🚀 go-pugleaf

> Consider the following in C, however:

>	. . . { T (x); . . . }
 
> How should the construct be parsed?

It would have to be an expression, since anything within brackets ain't a
decleration. This isn't Pascal kiddees, all functions are global.

> Consider instead the one that someone else posted:
> 
> 	{ T * x; . . . }

This just shows that your grammar is poor. Obviously it has to be the
interpreter's job to distinguish between decleration and expression.

I don't see any qualitative difference between typedef's and variable
decleration. An identifier is an identifier. Are you going to tell me
that Pascal isn't CF because A*B could have diffent meanings, because
multiplying integers creates different code than multiplynig reals?

Another thing. No language is a CFG. A language just isn't a grammar.
In order to prove that C isn't A CFL, you'ld have to prove that there
doesn't exist a CFG that only accepts valid C programs. A grammar
doesn't care what you do with each statement, so who cares how it would
be parsed? You'll just get a very complicated interpreter, since very
little could be said about statement types in advance.
-- 
					Micha Berger				     
Disclaimer: All opinions expressed here are my own. The spelling, noone's.
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