Randomic functions ?

[Cris]

I would like to use a function that returns a integer value between two values. I met the rand() function, but I need a value in a range. []'s Cristiano.

[Nish Nishant]

(rand()/RAND_MAX)*5 for [between 0-5] (rand()/RAND_MAX)*n for [between 0-n]

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Author of the romantic comedy Summer Love and Some more Cricket [New Win] Review by Shog9 Click here for review[NW]

[Lost User]

Or easier, rand () % n If you want a range between A and B then just shift everything along: int value = a + (rand () % (b - a));

[Cris]

This not work corretly. The returned value is zero. []'s

[Lost User]

The following will print 100 random values between 120 and 320 int getRange (int low, int high) { return low + (rand () % (high - low)); } int main (int argc, char *argv[]) { srand (time (NULL)); for (int x = 0; x < 100; x++) printf ("Random Value %d = %d\n", x, getRange (120, 320)); return 0; }

[CoY0te]

Almost... The problem is: int rand(void); so it should be: rand()*n/RAND_MAX or ((double)rand()/RAND_MAX)*n Railgun is like a Gilette Mach 3 - it does the job with one, easy stroke.

[Joaquin M Lopez Munoz]

This is not correct. The expression yields a value between 0 and n, but the probability of n is miniscule (1/(RAND_MAX+1), to be precise), which is not the probability for the other values. The following return a random value between 0 and n with uniform distribution:

int randn(int n)
{
int r;
while((r=rand())==RAND_MAX);
return ((double)r/RAND_MAX)*(n+1);
}

Joaquín M López Muñoz Telefónica, Investigación y Desarrollo