🚀 go-pugleaf

In article <4es8rp$98n@usenet.pa.dec.com>, diamond@tko.dec.com (Norman Diamond) writes:
:> In article <TANMOY.96Feb1102025@qcd.lanl.gov>, tanmoy@qcd.lanl.gov (Tanmoy Bhattacharya) writes:
:> >In article <DM3MC3.Atv@scr.siemens.com> mlg@scr.siemens.com (Michael
:> >Greenberg) writes: 
:> >>Is the following program legal?
:> >>#define foo(unused)
:> >>#define bar(x,y)
:> >>foo(bar(1))
:> 
:> >I read it as being legal.
:> 
:> Mr. Bhattacharya, I am shocked.
:>
[Snip]

Mr. Diamond, why are you shocked? Macro 'bar' is not invoked. Read 6.8.3.1
again: we do

   - identify the parameters
   
     Ok, 'foo' has one parameter, "bar(1)".
     
   - replace occurrences of the formal parameter ('unused') in the
     replacement list by the corresponding argument after macro re-
     placement within this argument.
     
     Well, 'foo's replacement list is empty, so this never happens.
     In other words, since there's no 'unused' in the replacement
     list, we'll never try to substitute it by the argument "bar(1)",
     thus we'll never try to perform macro replacement on "bar(1)"
     and therefore 'bar' is not invoked.
     
Your quote from 6.8.3 (3.8.3. in ANSI Classic) does not apply:
     
:>       lines 14 to 15:  "The number of arguments in an INVOCATION of a
:>       function-like macro shall agree with the number of parameters in
:>       the macro definition, and [...]"

(Emphasis added, TW)

Above quote applies only to *invocations*, but since 'bar' is not invoked,
it's irrelevant in this example.

Regards,

  Thomas
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Swiss Federal Institute of Technology | Thomas Wolf
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