How can I get the value of the least significant bit in a number?

Question

I'm working on a programming project and one of things I need to do is write a function that returns a mask that marks the value of the least significant 1 bit. Any ideas on how I can determine the value using bitwise operators?

ex: 
0000 0000 0000 0000 0000 0000 0110 0000 = 96
What can I do with the # 96 to turn it into:
0000 0000 0000 0000 0000 0000 0010 0000 = 32

I've been slamming my head against the wall for hours trying to figure this out any help would be greatly appreciated!

It's not a duplicate. I forgot to mention that it has to be using bitwise operators — Riptyde4, Sep 14 '13 at 21:31
Not a duplicate. The goal is not to get the *index* of the lowest bit, but the value. — R.. GitHub STOP HELPING ICE, Sep 14 '13 at 21:31
@R..: Getting the value from the index is trivial (`1 << index`). — cHao, Sep 14 '13 at 21:36
Yes, but getting the value via the index is grossly inefficient. You're performing an expensive operation the inverting it rather than solving the problem directly. — R.. GitHub STOP HELPING ICE, Sep 14 '13 at 21:52

score 60 · Accepted Answer · answered Sep 14 '13 at 21:32

60

x &= -x; /* clears all but the lowest bit of x */

answered Sep 14 '13 at 21:32

R.. GitHub STOP HELPING ICE

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Can you explain this? I'd like to understand how it works. and whats the difference between using x & -x? and x &= -x? – Riptyde4 Sep 14 '13 at 21:35
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For an unsigned type, -x is equivalent to TYPE_MAX - x + 1, which is equivalent to ~x + 1. It's easier to see why `x &= ~x + 1` works. – Tavian Barnes Sep 14 '13 at 21:37
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See Tavian's comment for the explanation. – R.. GitHub STOP HELPING ICE Sep 14 '13 at 21:55
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Mod down - question was to find the "position" - this function simply returns the value of that bit – Brad Oct 22 '17 at 21:49
@Brad OP gives an example of the intended output – M.M Dec 13 '17 at 03:15
Down vote for: 1. wrong resulting code in answer. @TavianBarnes code is correct. 2. Brad is correct too. – Ivy Growing Feb 05 '21 at 09:31

Seth · Answer 2 · 2020-11-29T23:41:13.013

To be sure you get the right bit/value:

The value at the least significant bit position = x & 1
The value of the isolated least significant 1 = x & -x
The zero-based index of the isolated least significant 1 = log2(x & -x)

Here's how it looks in JavaScript:

let x = 0b1101000;

console.log(x & 1);            // 0 (the farthest-right bit)
console.log(x & -x);           // 8 (the farthest-right 1 by itself)
console.log(Math.log2(x & -x); // 3 (the zero-based index of the farthest-right 1)

score 2 · Answer 3 · edited Dec 13 '17 at 03:08

2

A more readable code:

int leastSignificantBit(int number)
{
    int index = 0;

    while ((~number) & 1) {
        number >>= 1;
        index++;
    }
    return 1 << index;
}

edited Dec 13 '17 at 03:08

Craig McQueen

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answered Sep 14 '13 at 21:34

MasterID

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The text "a more readable code" is misleading. This is an alternate approach that performs an expensive operation (bit search, essentially a type of log) then inverts it to get the answer, rather than just computing the answer directly. – R.. GitHub STOP HELPING ICE Sep 14 '13 at 21:54
2

I understand. But this doesn't require knowledge about two's complement, this is why its "simpler". But I agree with you, it is much more expensive. – MasterID Sep 14 '13 at 22:01

How can I get the value of the least significant bit in a number?

3 Answers3

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