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lang-bash
"$@"is the same as"$*"in a scalar assignment. It is not. Thebashshell would not use the first character of$IFSas delimiter with"$@", but space (always).IFS:IFS=_$IFS(do stuff)IFS=${IFS#?}. (This sets the first character ofIFSto an_, and then removes it).args="$@". From a quick test, bash and ksh use "`", but zsh and dash use the first char ofIFS`. Fortunately, this isn't really something you should do, so not an incompatibility I'm going to worry about."$@"in a scalar context doesn't really make much sense. That's where you use"$*"after all.unset savedIFS; [ "${IFS+set}" ] && savedIFS=$IFS;, then do your thing, and thenunset IFS; [ "${savedIFS+set}" ] && IFS=$savedIFS;...