CODEWITHSUNDEEP
bashfunctionpointersreferenceglobal-variables
Libertad Pantoja
Libertad Pantoja

Reputation: 69

Change a referenced variable in BASH

I am intending to change a global variable inside a function in BASH, however I don't get a clue about how to do it. This is my code:

CANDIDATES[5]="1 2 3 4 5 6"
random_mutate()
{
a=$1 #assign name of input variable to "a"
insides=${!a} #See input variable value
RNDM_PARAM=`echo $[ 1 + $[ RANDOM % 5 ]]` #change random position in input variable
NEW_PAR=99 #value to substitute
ARR=($insides) #Convert string to array
ARR[$RNDM_PARAM]=$NEW_PAR #Change the random position
NEW_GUY=$( IFS=$' '; echo "${ARR[*]}" ) #Convert array once more to string 
echo "$NEW_GUY" 
### NOW, How to assign NEW_GUY TO CANDIDATES[5]?
}

random_mutate CANDIDATES[5]

I would like to be able to assign NEW_GUY to the variable referenced by $1 or to another variable that would be pointed by $2 (not incuded in the code). I don't want to do the direct assignation in the code as I intend to use the function for multiple possible inputs (in fact, the assignation NEW_PAR=99 is quite more complicated in my original code as it implies the selection of a number depending the position in a range of random values using an R function, but for the sake of simplicity I included it this way).

Hopefully this is clear enough. Please let me know if you need further information.

Thank you,

Libertad

Upvotes: 0

Views: 92

Answers (1)

choroba
choroba

Reputation: 241818

You can use eval:

eval "$a=\$NEW_GUY"

Be careful and only use it if the value of $a is safe (imagine what happens if $a is set to rm -rf / ; a).

Upvotes: 1

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