Yourselffff

Rust闭包返回引用时的规则及与变量本身的关系是什么？

fn main() { let mut my_string = String::from("Hello, Rust!"); let mut my_s_ref = &mut my_string; let consume_string = || { my_s_ref }; consume_string(); // consume_string(); println!("闭包外部消费字符串: {}", my_string); } fn main() { let my_string = String::from("Hello, Rust!"); let my_s_ref = & my_string; let consume_string = || { // my_s_ref.push_str("string"); my_s_ref }; consume_string(); consume_string(); println!("闭包外部字符串: {}", my_string); } 以上两段代码，一个返回String的可变引用，一个返回不可变引用，为什么返回可变引用的闭包实现了FnOnce,返回不可变引用的闭包只实现了Fn呢？ rust返回引用时，遵循什么规则？和my_s_ref本身有关还是my_string有关？

HAO起起

what is python dangling circular ref?

下面是wsgi web server的片段demo def start_response(status, response_headers, exc_info=None): if exc_info: try: if headers_sent: # Re-raise original exception if headers sent raise exc_info[1].with_traceback(exc_info[2]) finally: exc_info = None # avoid dangling circular ref elif headers_set: raise AssertionError("Headers already set!") headers_set[:] = [status, response_headers] # Note: error checking on the headers should happen here, # *after* the headers are set. That way, if an error # occurs, start_response can only be re-called with # exc_info set. return write 我想问: "exc_info = None # avoid dangling circular ref" 这里有什么必要这样操作？ 这和迷途指针有什么关系吗？