Hi > I'm writing a simple interpretter for a small extended-lambda-calculus sort > of language. And I'd just like to say... RECURSIVE LET-BINDS! GAAAAH!!! >_<
Agreed :-) > To illustrate: > > let x = f x; y = 5 in x y > > A simple-minded interpretter might try to replace every occurrance of "x" > with "f x". This yields > > let y = 5 in (f x) y That's wrong, its a two step transformation. You just substitute all occurances of x for f x: let x = f (f x); y = 5 in (f x) y For the case let x = 5 in x, you do the same thing: let x = 5 in 5 Now as a second step you hoover up unused let bindings and disguard: 5 You seem to be combining the substitution and the removing of dead let bindings, if you separate them you should have more luck. Thanks Neil _______________________________________________ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe