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 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

for i in $(seq 1 $END); do echo $i; done

edit: I prefer seq over the other methods because I can actually remember it ;)

Answer 2

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 3

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).

Answer 4

Here is why the original expression didn't work.

From man bash:

Brace expansion is performed before any other expansions, and any characters special to other expansions are preserved in the result. It is strictly textual. Bash does not apply any syntactic interpretation to the context of the expansion or the text between the braces.

So, brace expansion is something done early as a purely textual macro operation, before parameter expansion.

Shells are highly optimized hybrids between macro processors and more formal programming languages. In order to optimize the typical use cases, the language is made rather more complex and some limitations are accepted.

Recommendation

I would suggest sticking with Posix¹ features. This means using for i in <list>; do, if the list is already known, otherwise, use while or seq, as in:

#!/bin/sh

limit=4

i=1; while [ $i -le $limit ]; do
  echo $i
  i=$(($i + 1))
done
# Or -----------------------
for i in $(seq 1 $limit); do
  echo $i
done

---

^{1. Bash is a great shell and I use it interactively, but I don't put bash-isms into my scripts. Scripts might need a faster shell, a more secure one, a more embedded-style one. They might need to run on whatever is installed as /bin/sh, and then there are all the usual pro-standards arguments. Remember shellshock, aka bashdoor?}

Answer 5

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 6

You can use

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

Answer 7

Another layer of indirection:

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

Answer 8

I've combined a few of the ideas here and measured performance.

TL;DR Takeaways:

1. seq and {..} are really fast
2. for and while loops are slow
3. $( ) is slow
4. for (( ; ; )) loops are slower
5. $(( )) is even slower
6. Worrying about N numbers in memory (seq or {..}) is silly (at least up to 1 million.)

These are not conclusions. You would have to look at the C code behind each of these to draw conclusions. This is more about how we tend to use each of these mechanisms for looping over code. Most single operations are close enough to being the same speed that it's not going to matter in most cases. But a mechanism like for (( i=1; i<=1000000; i++ )) is many operations as you can visually see. It is also many more operations per loop than you get from for i in $(seq 1 1000000). And that may not be obvious to you, which is why doing tests like this is valuable.

Demos

# show that seq is fast
$ time (seq 1 1000000 | wc)
 1000000 1000000 6888894

real    0m0.227s
user    0m0.239s
sys     0m0.008s

# show that {..} is fast
$ time (echo {1..1000000} | wc)
       1 1000000 6888896

real    0m1.778s
user    0m1.735s
sys     0m0.072s

# Show that for loops (even with a : noop) are slow
$ time (for i in {1..1000000} ; do :; done | wc)
       0       0       0

real    0m3.642s
user    0m3.582s
sys 0m0.057s

# show that echo is slow
$ time (for i in {1..1000000} ; do echo $i; done | wc)
 1000000 1000000 6888896

real    0m7.480s
user    0m6.803s
sys     0m2.580s

$ time (for i in $(seq 1 1000000) ; do echo $i; done | wc)
 1000000 1000000 6888894

real    0m7.029s
user    0m6.335s
sys     0m2.666s

# show that C-style for loops are slower
$ time (for (( i=1; i<=1000000; i++ )) ; do echo $i; done | wc)
 1000000 1000000 6888896

real    0m12.391s
user    0m11.069s
sys     0m3.437s

# show that arithmetic expansion is even slower
$ time (i=1; e=1000000; while [ $i -le $e ]; do echo $i; i=$(($i+1)); done | wc)
 1000000 1000000 6888896

real    0m19.696s
user    0m18.017s
sys     0m3.806s

$ time (i=1; e=1000000; while [ $i -le $e ]; do echo $i; ((i=i+1)); done | wc)
 1000000 1000000 6888896

real    0m18.629s
user    0m16.843s
sys     0m3.936s

$ time (i=1; e=1000000; while [ $i -le $e ]; do echo $((i++)); done | wc)
 1000000 1000000 6888896

real    0m17.012s
user    0m15.319s
sys     0m3.906s

# even a noop is slow
$ time (i=1; e=1000000; while [ $((i++)) -le $e ]; do :; done | wc)
       0       0       0

real    0m12.679s
user    0m11.658s
sys 0m1.004s

Answer 9

If you need it prefix than you might like this

 for ((i=7;i<=12;i++)); do echo `printf "%2.0d\n" $i |sed "s/ /0/"`;done

that will yield

07
08
09
10
11
12

Answer 10

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

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

Answer 11

This works fine in bash:

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

Answer 12

There are many ways to do this, however the ones I prefer is given below

Using seq

Synopsis from man seq

$ seq [-w] [-f format] [-s string] [-t string] [first [incr]] last

Syntax

Full command
seq first incr last

• first is starting number in the sequence [is optional, by default:1]
• incr is increment [is optional, by default:1]
• last is the last number in the sequence

Example:

$ seq 1 2 10
1 3 5 7 9

Only with first and last:

$ seq 1 5
1 2 3 4 5

Only with last:

$ seq 5
1 2 3 4 5

Using {first..last..incr}

Here first and last are mandatory and incr is optional

Using just first and last

$ echo {1..5}
1 2 3 4 5

Using incr

$ echo {1..10..2}
1 3 5 7 9

You can use this even for characters like below

$ echo {a..z}
a b c d e f g h i j k l m n o p q r s t u v w x y z

Answer 13

I know this question is about bash, but - just for the record - ksh93 is smarter and implements it as expected:

$ ksh -c 'i=5; for x in {1..$i}; do echo "$x"; done'
1
2
3
4
5
$ ksh -c 'echo $KSH_VERSION'
Version JM 93u+ 2012-02-29

$ bash -c 'i=5; for x in {1..$i}; do echo "$x"; done'
{1..5}

Answer 14

This is another way:

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

Answer 15

If you want to stay as close as possible to the brace-expression syntax, try out the range function from bash-tricks' range.bash.

For example, all of the following will do the exact same thing as echo {1..10}:

source range.bash
one=1
ten=10

range {$one..$ten}
range $one $ten
range {1..$ten}
range {1..10}

It tries to support the native bash syntax with as few "gotchas" as possible: not only are variables supported, but the often-undesirable behavior of invalid ranges being supplied as strings (e.g. for i in {1..a}; do echo $i; done) is prevented as well.

The other answers will work in most cases, but they all have at least one of the following drawbacks:

• Many of them use subshells, which can harm performance and may not be possible on some systems.
• Many of them rely on external programs. Even seq is a binary which must be installed to be used, must be loaded by bash, and must contain the program you expect, for it to work in this case. Ubiquitous or not, that's a lot more to rely on than just the Bash language itself.
• Solutions that do use only native Bash functionality, like @ephemient's, will not work on alphabetic ranges, like {a..z}; brace expansion will. The question was about ranges of numbers, though, so this is a quibble.
• Most of them aren't visually similar to the {1..10} brace-expanded range syntax, so programs that use both may be a tiny bit harder to read.
• @bobbogo's answer uses some of the familiar syntax, but does something unexpected if the $END variable is not a valid range "bookend" for the other side of the range. If END=a, for example, an error will not occur and the verbatim value {1..a} will be echoed. This is the default behavior of Bash, as well--it is just often unexpected.

Disclaimer: I am the author of the linked code.

Answer 16

Replace {} with (( )):

tmpstart=0;
tmpend=4;

for (( i=$tmpstart; i<=$tmpend; i++ )) ; do 
echo $i ;
done

Yields:

0
1
2
3
4

Answer 17

These are all nice but seq is supposedly deprecated and most only work with numeric ranges.

If you enclose your for loop in double quotes, the start and end variables will be dereferenced when you echo the string, and you can ship the string right back to BASH for execution. $i needs to be escaped with \'s so it is NOT evaluated before being sent to the subshell.

RANGE_START=a
RANGE_END=z
echo -e "for i in {$RANGE_START..$RANGE_END}; do echo \\${i}; done" | bash

This output can also be assigned to a variable:

VAR=`echo -e "for i in {$RANGE_START..$RANGE_END}; do echo \\${i}; done" | bash`

The only "overhead" this should generate should be the second instance of bash so it should be suitable for intensive operations.

Answer 18

If you're doing shell commands and you (like I) have a fetish for pipelining, this one is good:

seq 1 $END | xargs -I {} echo {}

Answer 19

if you don't wanna use 'seq' or 'eval' or jot or arithmetic expansion format eg. for ((i=1;i<=END;i++)), or other loops eg. while, and you don't wanna 'printf' and happy to 'echo' only, then this simple workaround might fit your budget:

a=1; b=5; d='for i in {'$a'..'$b'}; do echo -n "$i"; done;' echo "$d" | bash

PS: My bash doesn't have 'seq' command anyway.

Tested on Mac OSX 10.6.8, Bash 3.2.48

Answer 20

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"
}