Thread: Random number in certain interval without using any standard function(algorithm)

This took a very long time to explain
r = min + rand() % (max - min);
which skews the natural distribution of rand() - a standard function, by the way - to get results that you want.

A better way would be to split RAND_MAX into groups equal to the size of your accepted range, and then divide into your result. This way you are at least working with rand()'s own distribution, and the results should be better because of it.

Code:

int random(int lim) {
   int value;
   int divisor = RAND_MAX / (1 + lim);

   do {
      value = rand() / divisor;
   } while (value > lim);

   return value;
}

int r = min + random(max - min);

This does virtually the same thing as scaling a floating point number (i.e. ((double)lim) / RAND_MAX).

Originally Posted by LabedNadjib82

hello everyone, I want to share with you this simple and great idea to get a bounded random number algorithmically- follow this youtube video : [YouTube]

The pattern you'll see in most professional programs is

Code:

/* get a uniformly distributed number p = 0  to 1 - episilon */
#define uniform() (rand()/(RAND_MAX + 1.0))

/* we can now use it to obtain integers in any bounds, like this */
int randletter = 'a' + (int) (uniform() * 26);

/* or if we need a lot of them, write a second macro */

#define rnd(low, high)  ((int) (((high)-(low)+1)*uniform()) + (low))

If you need better random numbers, you need to rewrite
uniform() to use a better generator than rand().
Often you want gaussian random numbers, however, with
a bell-curve distribution.