Re: Generating a simple hand-coded like recursive descent parser

On Sun, 17 Dec 2006, Robert A Duff wrote:
> Hans-Peter Diettrich <DrDiettrich1@aol.com> writes:
>> Tommy Thorn wrote:
>>> x * y;
>>> is a declation of a pointer to x or an expression multiplying x and y.
>>
>> A more practical example, found in this group already:
>> a = (b)-c;
>> is (b) a type cast, or an expression?
[...]
> P.S. Why do you say that the "a = (b)-c;" ambiguity is more practical
> than the "x * y;" ambiguity? They seem like more-or-less the same
> thing, to me: the parser wants to know whether each identifier is a type
> name.


      They are the same thing, to the parser. However,


          a = (b)-c;


might just possibly crop up in a program written by a human being, in
either of its meanings (if (b) were the result of macro expansion,
perhaps); whereas no human would ever write


          x * y;


in its meaning of "multiply x and y, and then discard the result".


(Someone with less respect for the C standard's formal grammar might
say, "Why don't we just make x * y; always declare a new pointer, and
ignore that other case that never comes up in practice?" They can't
say that about a = (b)-c; .)


-Arthur

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