How do I iterate over a range of numbers defined b

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 documentation):

 for i in {1..5}; do echo $i; done

Which gives:

1
2
3
4
5

Yet, how can I replace either of the range endpoints with a variable? This doesn't work:

END=5
for i in {1..$END}; do echo $i; done

Which prints:

{1..5}

Answer 1

This works in Bash and Korn, also can go from higher to lower numbers. Probably not fastest or prettiest but works well enough. Handles negatives too.

function num_range {
   # Return a range of whole numbers from beginning value to ending value.
   # >>> num_range start end
   # start: Whole number to start with.
   # end: Whole number to end with.
   typeset s e v
   s=${1}
   e=${2}
   if (( ${e} >= ${s} )); then
      v=${s}
      while (( ${v} <= ${e} )); do
         echo ${v}
         ((v=v+1))
      done
   elif (( ${e} < ${s} )); then
      v=${s}
      while (( ${v} >= ${e} )); do
         echo ${v}
         ((v=v-1))
      done
   fi
}

function test_num_range {
   num_range 1 3 | egrep "1|2|3" | assert_lc 3
   num_range 1 3 | head -1 | assert_eq 1
   num_range -1 1 | head -1 | assert_eq "-1"
   num_range 3 1 | egrep "1|2|3" | assert_lc 3
   num_range 3 1 | head -1 | assert_eq 3
   num_range 1 -1 | tail -1 | assert_eq "-1"
}

Answer 2

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.

Answer 3

The seq method is the simplest, but Bash has built-in arithmetic evaluation.

END=5
for ((i=1;i<=END;i++)); do
    echo $i
done
# ==> outputs 1 2 3 4 5 on separate lines

The for ((expr1;expr2;expr3)); construct works just like for (expr1;expr2;expr3) in C and similar languages, and like other ((expr)) cases, Bash treats them as arithmetic.

Answer 4

If you're on BSD / OS X you can use jot instead of seq:

for i in $(jot $END); do echo $i; done

Answer 5

This is another way:

end=5
for i in $(bash -c "echo {1..${end}}"); do echo $i; done

Answer 6

discussion

Using seq is fine, as Jiaaro suggested. Pax Diablo suggested a Bash loop to avoid calling a subprocess, with the additional advantage of being more memory friendly if $END is too large. Zathrus spotted a typical bug in the loop implementation, and also hinted that since i is a text variable, continuous conversions to-and-fro numbers are performed with an associated slow-down.

integer arithmetic

This is an improved version of the Bash loop:

typeset -i i END
let END=5 i=1
while ((i<=END)); do
    echo $i
    …
    let i++
done

If the only thing that we want is the echo, then we could write echo $((i++)).

ephemient taught me something: Bash allows for ((expr;expr;expr)) constructs. Since I've never read the whole man page for Bash (like I've done with the Korn shell (ksh) man page, and that was a long time ago), I missed that.

So,

typeset -i i END # Let's be explicit
for ((i=1;i<=END;++i)); do echo $i; done

seems to be the most memory-efficient way (it won't be necessary to allocate memory to consume seq's output, which could be a problem if END is very large), although probably not the “fastest”.

the initial question

eschercycle noted that the {a..b} Bash notation works only with literals; true, accordingly to the Bash manual. One can overcome this obstacle with a single (internal) fork() without an exec() (as is the case with calling seq, which being another image requires a fork+exec):

for i in $(eval echo "{1..$END}"); do

Both eval and echo are Bash builtins, but a fork() is required for the command substitution (the $(…) construct).