The differece between $@ and $*

johnny_jiang · 02-18-2006, 09:44 AM

Hello,everyone

I'm a linux newbie from Shanghai.There's a question regarding $@ and $* which has confused me for a long time.Actually I have read the bash manual and some bibliographies,I found that they just talk about these two positional parameters' difference when they're double quoted.So what if they're assigned to a variable?
I've done some tests below.

Code:

set -- "First Arg" "Second" "Third:Arg" "" "Fifth: :Arg"

IFS=:

var=$@
var1=$*

echo "$var"  # First Arg Second Third Arg  Fifth   Arg
echo "$var1" # First Arg:Second:Third:Arg::Fifth: :Arg

IFS=

var=$@
var1=$*

echo "$var"  # First Arg Second Third:Arg  Fifth: :Arg
echo "$var1" # First Arg Second Third:Arg   FFiifftthh::  ::AArrg

It's too weird for me to understand.

unSpawn · 02-18-2006, 11:42 AM

I'm a linux newbie from Shanghai.
Doesn't mean nothing, Im a noob from Hell when it comes to open heart surgery :-]

AFAIK \* is the concatenation (whole string) while @ the IFS-sep'ped parts:

]$ tellMeIWannaKnow() { i=("$*"); echo ${#i[@]}; }
]$ tellMeIWannaKnow onetwothree four
1

]$ tellMeIWannaKnow() { i=("$@"); echo ${#i[@]}; }
]$ tellMeIWannaKnow onetwothree four
2

johnny_jiang · 02-18-2006, 10:51 PM

Thank you,unSpawn!

I understand what you're talking about."$@" will seperate arguments with IFS.But there's one thing I am also a little fuzzy on it.

Code:

var=$@

echo "$var"  # First Arg Second Third Arg  Fifth   Arg

How to explain how that runs?Why do the colons disappear?