How to generate random numbers between 0.1 and 0.01

Question

I know its possible to generate integer random number using

(rand() % (difference_between_upper_and_lower_limit)) + lower_limit

I'm wondering if theres a way to generate a number between 0.1 and 0.01. I've tried double speed = (rand() % 10)/100; But it always just gives me 0.0000000; Thanks in Advance!`

`/100` is integer division. `/100.0` is floating point division. — user3386109, Aug 31 '18 at 01:59
Have you looked at `drand48`? It won't be perfect (i.e., the manipulations you would have to perform upon its result) but it might be adequate for what you need. Have a look [here](https://stackoverflow.com/questions/33058848/generate-a-random-double-between-1-and-1) for some ideas. — Jeff Holt, Aug 31 '18 at 02:00
There are an infinite amount of numbers between 0.1 and 0.01. Do you mean that you want to generate a random number from the ten numbers in this set (0.01, 0.02, 0.03, 0.04, 0.05, 0.06, 0.07, 0.08, 0.09, 0.10)? If not what do you mean? — Stuart, Aug 31 '18 at 02:08
Possible duplicate of [Integer division always zero](https://stackoverflow.com/questions/9455271/integer-division-always-zero) — phuclv, Aug 31 '18 at 03:59

score 2 · Answer 1 · edited Aug 31 '18 at 13:22

2

I think, you missed the type-casting part ==> ((double)(rand() % 10))/100;

Try this code.

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

int main(void) {
  // If you don't set the seed-value, you'll always get the same random numbers returned in every execution of the code
  srand(time(0));

  double speed = ((double)(rand() % 10))/100;
  printf("%f\n", speed );

  speed = ((double)(rand() % 10))/100;
  printf("%f\n", speed );

  speed = ((double)(rand() % 10))/100;
  printf("%f\n", speed );

  speed = ((double)(rand() % 10))/100;
  printf("%f\n", speed );

  return 0;
}

edited Aug 31 '18 at 13:22

chux - Reinstate Monica

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answered Aug 31 '18 at 02:12

Prabhakar

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Why use `float` when [0.01...0.1] (`double` constants) was specified and OP used `double speed`? – chux - Reinstate Monica Aug 31 '18 at 04:02
My mistake.... I thought OP has 'float'. Updated my answer. Thanks for pointing it out :) – Prabhakar Aug 31 '18 at 09:09
Note this will generate values [0 to 0.09] rather than [0.01 to 0.1]. `speed = (rand() % 10)/100.0;` would be easier to read. – chux - Reinstate Monica Aug 31 '18 at 13:29

Osiris · Accepted Answer · 2018-08-31T02:37:52.493

2

You can create an uniform distribution on an arbitrary interval with

((double)rand()/RAND_MAX)*(i_end-i_start)+i_start

where i_start and i_end denote the start and end of the interval.

In your case try

((double)rand()/RAND_MAX)*0.09+0.01

edited Aug 31 '18 at 02:37

answered Aug 31 '18 at 02:20

Osiris

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Let us call this uniform-ish. OK for basic programs. – chux - Reinstate Monica Aug 31 '18 at 03:32
@chux Yes, it gives at maximum `RAND_MAX+1` different results, which is most likely less then doubles in that interval. There are probably better functions to use, like `drand48()`, but i wanted to stick with the random function he mentioned in his question. – Osiris Aug 31 '18 at 03:57

selbie · Answer 3 · 2018-08-31T02:41:58.297

1

double upper_limit = .1;
double lower_limit = .01;
double value = ((upper_limit-lower_limit)*rand())/RAND_MAX + lower_limit;

edited Aug 31 '18 at 02:41

answered Aug 31 '18 at 02:34

selbie

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How to generate random numbers between 0.1 and 0.01

3 Answers3