CODEWITHSUNDEEP
operatorsbit-manipulationmodulusand-operator
MichaelMitchell
MichaelMitchell

Reputation: 1167

Is Modulus more efficient than And operator for finding halfs

I started playing with bitwise operators today I noticed that the & operator could be used to find if something is even or not.

100 & 1;//Gives 0
101 & 1;//Gives 1

Zeros for even and ones for odds.

So, I am currently wondering which is a more efficient method of finding if something is even:

A)if (n & 1 == 0) isEven = true;

or

B)if(n % 2 == 0) isEven = true;

Also, is there a more efficient way to find if something is even?

P.S. There is no reason that I need for an efficient way of finding if something is an even number, I am just genuinely curious.

Upvotes: 0

Views: 41

Answers (2)

olegarch
olegarch

Reputation: 3891

More efficient way:

isEven = !(n & 1);

Regarding using modulto operation %2 or so on:

On some compilers, it can generate longest/slowest code for signed integers, because of, for example: 

-5 % 2 == -1

For unsigned "n", usually compiler generate same code for both operations:

n % 2
n & 1

Upvotes: 1

Joachim Isaksson
Joachim Isaksson

Reputation: 180917

Compilers are pretty smart; if either is actually faster in practice, both &1 and %2 will quite possibly compile to the same - faster - code.

With a stupid compiler, I'd place a bet on &1 being at least as fast as the alternative since it maps to bit operations which is a very basic operation for pretty much any CPU.

Modulo may be as fast on some CPUs, but definitely not so on low cost embedded CPUs.

Upvotes: 2

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