How do I iterate over a range of numbers 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:

 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

My version of bash doesn't seem to support the curly brace notation at all.

Can you do this?

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

Update: I found a bash that supports the curly braces; the above does not work.

Answer 2

You can use

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

Answer 3

for i in `seq 1 $END`; do echo $i; done

Answer 4

This works for me in bash:

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

Answer 5

discussion

Using seq is fine, as Jim Robert 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 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 cleanest way, and possibly the "fastest"; it sure won't be necessary to allocate memory to consume seq's output, which could be a problem if END is very large.

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).

Answer 6

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