+ 1

# Odd & Even

Please friends write any example about this! I'm a student & it was my 2nd day today in my college, so I need your help to tell me.

11 Answers

+ 4

if(num%2==0) cout << "even";
else cout << "odd";

+ 3

A%2 or A&1
A%2 means A modulo 2 so it is 0 when even and 1 when odd
A&1 means A bitwise and 1 making it get only the binary bit that means 1, then 0 when even and 1 when odd

+ 3

Dorian, I thought A&1 was faster, and many many people are not aware of bitwise and operator :P

+ 2

there isn't any question...

+ 2

By the way, an even number has its modulo 2 equal to 0 and an odd one a modulo 2 equal to 1.

+ 2

Modulo(%) gives remainder so if we perform num%2 then it will give (0 or 1) only . so
if we get 0 then num is even otherwise num is odd.

+ 2

GG for the bitwise! Do you have some benchmark about gain (& vs %)?

+ 2

#include <iostream>
using namespace std;
int main ()
{
int no;
cout <<"enter a number"<<endl;
cin>>no;
if (no%2==0)
cout <<"number is even"<<endl;
else
cout <<"number is odd"<<endl;
return 0;
}

+ 1

bro you may use this code for checking odd even number enjoy coding

+ 1

bitwise ftw

0

no&1==0 is faster than no%2==0