Issue 3030: Who shall meet the requirements of try_lock?

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3030. Who shall meet the requirements of `try_lock`?

Section: 32.6.6 [thread.lock.algorithm] Status: C++20 Submitter: Jonathan Wakely Opened: 2017-11-07 Last modified: 2021-02-25

Priority: 0

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Discussion:

32.6.6 [thread.lock.algorithm] says:

"If a call to try_lock() fails, unlock() shall be called for all prior arguments and there shall be no further calls to try_lock()."

We try to use "shall" for requirements on the user (e.g. as in the previous paragraph) which is absolutely not what is meant here.

[2017-11 Albuquerque Wednesday night issues processing]

Moved to Ready

[2018-3-17 Adopted in Jacksonville]

Proposed resolution:

This wording is relative to N4700.

Change 32.6.6 [thread.lock.algorithm] as indicated:
```
template <class L1, class L2, class... L3> int try_lock(L1&, L2&, L3&...);
```
-1- Requires: […]
-2- Effects: Calls try_lock() for each argument in order beginning with the first until all arguments have been processed or a call to try_lock() fails, either by returning false or by throwing an exception. If a call to try_lock() fails, unlock() ~~shall be~~is called for all prior arguments ~~and there shall be~~with no further calls to try_lock().
[…]
```
template <class L1, class L2, class... L3> void lock(L1&, L2&, L3&...);
```
-4- Requires: […]
-5- Effects: All arguments are locked via a sequence of calls to lock(), try_lock(), or unlock() on each argument. The sequence of calls ~~shall~~does not result in deadlock, but is otherwise unspecified. [Note: A deadlock avoidance algorithm such as try-and-back-off must be used, but the specific algorithm is not specified to avoid over-constraining implementations. — end note] If a call to lock() or try_lock() throws an exception, unlock() ~~shall be~~is called for any argument that had been locked by a call to lock() or try_lock().