If RANDOM only goes up to 32767, how can I generate a 9-digit random number?

Question

How to generate 9 digit random number in shell?

I am trying something like this but it only gave numbers below 32768.

#!/bin/bash
mo=$((RANDOM%999999999))
echo "********Random"$mo

Please help

output should be ********Random453351111

You'll need to generate multiple smaller numbers and append them to each other. `RANDOM` only goes up to 32,767. — Charles Duffy, Mar 25 '21 at 15:34
Does [this](https://stackoverflow.com/a/1195035/11261546) answer your questoin? — Ivan, Mar 25 '21 at 15:35
@Ivan, insofar as that doesn't describe how to generate numbers with a wider range than RANDOM itself has, I doubt that it does. — Charles Duffy, Mar 25 '21 at 15:35
Does this answer your question? [How to generate random number in Bash?](https://stackoverflow.com/questions/1194882/how-to-generate-random-number-in-bash) — phuclv, Mar 25 '21 at 15:55
It's not pure shell, but `perl -e'print int rand(1000000)'` will do it for you. — Andy Lester, Mar 25 '21 at 17:17

score 3 · Answer 1 · answered Mar 25 '21 at 16:01

3

In Linux with /dev/urandom:

$ rnd=$(tr -cd "[:digit:]" < /dev/urandom | head -c 9) && echo $rnd
463559879

answered Mar 25 '21 at 16:01

James Brown

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This works, but it's really unfortunate to be eating so much more data from `/dev/urandom` than you really need (since anything that isn't a digit just gets thrown out). A better way to use `/dev/urandom` would be to read only a single 64-bit integer _as 64 bits of data_, so the total amount of entropy consumed is constrained to 8 bytes -- though that's easier to write in Python or C than bash. – Charles Duffy Mar 25 '21 at 16:06

score 2 · Answer 2 · answered Mar 25 '21 at 15:42

2

I think this should make it

shuf -i 99999999-999999999 -n 1

answered Mar 25 '21 at 15:42

Ivan

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Even better with allowing leading zeros: `printf '%010d\n' $(shuf -i 0-999999999 -n 1)` – Léa Gris Mar 25 '21 at 17:26
1

`printf '%09d\n' $(shuf -i 0-999999999 -n 1)` for this particular case :) – Ivan Mar 25 '21 at 17:30

score 1 · Answer 3 · answered Mar 25 '21 at 15:39

Because of RANDOM's limited range, it can only be used to retrieve four base-10 digits at a time. Thus, to retrieve 9 digits, you need to call it three times.

If we don't care much about performance (are willing to pay process substitution costs), this may look like:

#!/usr/bin/env bash
get4() {
  local newVal=32768
  while (( newVal > 29999 )); do # avoid bias because of remainder
    newVal=$RANDOM
  done
  printf '%04d' "$((newVal % 10000))"
}

result="$(get4)$(get4)$(get4)"
result=$(( result % 1000000000 ))
printf '%09d\n' "$result"

If we do care about performance, it may instead look like:

#!/usr/bin/env bash
get4() {
  local newVal=32768 outVar=$1
  while (( newVal > 29999 )); do # avoid bias because of remainder
    newVal=$RANDOM
  done
  printf -v "$outVar" '%04d' "$((newVal % 10000))"
}

get4 out1; get4 out2; get4 out3
result="${out1}${out2}${out3}"
result=$(( result % 1000000000 ))
printf '%09d\n' "$result"

_nod_ -- the 1-digit-at-a-time way works, but it consumes more entropy than it needs; 4-at-a-time is faster (when you use the version that isn't slowed down by use of `$(...)`); and also this provides a more random number on account of rerolling things that don't divide evenly into 10,000. — Charles Duffy, Mar 25 '21 at 16:02

0stone0 · Accepted Answer · 2021-03-25T16:07:47.147

1

As a work around, we could just simply ask for 1 random integer, for n times:

rand=''
for i in {1..9}; do
    rand="${rand}$(( $RANDOM % 10 ))"
done

echo $rand

Try it online!

Note ^[1]: Since RANDOM's upper limit has a final digit of 7, there's a slightly lesser change for the 'generated' number to contain 8 or 9's.

edited Mar 25 '21 at 16:07

answered Mar 25 '21 at 15:43

0stone0

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BTW, this is ever-so-slightly biased. Because `RANDOM` goes up to a number with a final digit of 7, that means 8 and 9 are slightly less likely than the digits below them. – Charles Duffy Mar 25 '21 at 16:03

score 0 · Answer 5 · answered May 15 '22 at 21:15

0

Use perl, as follows :

perl -e print\ rand | cut -c 3-11

Or

perl -MPOSIX -e 'print floor rand 10**9'

answered May 15 '22 at 21:15

MUY Belgium

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If RANDOM only goes up to 32767, how can I generate a 9-digit random number?

5 Answers5