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, 11 deletions

diff --git a/crates/ra_assists/src/assists/add_explicit_type.rs b/crates/ra_assists/src/assists/add_explicit_type.rs
index 562a09685..eeb4ff39f 100644
--- a/crates/ra_assists/src/assists/add_explicit_type.rs
+++ b/crates/ra_assists/src/assists/add_explicit_type.rs
@@ -1,4 +1,4 @@
-use hir::{db::HirDatabase, HirDisplay, Ty};
+use hir::{db::HirDatabase, HirDisplay};
 use ra_syntax::{
     ast::{self, AstNode, LetStmt, NameOwner},
     T,
@@ -43,7 +43,7 @@ pub(crate) fn add_explicit_type(ctx: AssistCtx<impl HirDatabase>) -> Option<Assi
     let analyzer = ctx.source_analyzer(stmt.syntax(), None);
     let ty = analyzer.type_of(db, &expr)?;
     // Assist not applicable if the type is unknown
-    if is_unknown(&ty) {
+    if ty.contains_unknown() {
         return None;
     }
 
@@ -53,15 +53,6 @@ pub(crate) fn add_explicit_type(ctx: AssistCtx<impl HirDatabase>) -> Option<Assi
     })
 }
 
-/// Returns true if any type parameter is unknown
-fn is_unknown(ty: &Ty) -> bool {
-    match ty {
-        Ty::Unknown => true,
-        Ty::Apply(a_ty) => a_ty.parameters.iter().any(is_unknown),
-        _ => false,
-    }
-}
-
 #[cfg(test)]
 mod tests {
     use super::*;