|[6 earlier articles]|
|Re: Why put type information into syntax? firstname.lastname@example.org (Michael Spencer) (2000-04-05)|
|Re: Why put type information into syntax? email@example.com (Rodney M. Bates) (2000-04-05)|
|Re: Why put type information into syntax? firstname.lastname@example.org (Keith Thompson) (2000-04-11)|
|Re: Why put type information into syntax? email@example.com (Ira D. Baxter) (2000-04-14)|
|Re: Why put type information into syntax? firstname.lastname@example.org (Robert A Duff) (2000-04-14)|
|Re: Why put type information into syntax? maratb@CS.Berkeley.EDU (Marat Boshernitsan) (2000-04-15)|
|Re: Why put type information into syntax? email@example.com (Michael Spencer) (2000-04-15)|
|From:||Michael Spencer <firstname.lastname@example.org>|
|Date:||15 Apr 2000 21:40:40 -0400|
|References:||00-03-133 00-03-146 00-04-017 00-04-050 00-04-073|
Keith Thompson wrote:
> Michael Spencer <email@example.com> writes:
> > More importantly, in C++, the built-in type keywords prevent parse
> > conflicts. They do in my grammar. I am using a backtracking LR
> > parser generator to parse C++ (since I need infinite lookahead). So I
> > am guessing anyway on conflicts. But I take a slight penalty every
> > time I do so, so I do not want to guess more than I have to.
> In either C or C++, the fact that the names of the predefined types
> are keywords (or sequences of keywords) doesn't really make parsing
> any easier, since the names of user-defined types are *not* keywords.
Consider this example:
typedef int A, B, C;
void foo ()
A B (C);
int long (C);
Initially, A B and C are all types. B is then declared as an extern
function taking type C and returning type A. Then, because "int" and
"long" are built-in *keywords*, C is declared as an object of type
"long int". If you group built-in type keywords with user-defined
types, don't you get the wrong answer? And only one token lookahead
is needed ...
You do not have this problem in Java.
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