Path: utzoo!attcan!uunet!mcsun!hp4nl!star.cs.vu.nl!maart
From: maart@cs.vu.nl (Maarten Litmaath)
Newsgroups: comp.unix.questions
Subject: Re: How does #!/bin/sh work ? Why does it sometimes not ?
Message-ID: <7942@star.cs.vu.nl>
Date: 13 Oct 90 00:11:55 GMT
References: <7980005@hpopd.HP.COM> <397@inews.intel.com>
Sender: news@cs.vu.nl
Reply-To: maart@cs.vu.nl (Maarten Litmaath)
Organization: VU Dept. of Computer Science, Amsterdam, The Netherlands
Lines: 24

In article <397@inews.intel.com>,
	bhoughto@cmdnfs.intel.com (Blair P. Houghton) writes:
)...
)When execve(2) sees '#!' (hex 2321) it interprets the rest
)of the line as the pathname to an interpreter, and its
)arguments; it opens _that_ file as an executable, and pipes
)it the balance of the first file as input.

Wrong.  If the file `foo' has `#!/bin/sh' as its first line,
executing `foo' will result in `/bin/sh foo', i.e. the filename
is passed as an _argument_.

If the rest of the file were passed as input to the interpreter,
e.g. the following example would not work:

	#!/bin/sh

	echo 'Give the name:'
	read name
	echo "The name is: $name"

Issuing the command `sh < foo' would have the same unwanted effects.
--
Waiting for this to work: cat /internet/cs.vu.nl/finger/maart