Is there a bitwise operation that returns 1 of the bits are equal?

Question

I am trying to write a simple C bitwise operation that returns YES if the bits are equal. Truth table is the following:

I tried various combinations of AND,XOR and OR and no luck. Granted, I could scan each bit pair and ask the question using an "IF" statement but rather do it more elegantly.

This is an example of what I need:

x = 251 - 11111011
y = 18  - 00010010
r = 22  - 00010110

I think we need to see an example input, with expected output. It's not clear whether the output is supposed to be a single boolean, or an integer value containing all the resulting bits. — user3386109, Apr 27 '23 at 23:11
its `~(X ^ Y)` which is the same as `~X ^ Y` and `X ^ ~Y`; for reference: https://en.wikipedia.org/wiki/Logical_biconditional — Pignotto, Apr 27 '23 at 23:35
So if `X ^ ~Y` produces the right input then `^~` is the operator you wanted. Kind of like the `-->` (down to) operator https://stackoverflow.com/a/2220319/2193968. — Jerry Jeremiah, Apr 27 '23 at 23:49

score 7 · Accepted Answer · edited May 02 '23 at 01:39

7

@daniel, @Barmar and @pignotto have given the answer in comments, but to make it clear for you. Make a truth table for the XOR operator (which in C would be X ^ Y):

X	Y	R
0	0	0
0	1	1
1	0	1
1	1	0

This is the inverse of what you want - i.e. where R is 0, you want 1 and vice-versa.

So you want R = ~(X ^ Y).

The ~ operator inverts all the bits, turning 0 into 1 and 1 into 0.

Equivalent expressions are ~X ^ Y and X ^ ~Y. If X or Y is a constant or is used repeatedly where it does not vary (as in a loop), then ~X or ~Y could be computed in advance, leaving only XOR to be computed. (This is a minor optimization, and a modern compiler might make it automatically.)

edited May 02 '23 at 01:39

InSync

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answered Apr 27 '23 at 23:38

pmacfarlane

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`~X ^ Y` has the same result and should be mentioned in case `X` is constant, so `~X` can be computed at compile time (or as a loop invariant etc.), thus saving one operation at run-time. Although the compiler might recognize this anyway as part of optimization. – Eric Postpischil Apr 28 '23 at 00:29
@EricPostpischil Please edit my answer to that effect if you wish. – pmacfarlane Apr 28 '23 at 00:31

Is there a bitwise operation that returns 1 of the bits are equal?

1 Answers1