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From: goer@ellis.uchicago.edu (Richard L. Goerwitz)
Newsgroups: comp.lang.icon
Subject: Re: A word/line procedure
Message-ID: <1991May16.153040.9843@midway.uchicago.edu>
Date: 16 May 91 15:30:40 GMT
References: <124E207800E17F06@cc.Helsinki.FI>
Sender: news@midway.uchicago.edu (NewsMistress)
Distribution: inet
Organization: University of Chicago
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In article <124E207800E17F06@cc.Helsinki.FI> KKTK_KOTUS@cc.helsinki.fi writes:
>
>I have a following, obviously simple problem. I have made a nice
>Icon procedure, which divides a text file one word/line....:
>
>	procedure main()
>   		while line:=read() do write(map(line," ","\l"))
>	end 
>
>It works fine and fast, but it produces also empty lines which I cannot get rid
>off in the same procedure, although I try to filter them out with something
>like 
>
>	if *line > 0 then write(line)

I don't see how you can eliminate blank lines that are due to adjacent
spaces.  If all words are guaranteed to be separated by one space, and
no more, then naturally you could say

    ... do write("" ~== map(line, " ", "\l"))

This would eliminate lines with no words on them.  If, though, your files
distance words from each other using more than one blank space, you will
surely need to scan the lines by hand.  I don't know what was in the Icon
book, but I'd probably just use:

procedure main(a)

    separators := \a[1] | ',":<>,._+-=)(*&^%$#!@~`\'?/|\\][} {\t;'

    while read(&input) ? {
        tab(many(separators))
	if not pos(0) then {
            while write(tab(upto(separators))) do
	        tab(many(separators))
        }
	pos(0) | write(tab(0))
    }

end

If you add in control characters to your separator list, and add a test
for length, you'll have something like the UNIX strings command.

Quiz time:

    Why must the line "pos(0) | write(tab(0))" stand after the if..then
    expression?

If you are stumped, here is a hint (don't peek):






    Notice how I called "if...then" an "expression."  That means it pro-
    duces a result.  What is that result?

If you are still stumped, here is the (or rather one) answer:





    The read(&input) ? { if...then; junk } expression always succeeds,
    because "junk" (i.e. "pos(0) | write(tab(0))") always succeeds.  If
    junk were not present, then we'd have read(&input) ? { if...then }.
    If the if-condition fails, the if...then expression fails.  More-
    over, if it succeeds, the then { expression } is evaluated.  Since
    "expression" is a while loop, it will eventually fail, causing the
    entire if...then expression to fail.  When it fails, the read(&input)
    ? { etc. } expression will fail, which will cause termination of the
    while construct it is part of, and ultimately termination of the
    program.  The upshot is that junk must be present, and must always
    succeed.

I probably should have tested the above program, but I have to run.  If
there are any errors, then I apologize (though I really don't see where
any could be [famous last words]).

-- 

   -Richard L. Goerwitz              goer%sophist@uchicago.bitnet
   goer@sophist.uchicago.edu         rutgers!oddjob!gide!sophist!goer