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Path: utzoo!utgpu!water!watmath!orchid!jagardner
From: jagardner@orchid.UUCP
Newsgroups: comp.lang.c
Subject: Re: ANS C Macro Processing
Message-ID: <11323@orchid.waterloo.edu>
Date: Thu, 22-Oct-87 19:56:26 EDT
Article-I.D.: orchid.11323
Posted: Thu Oct 22 19:56:26 1987
Date-Received: Sat, 24-Oct-87 14:36:41 EDT
References: <167@sdti.UUCP>
Reply-To: datanguay@watbun.waterloo.edu (David Tanguay)
Organization: U. of Waterloo, Ontario
Lines: 40

In article <167@sdti.UUCP> turner%sdti@harvard.harvard.edu (Prescott K. Turner, Jr.) writes:
}in the context of       #define z +y
}                        #define y(b) z
}                        #define f(a) f(a(3))
}start with              
}   f(z)
}      identify invocation f(z)
}      process argument
}         z
}            identify invocation z
}            replace z
}         +y
}            rescan, with definition of z suppressed
}      replace f(z)
}   f(+y(3))
}      rescan, with definition of f suppressed
}      identify invocation y(3)
}      process argument
}         3
}      replace y(3)
}   f(+z)
}      rescan starting at "+", with definitions of f and y suppressed 
}      identify invocation z
}      replace z
}   f(++y)
}      rescan, starting at 2nd "+", with definitions of f, y, and z suppressed
}   f(++y)
}done

Just a little point about the above: 
The result is displayed as "f(++y)", but at no point was a single "++" token
ever generated. The question I have is whether rescanning implies that
preprocesing tokens are converted back into text and re-lexed when rescanned.
The difference is whether the above represents the token sequence
"f" "(" "++" "y" ")" or the sequence "f" "(" "+" "+" "y" ")". I think that
the tokens are lexed once, and that the latter sequence is the correct
result, but I don't think I could make a good case for it (or rather, a good
case forbidding the other).

David Tanguay, Software Development Group, University of Waterloo