Thread: Simple Challenge

Complete the function IsEven() with just 1 line of code and without using Mod (%), +,*,-, or any built in functions.

NB. A line of code ends with a ';'

Good luck =D

Code:

#include <iostream>
using namespace std;

bool isEven(int f)
{
    //code here
}

int main (int argc, char * argv[])
{
    int f;
    cout << "Enter # ";
    cin >> f;
    cout << f << ( isEven(f)  ?  " is even" : " is odd" ) << endl;
    return 0;
}

Originally Posted by whiteflags

Even if you know the trick, it isn't guaranteed to work like mod... Mod works no matter what your system does to represent negative numbers.

that's true. at least you know the 'trick'.

oh whiteflags it wasn't intended to work like mod. every even number's last bit is 0 and odd's are 1. so by ANDing the last bit you can figure out which of the two it is. it doesn't find the remainder if that's what u thought.

Code:

bool IsEven(int f)
{
    return !(((f < 0) ? -f : f) & 1);
}

That trick solves the negative-number-representation problem.



EDIT: Unless the representation is 1s Complement and f == -0 ( Yes, in 1s Complement 0 and -0 are different )
EDIT^2: Anyway, what CPU uses 1s Compement nowadays?

Originally Posted by bobknows

couldnt u just make another function to do it with several lines and put it in as

Code:

bool iseven(int f)
{
      return eventest(f);
}

where eventest returns if it is even or not?

that is just bending the rule to the breaking point.