If I understand this correctly...
That means it's never possible to lock two such mutexes at once. You can't await the second one while holding the first one.
Doesn't that make it impossible to express a bunch of good old CS algorithms?
If I understand this correctly...
That means it's never possible to lock two such mutexes at once. You can't await the second one while holding the first one.
Doesn't that make it impossible to express a bunch of good old CS algorithms?