Pergunta de entrevista da empresa Kinaxis

Come up a most efficient way to determine if a integer is odd or even.

Respostas da entrevista

Sigiloso

2 de out. de 2017

int number = ...; if(number % 2) { odd } else { even }

2

Sigiloso

4 de fev. de 2018

Compilers are smart enough to understand the result of n%2. Don't over-complicate it.