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

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

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

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

END=5for ((i=1;i<=END;i++)); do    echo $idone# ==> 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.

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 ENDlet END=5 i=1while ((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 explicitfor ((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).