Clearing least significant bit

0

Question

How would I change the least significant bit to 0 without affecting the rest of the value, for example: 0xffffffff , for example in C or Python? I am not a mathematical genius, so for me it's a hassle to find out how to achieve this.

Solution

n = 0 // clear this bit
0xffffffff &= ~(1U << n) = 0xfffffffe
c
bit
asked on Stack Overflow Apr 7, 2014 by maikovich • edited Apr 7, 2014 by maikovich

4 Answers

2

Try this to clear the least significant bit:

int x = 0xffffffff;

...

/* clear least significant bit */
x &= ~1;

To set the LSB

int x = 0xffffffff;

...

/* set least significant bit */
x |= 1;

To change the LSB

int x = 0xffffffff;

...

/* change least significant bit */
x ^= 1;

If you want to extent that to set / clear / change bit n, change the 1 to (1U<<n).

answered on Stack Overflow Apr 7, 2014 by abligh • edited May 13, 2019 by Community
1
 unsigned x = (x>>1U)    
          x = (x<<1U)  

This is easy to understand ....
example :
Let x = 10010101011

step 1 : x = 0 1001010101 // x>>1
step 2 : x = 1001010101 0 // x<<1

compare :

10010101011
10010101010

Another method can be unsigned x = x & 0Xfffffffe which is equal to 11111111111111111111111111111110 i.e 31 bits 1 to keep bit 31 to 1 as it occours in x .

Hope this helps !

answered on Stack Overflow Apr 7, 2014 by Aseem Goyal • edited Apr 7, 2014 by Aseem Goyal
1

Using bitwise NOT operator ~0 evaluates to all bits set to 1. Assign this to an unsigned int variable and shift it left by 1 bit so that all bits are 1 except the least significant bit. The use bitwise AND operator & on x to toggle the least significant bit.

unsigned int x = 0xffffffff;
unsigned int y = 0;
x = x & (~0 << 1);
answered on Stack Overflow Apr 7, 2014 by ajay
0

If you don't want to dig into bit ops, just do i = (i / 2) * 2;. This will clear bit 0 (why?).

answered on Stack Overflow Apr 7, 2014 by mfro

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