Finding the last arg

[John Chambers]

> What ALLWAYS works in the Bourne-Shell is this:
> 
> 	for last do :; done

Wow! A one-liner that works for more than 9 args!  Of course, there's 
the question as to whether this loop is actually faster than starting 
a subprocess that just does puts(argv[artc-1]), but at least there's
a way to do it that is portable.

That comment isn't worth wasting the bandwidth, of course; my motive
for this followup is a bit of bizarreness that I discovered while
testing this command.  The usual format of a for loop is 3 lines:
	for last
	do :
	done
Usually when I want to collapse such vertical code into a horizontal
format, I follow the rule "Replace the newlines with semicolons", and
it works.  For instance,
	if [ <test> ]
	then <stuff>
	else <stuff>
	fi
reduces to
	if [ <test> ];then <stuff>;else <stuff>;fi
which I can do in vi via a series of "Jr;" commands.  With the above 
for-loop, this gives
	for last;do :;done
which doesn't work.  The shell gives a syntax error, complaining about
an unexpected ';' in the line.  Myself, I found this to be a somewhat 
unexpected error message.  It appears my simple-minded algorithm for 
condensing code doesn't work in this case.

So what's going on here?  What the @#$^&#( is the shell's syntax that 
makes the semicolon not only unneeded, but illegal in this case?

One of the real hassles I keep finding with /bin/sh (and /bin/csh is
even worse ;-) is that the actual syntax regarding things like white
space, newlines, and semicolons seems to be a secret.  It often takes 
a lot of experimenting to find a way to get these syntax characters 
right.  Is there any actual documentation on sh's syntax?  Is it truly 
as ad-hoc as the above example implies?  Is there perhaps some logical 
structure underlying it all that would explain why
 	for last do :; done
and
	for last
	do :
	done
both work but
	for last;do :;done
doesn't?
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