use of: index & 0x01

Question

I want to know about what is the use of that (index & 0x01) in the code?

if(((arr[index] >= 0) && (!(index & 0x01))) 
                || ((arr[index] < 0) && (index & 0x01))) 
{ 
    outofplace = index; 
}

Is this your question https://stackoverflow.com/questions/6533582/advantage-of-using-0x01-instead-of-1-for-an-integer-variable ? — Vinay Somawat, Jun 26 '20 at 07:15
Does this answer your question? [Advantage of using 0x01 instead of 1 for an integer variable?](https://stackoverflow.com/questions/6533582/advantage-of-using-0x01-instead-of-1-for-an-integer-variable) — kesarling He-Him, Jun 26 '20 at 07:15

score 6 · Accepted Answer · answered Jun 26 '20 at 07:20

6

A number is odd if and only if its last digit is odd, regardless of the base. So if we want to know the number's oddity, it's enough to check if the last bit is set.

index & 0x01

will be 1 if and only if index is odd.

answered Jun 26 '20 at 07:20

bereal

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brc-dd · Answer 2 · 2020-06-26T07:57:52.990

If we have to deduce a general rule, we can say that for any non-negative number x,

x % y == (x & (y - 1))

provided that y is a positive power of 2.

This is a common hack in competitive coding. It is used because the competitive programmers think that bit-wise AND works faster than modulo.

In modern compilers, there is no performance difference at all. Read this thread.

There is no special reason in writing it as 0x01 instead of 1. Both compile to give the same assembly! Almost everyone (who uses this hack),= uses 1, because we have to type 3 characters extra in 0x01. :P

Here in this case, index & 0x1 is equivalent to index % 2 which is simply a condition to check if the number is odd. (Array indexes in C++ are always positive, unless you are going out of bound.)

Bob__ · Answer 3 · 2020-06-26T09:26:28.403

As the other answers pointed out, while this is a well known pattern (see also this Q&A about that mask), it can be considered a premature optimization.

I'd like to suggest the following alternative to the posted code, which I find more readable. Your mileage may vary.

// Give a meaningful name.
constexpr auto is_odd = [] (auto x) -> bool {
    return x % 2;
}

// Use it to simplify the condition.
if ( (arr[index] < 0) == is_odd(index) ) {
    // Do something
}

use of: index & 0x01

3 Answers3