CODEWITHSUNDEEP
bashfor-loopbrace-expansion
spraff
spraff

Reputation: 33445

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

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?

Upvotes: 49

Views: 54506

Answers (5)

Jonathan Cross
Jonathan Cross

Reputation: 715

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

Upvotes: 2

elcaro
elcaro

Reputation: 2317

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.

Upvotes: 1

kev
kev

Reputation: 161914

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++))

Upvotes: 16

jordanm
jordanm

Reputation: 34974

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.

Upvotes: 70

Oliver Charlesworth
Oliver Charlesworth

Reputation: 272687

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.

Upvotes: 27

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