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by kajaktum 1173 days ago
Wait why is this a footgun?
2 comments

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.

You 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.

Holding the reference to a field that is protected by a mutex implies there is another thread out there that will race with your reference in either reading or writing it.

Even just a read is racy, as there is no “atomic read” of any size value if it is not already wrapped as atomic.