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.

Answer 1

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.

Answer 2

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...