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Shell Programming and Scripting
Bash IFS
Post 302697353 by Corona688 on Thursday 6th of September 2012 03:40:12 PM
09-06-2012
That's not the code you ran. You must have been running echo "$*" quotes and all, otherwise you'd have never seen the pipes.
$* after the set didn't remove the equals, it was already gone. If you ran ./script a b --arg=c, $* on line 3 wopuld be the string a|b|--arg=c and split apart on "|=" into the arguments "a", "b", "--arg", and "c" The equals sign is already gone, deleted.
So when you do $* after that, it just smashes $1 $2 ... all together with | inbetween: a|b|--arg|c
Now remember, $* always, always does this, even if you don't quote it.
You just don't see it without quotes, because the shell splits it. It'd see a|b|--arg|c, give "a" as echo's first argument, "b" as echo's second argument, "--arg" as echo's third argument, and "c" as echo's fourth argument. (echo doesn't split the arguments; that's the shell's job.) echo is not controlled by IFS, and will just put spaces inbetween...
$* after the set didn't remove the equals, it was already gone. If you ran ./script a b --arg=c, $* on line 3 wopuld be the string a|b|--arg=c and split apart on "|=" into the arguments "a", "b", "--arg", and "c" The equals sign is already gone, deleted.
So when you do $* after that, it just smashes $1 $2 ... all together with | inbetween: a|b|--arg|c
Now remember, $* always, always does this, even if you don't quote it.
You just don't see it without quotes, because the shell splits it. It'd see a|b|--arg|c, give "a" as echo's first argument, "b" as echo's second argument, "--arg" as echo's third argument, and "c" as echo's fourth argument. (echo doesn't split the arguments; that's the shell's job.) echo is not controlled by IFS, and will just put spaces inbetween...Last edited by Corona688; 09-06-2012 at 04:52 PM..
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nsd(1), info(n) KEYWORDS
AOLserver 4.0 Ns_Sock(3aolserver)