On Wed, May 20, 2015 at 12:24 PM, Matt Oliveri <atma...@gmail.com> wrote:
> On Wed, May 20, 2015 at 1:44 PM, Jonathan S. Shapiro <s...@eros-os.org> > wrote: > > > Last option, which I think does not really work, and cannot really be > done > > in F#, O'Caml, or Haskell, is to exploit the fact that a union leg type > > actually *is* a type, and use parameterization. I don't have a way to > > syntactically express this in these languages, but the general idea is > that > > the metadata node and the AST type would be mutually recursive types, and > > the metadata node would be of the form MetaASTNode<LegType-or-group-type> > > I can't figure out what you're getting at here. > > Don't you just want something like: > > type Node<T> = MetaData * T > > type AST = > ... > and Expr = > | MulExpr of Node<Expr> * Node<Expr> > More or less, yes. Though it's a bit curious here that Expr is considered a type where MulExpr is not. One of the conclusions from BitC is that it's actually useful to allow MulExpr to be a type that can be used and declared. I confess that I don't fully understand the syntax you (and F#) are using here. Are Expr and AST two independent type declarations that are co-recursive, or is Expr a subtype of AST here? I'm not sure what this syntax means, and I'd appreciate a quick explanation. I think this approach is usefully carried a step further, such that MulExpr is also a type, and we can then declare a Node<MulExpr>. For Expr there aren't that many cases to match. For Type ASTs, on the other hand.... shap
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