Function is_face_source 

pub fn is_face_source(source: &str, marker: &str) -> bool

Whether source declares a registration face. 源码是否声明了注册面。

A generated-marker line short-circuits the answer: a snapshot this tooling wrote is a face by construction, and parsing it again is wasted work. For every other file the parse decides. 生成标记行会短路答案：本工具写出的快照按构造就是注册面，再解析一遍是白费。其余 文件一律由解析裁决。