In a program, I need to set a variable to a random value of either 0 or 1.

I can't figure out how to do it and Google has failed me.

4 Answers

One easy method is to use $RANDOM to retrieve a pseudorandom 16 bit integer number in the range [0; 32767]. You can simply convert that to [0; 1] by calculating modulo 2 of the random number:

echo $(( $RANDOM % 2 ))

More information about Bash's $RANDOM:

With that simple construct you can easily build powerful scripts using randomness, like in this comic...

6

You could use shuf

DESCRIPTION Write a random permutation of the input lines to standard output. -i, --input-range=LO-HI treat each number LO through HI as an input line -n, --head-count=COUNT output at most COUNT lines

Example:

$ foo=$(shuf -i0-1 -n1)
$ echo $foo
1
$ foo=$(shuf -i0-1 -n1)
$ echo $foo
0
$ foo=$(shuf -i0-1 -n1)
$ echo $foo
0
$ foo=$(shuf -i0-1 -n1)
$ echo $foo
1

How about:

#!/bin/bash
r=$(($RANDOM % 2))
echo $r

Or even:

r=$(($(od -An -N1 -i /dev/random) % 2))

Or perhaps:

r=$(seq 0 1 | sort -R | head -n 1)

Or more hackily:

r=$(($(head -128 /dev/urandom | cksum | cut -c1-10) % 2))

And also:

r=$(apg -a 1 -M n -n 1 -m 8 -E 23456789 | cut -c1)

As well as:

r=$((0x$(cut -c1-1 /proc/sys/kernel/random/uuid) % 2))

This script has no benefits over existing answers. Just for entertainment purposes...

Get one byte from /dev/urandom (although in general sending arbitrary binary characters to console is not recommended™ because it might give unexpected/confusing results):

head -c 1 /dev/urandom

And turn in into decimal number:

head -c 1 /dev/urandom | od -An -t u1

And get the remainder of its division by 2:

echo $((`head -c 1 /dev/urandom | od -An -t u1` % 2))

0

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