Thread: Proper use of srand()

The exercise was to generate 1000 random numbers between 0-9 and to print how many of each number were generated - and to repeat the task with ten different seeds. I used srand() with time() and yet I get the same distribution of numbers each time. Why is this?

Code:

#include <stdio.h>
#include <stdlib.h>
#include <time.h>

int main(void)
{
   int nums[10] = {0, 0, 0, 0, 0, 0, 0, 0, 0, 0};
    int i,j;
    printf("1     2     3     4     5     6     7     8     9     10\n");
  
    for (i = 0; i < 10; i++)
    {
        srand((unsigned int) time(0));
        for (j = 0; j < 1000; j++)
            nums[(rand() % 10)]++;

        for (j = 0; j < 10; j++)
        {
            printf("%-6d", nums[j]);
            nums[j] = 0;
        }
        printf("\n");
    }
    getch();
    return 0;
}

time() only has second precision. Chances are your program finishes well within 1 second, so all calls to srand have the same seed.

Only probable explanation is that the random number seed isn't changing and that's why you are getting a uniform distribution.
Perhaps sleep() for a while and see if that changes it to a random distribution. Just my 2c.

If the exercise simply calls for the use of ten seeds, you can do something simple like:

Code:

for(int i = 0; i < 10; i++)
{
  srand(i);
  /* ... */
}

It sounds like the point is to see how the PRNG works, so it would hardly matter which seeds are used.

My C library treats srand(0) and srand(1) identically; perhaps it would make sense to use srand(i + 1).

Aha! Using i for the argument to srand() does the trick. I didn't know time() was only to the second! I thought it was like the timer on my old CBM-64....