Path: utzoo!mnetor!uunet!lll-winken!lll-tis!ames!mailrus!tut.cis.ohio-state.edu!bloom-beacon!gatech!udel!rochester!PT.CS.CMU.EDU!cadre!pitt!darth!formtek!ditka!csanta!greg
From: greg@csanta.UUCP (Greg Comeau)
Newsgroups: comp.unix.xenix
Subject: Re: shell problem
Message-ID: <120@csanta.UUCP>
Date: 5 Apr 88 04:52:11 GMT
References: <1171@bc-cis.UUCP>
Reply-To: greg@csanta.UUCP (Greg Comeau)
Organization: Comeau Computing, Richmond Hill, NY
Lines: 13

>An application I wrote in shell should accept a unix command 
>from a user and execute it. The solution should be:
>	read l
>	exec $l
>My solution was;
>	read l
>	echo $l>tmp$$
>	exec tmp$$

The ls -l should work, but the ls -l;who will not because the ';' is a shell
construct that needs to be intepreted (that is, $l only results in a string,
what you want to do is force the shell to scan the line again after
substitutions).  You can make this happen by using:
        eval $l  (or eval exec $l)