Question

I can’t find any solution to generate a random float number in the range of [0,a], where a is some float defined by a user.

I have tried the following, but it doesn’t seem to work correctly.

float x=(float)rand()/((float)RAND_MAX/a)

Answer 1

Try:

float x = (float)rand()/(float)(RAND_MAX/a);

To understand how this works consider the following.

N = a random value in [0..RAND_MAX] inclusively.

The above equation (removing the casts for clarity) becomes:

N/(RAND_MAX/a)

But division by a fraction is the equivalent to multiplying by said fraction’s reciprocal, so this is equivalent to:

N * (a/RAND_MAX)

which can be rewritten as:

a * (N/RAND_MAX)

Considering N/RAND_MAX is always a floating point value between 0.0 and 1.0, this will generate a value between 0.0 and a.

Alternatively, you can use the following, which effectively does the breakdown I showed above. I actually prefer this simply because it is clearer what is actually going on (to me, anyway):

float x = ((float)rand()/(float)(RAND_MAX)) * a;

Note: the floating point representation of a must be exact or this will never hit your absolute edge case of a (it will get close). See this article for the gritty details about why.

Sample

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

int main(int argc, char *argv[])
{
    srand((unsigned int)time(NULL));

    float a = 5.0;
    for (int i=0;i<20;i++)
        printf("%f\n", ((float)rand()/(float)(RAND_MAX)) * a);
    return 0;
}

Output

1.625741
3.832026
4.853078
0.687247
0.568085
2.810053
3.561830
3.674827
2.814782
3.047727
3.154944
0.141873
4.464814
0.124696
0.766487
2.349450
2.201889
2.148071
2.624953
2.578719