This item was added on: 2003/01/28
This can be done in various ways. Here are a couple:
Division by 2
A number is even if it can be evenly divided by two. So if after division by 2 the remainder is 0, the number is even. One can find the remainder of integer division in C or C++ with the remainder operator (%):
#include <stdio.h>
int is_even ( int i )
{
return i % 2 == 0;
}
int main ( void )
{
int i;
for ( i = 0; i < 10; i++ ) {
printf ( "Testing %d...", i );
if ( is_even ( i ) )
printf ( "Yep, it's even\n" );
else
printf ( "Nope, it's odd\n" );
}
return 0;
}
Bitwise AND
For those familiar with binary numbers, the least significant bit always alternates between 0 and 1. Every other number has the same bit setting. Try it and see, I'll wait... Of better yet, here's a program that will show you:
#include <bitset>
#include <climits>
#include <iostream>
int main()
{
for ( int i = 0; i <= UCHAR_MAX; i++ )
std::cout<< i <<": "<< std::bitset<8> ( i ) <<'\n';
}
Notice how all the least significant bit in all of the even numbers is 0? That's the revelation required to figure out this little trick for finding an even number:
#include <stdio.h>
int is_even ( int i )
{
return ( i & 1 ) == 0;
}
int main ( void )
{
int i;
for ( i = 0; i < 10; i++ ) {
printf ( "Testing %d...", i );
if ( is_even ( i ) )
printf ( "Yep, it's even\n" );
else
printf ( "Nope, it's odd\n" );
}
return 0;
}
For more on bitwise operators, check out bitwise operators in C++ or bit manipulation in C++.
Restrictions
Of course, both of these tricks fail to work on floating-point values. The first fails because the remainder operator is not defined for floating-point; you have to use the fmod function instead to take the remainder of floating-point division. The second also fails because bitwise operations are not defined for floating-point. Why you would want to do this on a floating-point value is another question entirely.
Credit: Prelude
