In bash, is it possible to use an integer variable in a brace expansion

Question

I have the following bash script:

#!/bin/bash

upperlim=10

for i in {0..10}
do
echo $i
done

for i in {0..$upperlim}
do
echo $i
done

The first for loop (without the variable upperlim in the loop control) works fine, but the second for loop (with the variable upperlim in the loop control) does not. Is there any way that I can modify the second for loop so that it works? Thanks for your time.

Answer 1

The reason for this is the order in which things occur in bash. Brace expansion occurs before variables are expanded. In order to accomplish your goal, you need to use C-style for loop:

upperlim=10

for ((i=0; i<=upperlim; i++)); do
   echo "$i"
done

Answer 2

To complete this in your style using nothing but built-ins you'd have to use eval:

d=12

for i in `eval echo {0..$d}`
do
echo $i
done

But with seq:

lowerlimit=0
upperlimit=12

for i in $(seq $lowerlimit $upperlimit)
do
echo $i
done

Personally I find the use of seq to be more readable.

Answer 3

The POSIX way

If you care about portability, use the example from the POSIX standard:

i=2
END=5
while [ "$i" -le "$END" ]; do
    echo "$i"
    i=$(($i+1))
done

Output:

2
3
4
5

Things which are not POSIX:

• (( )) without dollar, although it is a common extension as mentioned by POSIX itself.

• [[. [ is enough here. See also: https://stackoverflow.com/questions/13542832/bash-if-difference-between-square-brackets-and-double-square-brackets

• for ((;;))

• seq

• {start..end}, and that cannot work with variables as mentioned by the Bash manual.

• let i=i+1: POSIX 7 2. Shell Command Language does not contain the word let, and it fails on bash --posix 4.3.42

• the dollar at i=$i+1 might be required, but I'm not sure. POSIX 7 2.6.4 Arithmetic Expansion says:

  If the shell variable x contains a value that forms a valid integer constant, optionally including a leading plus or minus sign, then the arithmetic expansions "$((x))" and "$(($x))" shall return the same value.

  but reading it literally that does not imply that $((x+1)) expands since x+1 is not a variable.

Answer 4

Your approach will not work since in bash brace-expansion occurs before parameter expansion. You need to expand the variable before.

You can work around with eval:

upperlim=10
eval '
        for i in {0..'"$upperlim"'}
        do
                echo $i
        done
'

With While loop:

upperlim=10
#with while
start=0
while [[ $start -le $upperlim ]]
do
    echo "$start"
    ((start = start + 1))
done

Also you can do it with seq command:

upperlim=10
#seq
for i in $(seq "$upperlim"); do
  echo "$i"
done

If you want to run with for i in {0..$upperlim} you will need to use kornshell. eg:

#!/bin/ksh
upperlim=10

for i in {0..$upperlim}
do
        echo $i
done

Answer 5

for i in $(seq ${1:-1} ${2:-10}); do                                            
        echo $i                                                                 
done 

                                                                       
                                                                            
i=${1:-1}                                                                       
while (( $i <= ${2:-10} )); do                                                  
        echo $i                                                                 
        i=$(($i + 1))                                                           
done