GDScript: Call implicit ready on base script first

It is generally expected that the base class is called before the
inherited clas. This commit implements this behavior for the implicit
ready function (`@onready` annotation) to make it consistent with the
expectations.

1 files changed, 13 insertions, 10 deletions

diff --git a/modules/gdscript/gdscript.cpp b/modules/gdscript/gdscript.cpp
index f238958f25..73abf71bde 100644
--- a/modules/gdscript/gdscript.cpp
+++ b/modules/gdscript/gdscript.cpp
@@ -1958,19 +1958,22 @@ int GDScriptInstance::get_method_argument_count(const StringName &p_method, bool
 	return 0;
 }
 
+void GDScriptInstance::_call_implicit_ready_recursively(GDScript *p_script) {
+	// Call base class first.
+	if (p_script->_base) {
+		_call_implicit_ready_recursively(p_script->_base);
+	}
+	if (p_script->implicit_ready) {
+		Callable::CallError err;
+		p_script->implicit_ready->call(this, nullptr, 0, err);
+	}
+}
+
 Variant GDScriptInstance::callp(const StringName &p_method, const Variant **p_args, int p_argcount, Callable::CallError &r_error) {
 	GDScript *sptr = script.ptr();
 	if (unlikely(p_method == SNAME("_ready"))) {
-		// Call implicit ready first, including for the super classes.
-		while (sptr) {
-			if (sptr->implicit_ready) {
-				sptr->implicit_ready->call(this, nullptr, 0, r_error);
-			}
-			sptr = sptr->_base;
-		}
-
-		// Reset this back for the regular call.
-		sptr = script.ptr();
+		// Call implicit ready first, including for the super classes recursively.
+		_call_implicit_ready_recursively(sptr);
 	}
 	while (sptr) {
 		HashMap<StringName, GDScriptFunction *>::Iterator E = sptr->member_functions.find(p_method);