Generating a random, uniformly distributed real number in C [duplicate]

Answer 1

You can use:

#include <time.h>
srand(time(NULL)); // randomize seed
double get_random() { return (double)rand() / (double)RAND_MAX; }
n = get_random();

srand() sets the seed which is used by rand to generate pseudo-random numbers. If you don't call srand before your first call to rand, it's as if you had called srand(1) (serves as a default).

If you want to exclude [1] use:

(double)rand() / (double)((unsigned)RAND_MAX + 1);

Full solution:

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

double get_random() { return ((double)rand() / (double)RAND_MAX); }

int main()
{
    double n = 0;
    srand(time(NULL)); // randomize seed
    n = get_random(); // call the function to get a different value of n every time

    printf("%f\n", n);  // print your number
    return 0;
}

Every time you run it you will get a different number for n.

Answer 2

This shows how to get random real numbers in the range 0..1 but please note that they are not uniformly distributed. There are only (RAND_MAX+1) discrete values.

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

int main(void) {
    int i;
    double n;
    srand((unsigned)time(NULL));            // seed the random num generator ONCE only
    for(i = 0; i < 3; i++) {                // get 3 random numbers
        n = (double)rand() / RAND_MAX;      // in the range 0 ... 1
        printf("%f\n", n);                  // use correct format specifier for the var
    }
    return 0;
}

My program output:

0.622608
0.814081
0.878689