How do I iterate over a range of numbers defined by variables in Bash?

Question

How do I iterate over a range of numbers in Bash when the range is given by a variable?

I know I can do this (called \"sequence expression\" in the Bash documentatio

Answer 1

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: What is the difference between single and double square brackets in Bash?
• for ((;;))
• seq (GNU Coreutils)
• {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.