Bash Random Variables = Not so random?

Question

I wanted to generate a random decimal that was rounded to the 10ths place between .1 and 1.

I did this with the command

`echo "scale=1; $(( ( RANDOM % 10 ) + 1 ))/10" | bc -l`

Now if you set it as a variable, say

var=`echo "scale=1; $(( ( RANDOM % 10 ) + 1 ))/10" | bc -l`

and have your script echo var like so,

#!/bin/bash
var=`echo "scale=1; $(( ( RANDOM % 10 ) + 1 ))/10" | bc -l`
echo $var
echo $var
echo $var

It repeats the same decimal, say .4, but if you

#!/bin/bash
echo `echo "scale=1; $(( ( RANDOM % 10 ) + 1 ))/10" | bc -l`
echo `echo "scale=1; $(( ( RANDOM % 10 ) + 1 ))/10" | bc -l`
echo `echo "scale=1; $(( ( RANDOM % 10 ) + 1 ))/10" | bc -l`

It will give you three random numbers using the same command as

$var=`echo "scale=1; $(( ( RANDOM % 10 ) + 1 ))/10" | bc -l`

Why does Bash not generate three new numbers if given the same command but as a variable?

Actually $var=/=foo because it was defined at the top of the script. — Chirality
– Chirality, Commented Apr 21, 2015 at 1:16

John1024 · Accepted Answer · 2015-04-21 00:50:09Z

4

The following command sets a value for var:

var=`echo "scale=1; $(( ( RANDOM % 10 ) + 1 ))/10" | bc -l`

Once the value is set, it does not change. Thus, however many times you view it, it will be the same:

echo $var
echo $var
echo $var

The only way to get a new value is for bash to evaluate RANDOM again.

Using a function instead

You might prefer a function that would return a different random variable with each invocation:

$ var() { echo "scale=1; $(( ( RANDOM % 10 ) + 1 ))/10" | bc -l; }
$ var
.3
$ var
.4
$ var
1.0
$ var
.4

edited Apr 21, 2015 at 0:50

answered Apr 21, 2015 at 0:39

John1024

115k15 gold badges152 silver badges183 bronze badges

Sign up to request clarification or add additional context in comments.

7 Comments

Chirality Over a year ago

I see. I knew bash stored variables, but wasn't sure why it wouldn't execute the command on each new line! Thanks for the one-liner function idea. :)

Chirality Over a year ago

Is there any way of adding functions? I have #!/bin/bash var() { echo "scale=1; $(( ( RANDOM % 10 ) + 1 ))/10" | bc -l; } add=$((var+var+var+var+ var)) echo $add but I'm getting an operand error. Same thing with quotations around var.

Chirality Over a year ago

@CharlesDuffy, When I make my scripts I set it to #!/bin/bash and run with ./scriptname. My default interpreter is also bash not sh.

Charles Duffy Over a year ago

Oh. It's floating-point. Bash doesn't support floating-point math natively.

Charles Duffy Over a year ago

See BashFAQ #22: mywiki.wooledge.org/BashFAQ/022; this is why you're using bc rather than $(( )) alone in the first place.

|

Slizzered · Accepted Answer · 2015-04-21 11:02:52Z

2

It is because

var=`echo "scale=1; $(( ( RANDOM % 10 ) + 1 ))/10" | bc -l`

executes the echo command and stores its output into a variable. The command itself is only evaluated once by the bash interpreter.

If you want a compact way to generate a random number, I suggest using a function:

#!/bin/bash
myRandom(){
    echo "scale=1; $(( ( RANDOM % 10 ) + 1 ))/10" | bc -l
}

echo $(myRandom)
echo $(myRandom)
echo $(myRandom)

edited Apr 21, 2015 at 11:02

answered Apr 21, 2015 at 0:50

Slizzered

8992 gold badges10 silver badges23 bronze badges

3 Comments

cdarke Over a year ago

Using backticks? Not echo $(myRandom)?

Slizzered Over a year ago

@cdarke Both are possible. AFAIK, the $()-syntax was adopted later for readability (from Korn shell).

cdarke Over a year ago

If by later you mean about 20 years ago, then yes, but the backtick form is considered obsolete in bash and if you use ksh -n then it is flagged as deprecated. See wiki.bash-hackers.org/syntax/expansion/cmdsubst

Collectives™ on Stack Overflow

Bash Random Variables = Not so random?

2 Answers 2

Using a function instead

7 Comments

3 Comments

Your Answer

Hot Network Questions

Collectives™ on Stack Overflow

2 Answers 2

Using a function instead

7 Comments

3 Comments

Your Answer

Sign up or log in

Post as a guest

Related