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;
}

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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.

Quote:

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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.

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

I can think of 3 ways to do this:

binary & operator
left/right shifting
division/multiplication

Using bit shifting amounts to multiplication or division, but that should indeed work.

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.

O_o

That's clearly not what he thought.

He was saying that the "trick" may not work if your system doesn't use "Two's Complement" representation for negative numbers whereas using `mod' will.

Soma

Code:

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

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That trick solves the negative-number-representation problem.

:cool:

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?

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

Code:

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

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where eventest returns if it is even or not?

Quote:

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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);
}

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where eventest returns if it is even or not?

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that is just bending the rule to the breaking point.

yep :P also another way, use all the lines you want, just dont hit enter

Code:

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/* some code */; /* some more code*/; /*even more code*/;

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