C changing a bit of uint32_t not working

Question

I got

uint32_t bits = 0;

bits |= 1<< 31;
bits |= 1<< 15;
bits |= 1<< 14;
bits |= 1<< 13;

which gives me 10000000000000001110000000000000

and in another function I do *(bits) |= 0 << 15; but it doesn't do anything, it should change the 16th 1 to a 0, so the following should come: 10000000000000000110000000000000

any help?

`0 << 15` is `0`. No matter how many times you shift `0` to the left or right, you always end up with `0`. — Hristo Iliev, Nov 15 '13 at 13:25
Is the dereferencing in `*(bits) |= 0 << 15;` supposed to be there? Is the second `bits` not of type `uint32_t`? Reading through the question and answers this inconsistency seems a bit confusing although irrelevant to the problem. — , Dec 26 '13 at 14:14

score 5 · Accepted Answer · answered Nov 15 '13 at 13:26

5

The correct way to change a bit to 0 is to and the negation, e.g.

bits &= ~(1 << 15);

answered Nov 15 '13 at 13:26

dornhege

1,500
8
8

score 1 · Answer 2 · answered Nov 15 '13 at 13:27

1

You have to do:

bits &= ~(1 << 15)

answered Nov 15 '13 at 13:27

Sergey K.

24,894
13
106
174

score 1 · Answer 3 · answered Nov 15 '13 at 13:27

Clearing a bit is done by AND-ing with the complement, i.e.:

*(bits) &= ~(1 << 15);

(1 << 15) produces a number with bit 15 set to 1:

00000000000000001000000000000000

then the bitwise NOT operator inverts this and produces a number that is all ones except bit 15:

11111111111111110111111111111111

When you AND this number to the original one, the bit in position 15 is cleared:

XXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXX
               &
11111111111111110111111111111111
================================
XXXXXXXXXXXXXXXX0XXXXXXXXXXXXXXX

where X marks the original bits.

C changing a bit of uint32_t not working

3 Answers3