Even or Odd integer detection.

Someone once asked me in a sandbox if I knew how to detect even or odd. I couldn't think of it then, and I forgot the person's name, but here it is now. Hoping this can help anyone. If you want help adapting this for your script just IM me in world.

//Even/Odd from Chris Knox

float n;
float m;

default
{
    state_entry()
    {
        n = 0;
        
    }

    touch_start(integer total_number)
    {
        m = (n + 1);
        
        float number = n;
        
        float tell = llRound(number / 2);
        
        if ((tell * 2) == number)
        {
            llSay(0, (string)llRound(number) + " is even.");
        }
        else
        {
            llSay(0, (string)llRound(number) + " is odd.");
        }
        
        n = m;        
        
    }
}

you can also do it this way. my script is probably sloppy but the method seems to work alright.

string OddOrEven(integer n) {
    if (!n)   return "i dunno if zero is odd or even.";
    integer temp = n % 2;
    if (temp) return (string)n +" is odd.";
    else      return (string)n +" is even.";
}

default {
    state_entry() {
        // replace 11 with whatever number you want to check.
        llSay(0, OddOrEven(11));
    }
}

It's much easier:

if (number % 2 == 0) { llSay(0, (string)number + " is even."; }
if (number % 2 == 1) { llSay(0, (string)number + " is odd."; }

// Returns TRUE if testNumber is even, FALSE otherwise
integer IsEven(integer testNumber) {
if (testNumber % 2 == 0) {
return(TRUE);
}
return(FALSE);
}

Or, you could use the modulus operator:

(pseudo code)

if (some_integer % 2 == 0) then some_integer is even
else some_integer is odd

Xylor

Simpler method given 'integer n'


integer isOdd = n & 1;


This does a bitwise comparison; if the number is odd, the low bit (1) is set - no even number can have the low bit set, since the rest of the bits are all evens, and even+even number = even.

Try this... given 'integer n':

integer isOdd = n & 1;

This will do a bitwise comparison; if a number is odd, the low bit (the only odd bit - 1), will be set.

Try this... given 'integer n':

integer isOdd = n & 1;

This will do a bitwise comparison; if a number is odd, the low bit (the only odd bit - 1), will be set.

Even better... Given 'integer n':

integer isOdd = n & 1;

This will do a bitwise comparison; if a number is odd, it's lowest bit (1) is set.

Even better... Given 'integer n':

integer isOdd = n & 1;

This will do a bitwise comparison; if a number is odd, it's lowest bit (1) is set.

Given float n, cast it to an integer:

integer isOdd = (integer)n & 1;

Floats cast to integers drop the decimal. If you want it rounded to nearest:

integer isOdd = (integer)(n + 0.5) & 1;

In some languages, the "n" will be ANDed with the 1, and the result "isOdd" will be TRUE or FALSE but the value of "n" will still be changed. If "n" is a "throw away" variable, that's not a problem. The method used by Xylor does not change the variable. I do not mean to be picky, but if you were tight on memory, the Modulo version of "Even/Odd" discovery is best.

And of course this applies to any value. IE, is it divisible by x?

llWhisper(0, "Divisible by " + (string)x) && (number % x) == 0;

Hmmm well, a similar method like that works in PERL, =)

Just a minor point in response to the last post; but "n & 1" will not store the result back in n; the value of n is preserved.

To store the result back in n, you'd use n = n & 1; or n &= 1;.

Jack