Gameboy Development Forum

you could just keep generating random numbers untill its the one you want

unsigned short rand_num;

while(1){


rand(rand_num);//get random number (i dont know how you do it)

if(rand_num < 1000){//if less than 1000 then accept it
break;
}
else{
continue;
}

}

The simplest way to go about solving this problem is nothing more than a bitwise AND operation.
Simply define your value as for example:

int r = rand() & 7;

This will take the value and remove everything above the first 3 bits, giving us a max value of 1 + 2 + 4 (7)

More In-depth:
In case you don't know what a bitwise AND does, here's a little explanation.

AND in logic terms is a operator which takes 2 or more entries and checks if all of these are [On/ True/ 1], returning [On/ True/ 1] if such is the case and [Off/ False/ 0] if not.

example:

1 & 1 = 1
1 & 0 = 0
0 & 1 = 0
0 & 0 = 0

What a bitwise AND does, is it takes all the bits that makes up the two values you feed into it and returns a value consisting of the AND result of each pair.

for example, let's say we calculate a bitwise AND for the values 150 and 7:

01101001 = 150
AND
11100000 = 7
||||||||
01100000 = 6

No matter what we feed into the first value and second value, the result won't exceed either (as any AND against a 0 will return 0)
And thus you can use that to limit the range of your randomly generated number

Last edited by RoboTwo (2014-11-12 11:30:35)

It will return any number between 0 and the number you AND it with.
Care to explain what you mean by not work for even numbers? Do you want to round it to a even number or what?

Yes, if you AND with an even number it'll return a even number and if you AND with a uneven number it will return either a uneven or even based on the second number you input.
But given that you wanted a number in the range 0-7 you should be perfectly fine with this approach, as 7 is a uneven number and can thusly return both even and uneven numbers if used in a AND operation.

Crona, you are correct. RoboTwo's approach only works for numbers that are a power of two: 0, 1, 2, 4, 8, 16 etc.

Basically, RoboTwo's approach is a special case of G.H.O.O.N.'s approach. For N = 2^x, rand() % (N - 1) == rand() & (N - 1). So if you need N different random numbers, you can AND by N-1. Hence, rand & 7 will generate 0 thru 7, i.e. 8 (=N) states.
For all other numbers, you will have to use the actual formula with the modulo.

cYa,

Tauwasser