Setting nth bit of unsigned int

Question

Given a unsigned int x, I want to set the nth bit to y, and y can be either 0 or 1. Is it possible to create an expression using bitwise operators to do this while avoiding the use of any conditional statements? Thanks.

Please read [How to ask](http://stackoverflow.com/help/how-to-ask). — segarci, Feb 06 '15 at 07:26

minmaxavg · Accepted Answer · 2017-01-30T11:36:22.230

10

x = (x & (~(1 << n))) | (y << n)

Quite simple. (First, clear the nth bit, and set nth bit to 1 if y is 1.)

edited Jan 30 '17 at 11:36

answered Feb 06 '15 at 07:33

minmaxavg

686
6
21

Still this has a condition, even if it is in one line. – Eun Feb 06 '15 at 07:39
Thank you! I never considered doing it this way. – user95297 Feb 06 '15 at 07:39
@Eun By condition I meant without having to check the value of y using an if-else statement or something similar. – user95297 Feb 06 '15 at 07:40
3

...or, more cleaner: `x = (x | (1 << n)) & (y << n)` - this version sets `n`th bit to 1 in the first place. – minmaxavg Jun 12 '16 at 09:34

Emil Laine · Answer 2 · 2015-02-06T07:39:41.510

2

x ^= (-y ^ x) & (1 << n);

edited Feb 06 '15 at 07:39

answered Feb 06 '15 at 07:28

Emil Laine

41,598
9
101
157

This is not unconditional. OP wants this without any condition – Eun Feb 06 '15 at 07:30
I am aware of this, but I am wondering if it is possible to incorporate y into the expression so that both cases (with y being 0 or 1) can be handled with one statement. – user95297 Feb 06 '15 at 07:31
@Eun Sorry you're right, I missed that. Fixed my answer. – Emil Laine Feb 06 '15 at 07:41

Setting nth bit of unsigned int

2 Answers2