Bash RANDOM with seed

bashrandom

I've been using $((1 + RANDOM % 1000)) to generate a random number.

Is it possible to do something similar but provide a seed?

So that given the same seed the same random number will always be output?

Best Answer

Assign a seed value to RANDOM

$ bash -c 'RANDOM=640; echo $RANDOM $RANDOM $RANDOM'
28612 27230 24923
$ bash -c 'RANDOM=640; echo $RANDOM $RANDOM $RANDOM'
28612 27230 24923
$ bash -c 'RANDOM=640; echo $RANDOM $RANDOM $RANDOM'
28612 27230 24923
$

Notice that single quotes are used; double quotes run afoul shell interpolation rules:

$ bash -c 'RANDOM=42; echo $RANDOM $RANDOM $RANDOM'
19081 17033 15269
$ RANDOM=42
$ bash -c "RANDOM=640; echo $RANDOM"
19081
$ bash -c "RANDOM=640; echo $RANDOM"
17033
$ bash -c "RANDOM=640; echo $RANDOM"
15269
$

because $RANDOM is being interpolated by the parent shell before the child bash -c ... process is run. Either use single quotes to turn off interpolation (as shown above) or otherwise prevent the interpolation:

$ RANDOM=42
$ SEED_FOR_MY_GAME=640
$ bash -c "RANDOM=$SEED_FOR_MY_GAME; echo \$RANDOM"
28612
$ bash -c "RANDOM=$SEED_FOR_MY_GAME; echo \$RANDOM"
28612
$

This feature of RANDOM is mentioned in the bash(1) manual

   RANDOM Each time this parameter is referenced, a random integer between
          0 and 32767 is generated.  The sequence of random numbers may be
          initialized by assigning a value to RANDOM.  If RANDOM is unset,
          it loses its special properties,  even  if  it  is  subsequently
          reset.
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