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, 2 insertions, 2 deletions

diff --git a/crates/ra_ide_api/src/call_info.rs b/crates/ra_ide_api/src/call_info.rs
index 9beceb29c..d0283e410 100644
--- a/crates/ra_ide_api/src/call_info.rs
+++ b/crates/ra_ide_api/src/call_info.rs
@@ -26,8 +26,8 @@ pub(crate) fn call_info(db: &RootDatabase, position: FilePosition) -> Option<Cal
     );
     let (mut call_info, has_self) = match &calling_node {
         FnCallNode::CallExpr(expr) => {
-            //FIXME: don't poke into Ty
-            let (callable_def, _subst) = analyzer.type_of(db, &expr.expr()?)?.as_callable()?;
+            //FIXME: Type::as_callable is broken
+            let callable_def = analyzer.type_of(db, &expr.expr()?)?.as_callable()?;
             match callable_def {
                 hir::CallableDef::FunctionId(it) => {
                     let fn_def = it.into();