6 ms·
Because now you're holding a reference to `x` which is supposed to be protected by a mutex, even after the mutex is unlocked. With the lambda-only API, it's mu
by akling 3y ago
Because now you're holding a reference to `x` which is supposed to be protected by a mutex, even after the mutex is unlocked.
With the lambda-only API, it's much harder to make this mistake, since a temporary reference like this will still go out of scope at the end of the lambda expression.
- kajaktum 3y agoYou specifically mentioned that this is a footgun: > auto& x = state.locked()->x; But I don't see how the reference here is gonna make a difference unless i am reading the lifetime of the lock here incorrectly. For example, this is perfectly fine right? ``` { auto& x = state.locked()->x; } ``` This will only be a problem if you have an outside struct that holds a reference ``` auto &a = "something"; { auto& x = state.locked()->x; a = x; } ``` Which can still happen even if you use a lambda.