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+A type alias impl trait can only have its hidden type assigned
+when used fully generically (and within their defining scope).
+This means
+
+```compile_fail,E0792
+#![feature(type_alias_impl_trait)]
+
+type Foo<T> = impl std::fmt::Debug;
+
+fn foo() -> Foo<u32> {
+    5u32
+}
+```
+
+is not accepted. If it were accepted, one could create unsound situations like
+
+```compile_fail,E0792
+#![feature(type_alias_impl_trait)]
+
+type Foo<T> = impl Default;
+
+fn foo() -> Foo<u32> {
+    5u32
+}
+
+fn main() {
+    let x = Foo::<&'static mut String>::default();
+}
+```
+
+
+Instead you need to make the function generic:
+
+```
+#![feature(type_alias_impl_trait)]
+
+type Foo<T> = impl std::fmt::Debug;
+
+fn foo<U>() -> Foo<U> {
+    5u32
+}
+```
+
+This means that no matter the generic parameter to `foo`,
+the hidden type will always be `u32`.
+If you want to link the generic parameter to the hidden type,
+you can do that, too:
+
+
+```
+#![feature(type_alias_impl_trait)]
+
+use std::fmt::Debug;
+
+type Foo<T: Debug> = impl Debug;
+
+fn foo<U: Debug>() -> Foo<U> {
+    Vec::<U>::new()
+}
+```