use the definition of 2's complement negation: invert the bits, then add one: y = ~x + 1. That will give either all bits clear, or all bits set.

Then I'd and that directly with one variable y & b, its inverse with the other: ~y & c. That will give a 0 for one of the expressions, and the original variable for the other. When we or those together, the zero will have no effect, so we'll get the original variable, unchanged.