Shell argument handling: $* vs. $@

I never can recall the difference between $* and $@ in scripts so here's a little experiment which should help, hope that I remember this time :-)

First of all there is this little script to show all args of a script, call it args.sh

#!/bin/sh cnt=0 for arg do cnt=`expr $cnt + 1` printf "arg$cnt =\t$arg\n" done

args.sh 1 "2 3" "4 \"5\" 6" will show

arg1 = 1 arg2 = 2 3 arg3 = 4 "5" 6

i.e. the script counts and shows three arguments.

In order to test how $* and $@ are passed to sub processes there is a little script test.sh in four variations, once pure and once surrounded by quotes:

#!/bin/sh args.sh $*	#!/bin/sh args.sh "$*"	#!/bin/sh args.sh $@	#!/bin/sh args.sh "$@"
test.sh 1 "2 3" "4 \"5\" 6" will lead to
arg1 = 1 arg2 = 2 arg3 = 3 arg4 = 4 arg5 = "5" arg6 = 6	arg1 = 1 2 3 4 "5" 6	arg1 = 1 arg2 = 2 arg3 = 3 arg4 = 4 arg5 = "5" arg6 = 6	arg1 = 1 arg2 = 2 3 arg3 = 4 "5" 6

i.e. only the last invocation does what we want: it keeps and forwards the original arguments.

All other invocations either split the arguments by space or concatenate all arguments into one big single argument.

Another little trap I fell into: when trying to save the arguments in a variable neither invocation is successful

#!/bin/sh a=$* b="$*" c=$@ d="$@" args.sh $a args.sh $b args.sh $c args.sh $d	#!/bin/sh a=$* b="$*" c=$@ d="$@" args.sh "$a" args.sh "$b" args.sh "$c" args.sh "$d"
all lead to the complete split	all lead to a single argument
arg1 = 1 arg2 = 2 arg3 = 3 arg4 = 4 arg5 = "5" arg6 = 6	arg1 = 1 2 3 4 "5" 6

This is all in Bourne shell (if in doubt).