Swift: make type optional in TypeRepr

A type representation may not have a type in unresolved things, which
for example pop up in inactive `#if` clauses.

swift/codegen/schema.yml

@@ -184,7 +184,7 @@ Stmt:
 
 TypeRepr:
   _extends: AstNode
-  type: Type
+  type: Type?  # type can be absent on unresolved entities
 
 FunctionType:
   _extends: AnyFunctionType

swift/extractor/visitors/TypeVisitor.cpp

@@ -11,7 +11,7 @@ void TypeVisitor::visit(swift::TypeBase* type) {
 
 codeql::TypeRepr TypeVisitor::translateTypeRepr(const swift::TypeRepr& typeRepr, swift::Type type) {
   auto entry = dispatcher_.createEntry(typeRepr);
-  entry.type = dispatcher_.fetchLabel(type);
+  entry.type = dispatcher_.fetchOptionalLabel(type);
   return entry;
 }
 

swift/ql/lib/codeql/swift/generated/type/TypeRepr.qll

@@ -7,8 +7,10 @@ class TypeReprBase extends @type_repr, AstNode {
 
   Type getType() {
     exists(Type x |
-      type_reprs(this, x) and
+      type_repr_types(this, x) and
       result = x.resolve()
     )
   }
+
+  predicate hasType() { exists(getType()) }
 }

swift/ql/lib/swift.dbscheme

@@ -473,7 +473,12 @@ expr_types(  //dir=expr
 ;
 
 type_reprs(  //dir=type
-  unique int id: @type_repr,
+  unique int id: @type_repr
+);
+
+#keyset[id]
+type_repr_types(  //dir=type
+  int id: @type_repr ref,
   int type_: @type ref
 );