How to find a no is even or odd without using % operator or without using loop?

Answer 1

if LSB bit is set ,the given no is odd otherwise given no is even.

#include <stdio.h>
int num_odd_even(unsigned int num)
{
      if(num & 1)
          printf("Given no is odd\n");
    else
      printf("given no is even");
}

Answer 2

Try something like (assume a is your integer)

if (a & 1)
 printf("a is odd number");
else
 printf("a is even number");

Answer 3

#include<stdio.h>
void main()
{
    int i,j;
    printf("Enter a number:\t");
    scanf("%d",&i);
    j=i/2;
    if(i==j*2)
        printf("even");
    else
        printf("odd ");
}

//-------------------------------------OR---------------------------------------

#include<stdio.h>
int main()
{
int x = 19;
(x & 1)? printf("Odd"): printf("Even");
return 0;
}
// Output: Odd

Answer 4

main()
{
 int n,a;
 cin>>n;
 a=n&1;
 if(a==1)
  cout<<"odd";
 else
  cout<<"even";
}

Answer 5

#include<stdiio.h>


#include<conio.h>


void main()


{


int i,j;


  clrscr();


printf("


 Enter the number");


scanf("%d",&i);


j=i/2;


if(i==j)


printf("


 Entered number is even");


else


printf("


 Entered number is odd ");


getch();


}