Bash loop and function parameter expansion

Question

I have a trivial for loop in a bash script which is not working and I cannot see what is wrong.

#!/bin/bash
function deleteOut {

for index in {0..$1}
do
    echo "/bin/rm -rf ./OUT/Message_${index}.log"
    /bin/rm -rf ./OUT/Message_${index}.log
done
}

and called it using

deleteOut 5

I was expecting a loop like

    /bin/rm -rf ./OUT/Message_0.log
    /bin/rm -rf ./OUT/Message_1.log
    /bin/rm -rf ./OUT/Message_2.log
    /bin/rm -rf ./OUT/Message_3.log
    /bin/rm -rf ./OUT/Message_4.log
    /bin/rm -rf ./OUT/Message_5.log

Instead I got

/bin/rm -rf ./OUT/Message_{0..5}.log

I've solved the problem doing something like this instead

find ./OUT -name "Message_*.log" | xargs /bin/rm

but this is not the point. Why my first approach didn't work?

sehe · Accepted Answer · 2012-02-08 08:38:25Z

2

Use eval or subshell

Brace expansion comes before variable expansion

 for index in $(eval "echo {0..$q}")
 do 
     echo $index
 done

answered Feb 8, 2012 at 8:38

sehe

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Comments

l0b0 · Accepted Answer · 2012-02-08 12:21:18Z

1

You can use seq, which is in general safer than eval:

for index in $(seq $1)

Or even faster without any external program:

for ((index=1; index<=$1; index++))

answered Feb 8, 2012 at 12:21

l0b0

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2 Comments

Luixv Over a year ago

The second command would be correct if index=0. The first command is not usable because the range was 0..$1 and seq (as far as I know) start with 1. But actually you are answering how to loop over the files whereas my question is about the parameter expansion in the range. The answer from "sehe" is the correct one: "Brace expansion comes before variable expansion" (I didn't know it). Anyway, thanks for your answer.

l0b0 Over a year ago

The link explains the expansion issue and links to more information.

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