Coerce two FnDefs to fn pointers even if they are the same, if they are subtypes

That's because they can be the same function but still different substs, which mandates them to coerce to fn pointers in order to unify.

crates/hir-ty/src/infer/coerce.rs

@@ -125,7 +125,11 @@ impl CoerceMany {
         // pointers to have a chance at getting a match. See
         // https://github.com/rust-lang/rust/blob/7b805396bf46dce972692a6846ce2ad8481c5f85/src/librustc_typeck/check/coercion.rs#L877-L916
         let sig = match (self.merged_ty().kind(Interner), expr_ty.kind(Interner)) {
-            (TyKind::FnDef(x, _), TyKind::FnDef(y, _)) if x == y => None,
+            (TyKind::FnDef(x, _), TyKind::FnDef(y, _))
+                if x == y && ctx.table.unify(&self.merged_ty(), &expr_ty) =>
+            {
+                None
+            }
             (TyKind::Closure(x, _), TyKind::Closure(y, _)) if x == y => None,
             (TyKind::FnDef(..) | TyKind::Closure(..), TyKind::FnDef(..) | TyKind::Closure(..)) => {
                 // FIXME: we're ignoring safety here. To be more correct, if we have one FnDef and one Closure,

crates/hir-ty/src/tests/coercion.rs

@@ -942,3 +942,19 @@ fn main() {
 "#,
     )
 }
+
+#[test]
+fn regression_18626() {
+    check_no_mismatches(
+        r#"
+fn f() {
+    trait T {
+        fn f() {}
+    }
+    impl T for i32 {}
+    impl T for u32 {}
+    &[i32::f, u32::f] as &[fn()];
+}
+    "#,
+    );
+}