How to iterate over arguments in a Bash script

Question

I have a complex command that I\'d like to make a shell/bash script of. I can write it in terms of $1 easily:

foo $1 args -o $1.ext
         


        

Answer 1

Note that Robert's answer is correct, and it works in sh as well. You can (portably) simplify it even further:

for i in "$@"

is equivalent to:

for i

I.e., you don't need anything!

Testing ($ is command prompt):

$ set a b "spaces here" d
$ for i; do echo "$i"; done
a
b
spaces here
d
$ for i in "$@"; do echo "$i"; done
a
b
spaces here
d

I first read about this in Unix Programming Environment by Kernighan and Pike.

In bash, help for documents this:

for NAME [in WORDS ... ;] do COMMANDS; done

If 'in WORDS ...;' is not present, then 'in "$@"' is assumed.