4 ms·
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 a
by kajaktum 4y ago
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.