How can I use variable in for loop digits?
for example:
num="12"
for i in {0..$num}; do
...
done
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2 Answers
Brace expansion with variables doesn't work as one would expect (see Appendix B for juicy details), i.e. {0..$num} would only return {0..12} literally instead of a list of number.
Try seq instead like this:
num="12"
for i in $(seq 0 $num); do
echo $i
done
Appendix B: Juicy details
The bash manual saith,
The order of expansions is: brace expansion, tilde expansion, parameter, variable, and arithmetic expansion and command substitution (done in a left-to-right fashion), word splitting, and filename expansion.
At the time shell expands {0..$num} (brace expansion), $num isn't expanded (variable expansion) yet. The sequence expression a..b needs both a and b to be numbers to generate a sequence, but here we have one number and one non-number (the literal string $num). Failing this, the shell falls back to interpreting {0..$num} literally. Then variable expansion takes over, and finally we get {0..12}
3 Comments
jaypal singh
Brace expansion doesn't support variables not entirely true :)
doubleDown
@JS웃, yeah the wording is wrong. I added a more proper explanation.
jaypal singh
+1 for the update and added explanation. Though minor correction, $num will get expanded after a failed
brace expansion reporting as {1..12} instead of {1..$num} but I guess OP will now get the bigger picture.Bash does brace expansion before variable expansion, so you will get output like {1..12}. With use of eval you can get it to work.
Test:
$ num=5
$ for i in {1..$num}; do echo "$i"; done
{1..5}
$ for i in $(eval echo {1..$num}); do echo "$i"; done
1
2
3
4
5
Please note: eval is evil in disguise.
1 Comment
doubleDown
eval and evil even sound alike!