Making a particular bit "0" in C++

Question

I'm new at programming. Recently i came across a problem in which i have to make a particular bit 0 of a number.

For example :

I have a number p

p      = 73
binary = 1001001

Now i want to make 4th bit to 0, that is 1000001(2) = 65(10)

I did this in following way :

int p = 73;
int pos = 1<<3; // 4th bit
int max_bit = (1<<31) - 1; // making all bit to 1
int mask = pos ^ max_bit; // making 4th bit to 0 except others
p = p & mask; // changing 4th bit of p to 0
cout<<p<<endl;

Is there a better way to do this ?

just as a "did you know": that is called clearing a bit. Making it "1" is called setting a bit — bolov, Dec 30 '14 at 13:51

Ali Akber · Accepted Answer · 2014-12-30T13:58:14.413

19

Just use :

p = p & ~(1u<<3);

What happens here ?

 1. (1u<<3)       0...01000
 2. ~(1u<<3)      1...10111 // Invert the bits
 3. p & ~(1u<<3)  *****0*** // Here * means the bit representation of p

That's how the bit changes to 0.
Hope it helps :)

edited Dec 30 '14 at 13:58

answered Dec 30 '14 at 13:49

Ali Akber

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Maybe add an explanation for those not familiar with bit manipulation? – jpw Dec 30 '14 at 13:51
@jpw okay.. i'm adding an explanation – Ali Akber Dec 30 '14 at 13:52
I think, may be better is to write `p & ~(1u << n);` – Grijesh Chauhan Dec 30 '14 at 13:53
@GrijeshChauhan yes you are right. But OP using `int` so i think `1< – Ali Akber Dec 30 '14 at 13:55
no I means it can be undefined behaviour if singed bit set but I am not sure about C++ standards. Anyways this is simple and good answer. – Grijesh Chauhan Dec 30 '14 at 13:57
@GrijeshChauhan you are right :) – Ali Akber Dec 30 '14 at 14:00
Use `p &= ~(1u<<3);` instead. – S.S. Anne Jun 07 '19 at 16:11

Making a particular bit "0" in C++

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