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?

Solution 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.

Solution 2

In other words, you need a to have all bits set to 0, if a is false (i.e. 0), and have all bits set to 1, if a is true (i.e. a > 0).

For the former case, the work is already done for you; for the latter -- try to work out result of the expression ~!1.