Thread: Help with rand()

Originally Posted by brewbuck

Then I'd call it like this:

Code:

double x = round( RandRange( -100.0, 100.0 ) ) / 100.0;

I'm not sure, but at first glance I think that produces a bias against both -1.0 and 1.0.

Maybe

Code:

double x = round( RandRange( -100.4999, 100.4999 ) ) / 100.0;

or whatever number of 9s you like to reduce the bias.

Originally Posted by R.Stiltskin

You're getting integers in the range 0 - 199 so the mean will be 99.5, not 100. Divide by 100, subtract 1 and multiply by 100,000 and your mean *should* be -500. You probably want something like this:

Code:

for (int j = 0; j < 10000; j++) {
  double sum = 0;
  double frac = 201.0 / (unsigned int)(RAND_MAX + 1);
  for (int i = 0; i < 100000; i++) {
    double n =  (int)(rand() * frac) / 100.0 - 1;
    sum += n;
  }
}

I am disappointed I missed that one. Thanks a lot for all the help to everyone that posted.

Originally Posted by quzah

(rand() * 201), on anything more than (RAND_MAX/201) is going to end up with an overflow.

That's fairly obvious. I was careful to avoid doing that by storing the fraction first.

There's no overflow here:

Code:

#include<stdio.h>
#include<stdlib.h>
int main( void )
{
    int i;
    double n;
    double frac = 201.0 / ((unsigned int)RAND_MAX + 1);
    
    printf( "denom = %u\n", (unsigned int)RAND_MAX + 1 );
    printf( "frac = %e\n", frac);
    
    for( i = 10000000; i > 0 && i < RAND_MAX; i+=10000000 ) {
      n = (int)(i * frac) / 100.0 - 1;
      printf( "%f\n", n);
//      printf("%f\n", (int)(i * frac) / 100.0 - 1);
    }
    n = (int)(RAND_MAX * frac) / 100.0 - 1;
    printf( "finally %f\n", n );
    return 0;
}