Using a variable in brace expansion range fed to a for loop

Question

Here is myscript.sh

#!/bin/bash
for i in {1..$1};
do
    echo $1 $i;
done

If I run myscript.sh 3 the output is

3 {1..3}

instead of

3 1
3 2
3 3

Clearly $3 contains the right value, so why doesn't for i in {1..$1} behave the same as if I had written for i in {1..3} directly?

Minor thing: You wrote $3 ("Clearly $3 contains ...") but are using $1 in your script. Why not just remove the $1? It doesn't have that much relevance except to prove that variable expansion works. — Sybille Peters, Mar 05 '23 at 07:33

score 70 · Accepted Answer · answered Mar 28 '12 at 21:56

70

You should use a C-style for loop to accomplish this:

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

This avoids external commands and nasty eval statements.

answered Mar 28 '12 at 21:56

jordanm

33,009
7
61
76

Not working for me `test.sh: line 1: ((: i<=: syntax error: operand expected (error token is "<=")` – Temak Nov 03 '15 at 16:05
@Temak are you using `bash`? Is your shebang `#!/bin/bash` or `#!/bin/sh`? – jordanm Nov 03 '15 at 16:06
@jordann, file contains only these 3 lines. I'm using GNU bash, version 4.3.11(1)-release (x86_64-pc-linux-gnu). I run script `bash test.sh` – Temak Nov 03 '15 at 16:09
3

@Temak you need to run it like this: `bash test.sh 12`. The `$1` is a positional arg. – jordanm Nov 03 '15 at 16:11
Yes, I haven't noticed. Sorrry:) – Temak Nov 03 '15 at 16:13

score 27 · Answer 2 · answered Mar 28 '12 at 15:44

27

Because brace expansion occurs before expansion of variables. http://www.gnu.org/software/bash/manual/bashref.html#Brace-Expansion.

If you want to use braces, you could so something grim like this:

for i in `eval echo {1..$1}`;
do
    echo $1 $i;
done

Summary: Bash is vile.

answered Mar 28 '12 at 15:44

Oliver Charlesworth

267,707
33
569
680

2

Heh, +1 for the answer, -0.5 for the vile. – glenn jackman Mar 28 '12 at 16:54
1

@glennjackman: Wow, you're saying that Bash's myriad expansion and escaping rules aren't vile? – Oliver Charlesworth Mar 28 '12 at 17:03
1

I feel dirty for saying +1 on an answer involving `eval`. Re: vile: `bash` isn't vile, but it's a bit like drinking grain alcohol straight. Some people seem to like it but it's hard not to choke at first and the more you do it the less you're bothered by it. – sorpigal Mar 28 '12 at 17:15
1

@Sorpigal: It must just be me then. I use it pretty much every day, and it still bothers me... (but then I'm not good with straight alchohol either) – Oliver Charlesworth Mar 28 '12 at 17:40
This is yet working and solving the issue. Thank you! – Geppettvs D'Constanzo Jun 22 '17 at 02:07

kev · Answer 3 · 2012-03-29T00:39:07.093

16

You can use seq command:

for i in `seq 1 $1`

Or you can use the C-style for...loop:

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

edited Mar 29 '12 at 00:39

answered Mar 28 '12 at 15:45

kev

155,172
47
273
272

A C-style for loop is preferred over the external seq command. – jordanm Mar 28 '12 at 21:53

score 2 · Answer 4 · answered Jan 21 '15 at 12:46

2

Here is a way to expand variables inside braces without eval:

end=3
declare -a 'range=({'"1..$end"'})'

We now have a nice array of numbers:

for i in ${range[@]};do echo $i;done
1
2
3

answered Jan 21 '15 at 12:46

Jonathan Cross

675
8
17

score 1 · Answer 5 · answered Feb 04 '14 at 06:44

I know you've mentioned bash in the heading, but I would add that 'for i in {$1..$2}' works as intended in zsh. If your system has zsh installed, you can just change your shebang to zsh.

Using zsh with the example 'for i in {$1..$2}' also has the added benefit that $1 can be less than $2 and it still works, something that would require quite a bit of messing about if you wanted that kind of flexibility with a C-style for loop.

Using a variable in brace expansion range fed to a for loop

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