Write a c++ code to check wether a given number is odd or even?

Answer 1

AND the number with 0x1, if its true then its odd, if false, then even.

Answer 2

The above posters who mention bitwise AND have the most efficient/correct answer. Here's the corrected version for above poster's program.

#include
using namespace std;

int main()
{
int i = 0;
cout << "Enter a number: ";
cin >> i;
cout << i << " is an " << (i&1) ? "odd" : "even" << " number." << endl;
return 0;
}

C++ has been revised in 99, hence the iostream, not iostream.h. All the standard functions are in the standard namespace. Hence using namespace std;. Main returns an int, so it's int main, not void main. Drop the conio.h, it's non-standard.

Answer 3

The key is the modulo operator (%).

The following code will do:

#include
using namespace std;
int main() {
int x;
cin >> x;
if(x%2) {
cout << x << " is odd";
} else {
cout << x << " is even";
}
return 0;
}

Answer 4

Program that check whether a given number is odd or even:

#include
#include
#include
void main()
{
int i;
clrscr();
cout<<"Enter a Number :";
cin>>i;
if((i/2)==0)
{
cout<<"Its a even number";
}
else
{
cout<<"its a odd number";
}
getch();
}

Answer 5

Using bitwise AND (&) is MUCH faster than division or modulus.

#include
#include
#include
void main()
{
int i;
clrscr();
cout<<"Enter a Number :";
cin>>i;
cout << i << " is an " << (i&1) ? "odd" : "even" << " number."
getch();
}

Answer 6

Check and see if the modulus of being divided by two is one or zero. If it is one it is odd, if it is zero it is even.