Re: Why put type information into syntax?

Keith Thompson wrote:
>
> Michael Spencer <michael.prqa@indigo.ie> 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.


Mike
michael.prqa@indigo.ie

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