I know that the exit codes in scripting "$*" will returns all the parameters/arguments passwd to the script.
But I also know that "$@" will also returns the same. What is the difference between those two [ $* and $@ ] ?
"$*" and "$@" are not exit codes; they are the command line arguments.
Exit codes are stored in "$?".
When unquoted, $* and $@ are the same; each word in the argument list becomes a separate word.
When quoted, "$*" returns a single argument containing all the command-line arguments; "$@" presents each command-line argument as a separate argument.
For a demonstration, put the following into a script and exceute it with the arguments: a "b c" "d e f" g.
printf "\n%s\n" "Using \$\*:"
printf "%s\n" $*
printf "\n%s\n" "Using \$\@:"
printf "%s\n" $@
printf "\n%s\n" "Using \"\$\*\":"
printf "%s\n" "$*"
printf "\n%s\n" "Using \"\$\@\":"
printf "%s\n" "$@"
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