How to change values of bash array elements without loop

Question

array=(a b c d)

I would like to add a character before each element of the array in order to have this

array=(^a ^b ^c ^d)

An easy way to do that is to loop on array elements and change values one by one

for i in "${#array[@]}"
do
    array[i]="^"array[i]
done

But I would like to know if there is any way to do the same thing without looping on the array as I have to do the same instruction on all elements.

Thanks in advance.

choroba · Accepted Answer · 2015-06-24 14:53:31Z

102

Use Parameter Expansion:

array=("${array[@]/#/^}")

From the documentation:

${parameter/pattern/string}

Pattern substitution. The pattern is expanded to produce a pattern just as in pathname expansion. Parameter is expanded and the longest match of pattern against its value is replaced with string. If pattern begins with /, all matches of pattern are replaced with string. Normally only the first match is replaced. If pattern begins with #, it must match at the beginning of the expanded value of parameter. If pattern begins with %, it must match at the end of the expanded value of parameter. If string is null, matches of pattern are deleted and the / following pattern may be omitted. If parameter is @ or *, the substitution operation is applied to each positional parameter in turn, and the expansion is the resultant list. If parameter is an array variable subscripted with @ or *, the substitution operation is applied to each member of the array in turn, and the expansion is the resultant list.

edited Jun 24, 2015 at 14:53

answered Oct 5, 2012 at 10:19

choroba

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9 Comments

saloua Over a year ago

I have just another question. I would like to know if I can use this to change just a subset of an array because I obtain a syntax error when I run this array[@]:i:10=(${array[@]:i:10}/#/^})

Werner Lehmann Over a year ago

@tuxworker I think you need to assign your sub-range array to another array variable first, then add the prefixes.

Magne Over a year ago

NB: If you have IFS=$'\n' (to handle names with whitespace) then the command will not keep the original array structure. It will just make array contain only 1 item (a string list), with all the changed strings. As seen if you run: declare -p array afterwards.

DKebler Over a year ago

DIRS=("${DIRS[@]/%/\/module}") use % instead of # to substitute at the end. tldp.org/LDP/abs/html/string-manipulation.html

beliz Over a year ago

what if we want to add a number to all elements in an array?

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socketpair · Accepted Answer · 2014-01-06 18:48:30Z

10

This way also honor whitespaces in array values:

array=( "${array[@]/#/^}" )

Note, this will FAIL if array was empty and you set previously

set -u

I don't know how to eliminate this issue using short code...

answered Jan 6, 2014 at 18:48

socketpair

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3 Comments

x-yuri Over a year ago

Maybe array=("${array[@]:+${array[@]/#/^}}"), but I can't reproduce it (empty array + set -u) with neither bash-4.4.0, nor bash-4.1.5. In the latter you need this construct for iterating over an array with for, though.

mattalxndr Over a year ago

How would you mark start and end of the string? From this: declare -a arr=(12 123); echo "${arr[@]/12/00}" I want 00 123 but I'm getting 00 003.

DKebler Over a year ago

DIRS=("${DIRS[@]/%/\/module}") use % instead of # to substitute at the end. tldp.org/LDP/abs/html/string-manipulation.html

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