10 SC2051

scottdcoulter edited this page 2022-03-28 09:39:55 -04:00

Table of Contents

• Bash doesn't support variables in brace range expansions.
• Problematic code:
• Correct code:
• Rationale:
• Exceptions
• Related Resources:

Bash doesn't support variables in brace range expansions.

Problematic code:

for i in {1..$n}
do
  echo "$i"
done

Correct code:

for ((i=1; i <= n; i++))
do
  echo "$i"
done

Rationale:

In Bash, brace expansion happens before variable expansion. This means that brace expansion will not account for variables.

For integers, use an arithmetic for loop instead. For zero-padded numbers or letters, use of eval may be warranted:

from="a" to="m"
for c in $(eval "echo {$from..$to}"); do echo "$c"; done

or more carefully (if from/to could be user input, or if the brace expansion could have spaces):

from="a" to="m"
while IFS= read -d '' -r c
do
  echo "Read $c"
done <  <(eval "printf '%s\0' $(printf "{%q..%q}.jpg" "$from" "$to")")

Exceptions

None (if you're writing for e.g. zsh, you can use a directive to disable this check)

Related Resources:

• StackOverflow: Variables in bash seq replacement ({1..10})

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