Merge #2398

2398: WIP: introduce hir::Type r=matklad a=matklad

This introduces `hir::Type` wrapper over `hir::Ty`, with two purposes:

* bind `Ty` and it's corresponding environment
  * Am I correct that `Ty` without an env doesn't make much sense, because the meaning of type parameters is unclear
  * Am I correct that we can safely re-use the same environment for all types derived from the given type? 
* hide representation defails of `Ty`. Specifically, I want to change `Ty::Adt` to use `hir_def::AdtId` instead of `hir::Adt`, but IDE doesn't know about underlying IDs. More generally, I feel like IDE shouldn't know that `Ty` is enum.

@flodiebold what do you think about this?

Co-authored-by: Aleksey Kladov <[email protected]>

1 files changed, 4 insertions, 3 deletions

diff --git a/crates/ra_assists/src/assists/fill_match_arms.rs b/crates/ra_assists/src/assists/fill_match_arms.rs
index 8482897c5..b75bd44eb 100644
--- a/crates/ra_assists/src/assists/fill_match_arms.rs
+++ b/crates/ra_assists/src/assists/fill_match_arms.rs
@@ -83,10 +83,11 @@ fn resolve_enum_def(
 ) -> Option<ast::EnumDef> {
     let expr_ty = analyzer.type_of(db, &expr)?;
 
-    analyzer.autoderef(db, expr_ty).find_map(|ty| match ty.as_adt() {
-        Some((Adt::Enum(e), _)) => Some(e.source(db).value),
+    let res = expr_ty.autoderef(db).find_map(|ty| match ty.as_adt() {
+        Some(Adt::Enum(e)) => Some(e.source(db).value),
         _ => None,
-    })
+    });
+    res
 }
 
 fn build_pat(var: ast::EnumVariant) -> Option<ast::Pat> {