Path: utzoo!dptcdc!jarvis.csri.toronto.edu!mailrus!ames!elroy!ucla-cs!rutgers!att!pacbell!varian!zehntel!donw
From: donw@zehntel.zehntel.com (Don White)
Newsgroups: comp.sys.amiga
Subject: Re: Struggling through the "C"
Message-ID: <815@zehntel.UUCP>
Date: 15 Apr 89 00:47:08 GMT
References: <12000@louie.udel.EDU> <522@lzaz.ATT.COM>
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Reply-To: donw@zehntel.UUCP (Don White)
Organization: Zehntel, Inc. Walnut Creek, CA
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In article <522@lzaz.ATT.COM> hcj@lzaz.ATT.COM (HC Johnson) writes:
>In article <12000@louie.udel.EDU>, LIZAK98%SNYBUFVA.BITNET@cornellc.cit.cornell.edu (A SHOW OF HANDS) writes:
>> for(i=0;a[i]=!'\0';i++)
>The =! has meanings you are not ready for.
>Howard C. Johnson

     Sorry about this but, ... I HATE being told I'm not ready for something!
  Just in case (A SHOW OF HANDS) feels the same way ...

     Using =! instead of !=  in this case means that you are ASSIGNING the 
  logical negation of '\0'.  And '\0' equals 0. And !0 equals 1. So, a[i]
  is being set equal to 1. Since the value of the assignment is one (i.e.
  it is TRUE) and the phrase "a[i]=!'\0'" is the conditional part of a for 
  loop, the loop would never exit.

     There now. Isn't that special?                           :-)

     Don White
     zehntel!donw
     Box 271177 Concord, CA. 94527-1177