Double expansion of a parameter in Bash script

Question

I want to expand a parameter twice, and ${$PARAM} doesn't work.

For example, if I set two variables to file names and then want to loop through those variables and expand to their file names:

INPF_1=input1.inp
INPF_2=input2.inp

# copy the input files to the execution directory
for input_param in ${!INPF_*}; do
    echo ${$input_param}
done

How can I to access the file names from those parameters in the for loop?

I.e., expand to input1.inp and input2.inp.

This would be easier if you used a single array. `infp=( input1.inp input2.inp )` — Charles Duffy, Feb 06 '16 at 00:53
What you're trying to do is called indirect expansion, though. We already have questions and answers covering it. — Charles Duffy, Feb 06 '16 at 00:53

Benjamin W. · Answer 1 · 2016-02-06T01:39:27.693

2

You had it almost: just use "${!input_param}" instead of ${$input_param}.

The quoting doesn't do anything in this case, but it's a good habit.

Be aware of the difference to ${!INPF_@}, which – when used between double quotes – expands to a separate word for each variable name, whereas "${!INPF_*}" doesn't. You almost always want "${!INPF_@}". See the difference:

$ for var in  "${!INPF_*}"; do echo "$var"; done
INPF_1 INPF_2
$ for var in  "${!INPF_@}"; do echo "$var"; done
INPF_1
INPF_2

See the manual for the various parameter expansions.

edited Feb 06 '16 at 01:39

answered Feb 06 '16 at 00:38

Benjamin W.

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This doesn't get me the file names though. I've updated to make clearer. – salotz Feb 06 '16 at 00:44
@salotz Ah, I see. Updating. – Benjamin W. Feb 06 '16 at 00:44
I actually go this far, but I didn't realize the * vs @ difference. The duplicate answer gives the rest, or like @Charles Duffy says use an array like a normal programming language. – salotz Feb 08 '16 at 20:05

Double expansion of a parameter in Bash script

1 Answers1