Title
Consistent total ordering of pointers by comparison functors
Status
c++17
Section
[comparisons]
Submitter
Casey Carter

Created on 2015-11-18.00:00:00 last changed 89 months ago

Messages

Date: 2016-08-05.03:58:49

Proposed resolution:

This wording is relative to N4606.

  1. Change [comparisons] p14 as indicated:

    -14- For templates greater, less, greater_equal, and less_equalless, greater, less_equal, and greater_equal, the specializations for any pointer type yield a strict total order that is consistent among those specializations and is also consistent with the partial order imposed by , even if the built-in operators <, >, <=, >= do not. [Note: When a < b is well-defined for pointers a and b of type P, this implies (a < b) == less<P>(a, b), (a > b) == greater<P>(a, b), and so forth. — end note] For template specializations greater<void>, less<void>, greater_equal<void>, and less_equal<void>less<void>, greater<void>, less_equal<void>, and greater_equal<void>, if the call operator calls a built-in operator comparing pointers, the call operator yields a strict total order that is consistent among those specializations and is also consistent with the partial order imposed by those built-in operators.

Date: 2016-08-05.03:58:49

[ 2016-08 - Chicago ]

Thurs PM: Moved to Tentatively Ready

Date: 2016-08-04.00:00:00

[ 2016-08-04 Chicago LWG ]

LWG discusses and concludes that we are trying to accomplish the following:

  1. T* a = /* ... */;
    T* b = /* ... */;
    

    if a < b is valid, a < b == less<T*>(a, b), and analogously for >, <=, >=.

  2. less<void>(a, b) == less<T*>(a, b);
    less<T*>(a, b) == greater<T*>(b, a);
    

    etc.

  3. less<T*> produces a strict total ordering with which the other three function objects are consistent

  4. less<void> when applied to pointers produces a strict total ordering with which the other three are consistent

  5. less<void> when applied to pointers of the same type produces the same strict total ordering as less<T*>, and analogously for the other three

  6. we are not addressing less<void> (and the other three) when applied to pointers of differing types

Walter and Nevin revise Proposed Wording accordingly.

Date: 2016-05-15.00:00:00

[ 2016-05-20, Casey Carter comments and suggests revised wording ]

The new proposed wording is attempting to address the issue raised in the 2016-02-04 telecon.

The real issue I'm trying to address here is ensure that "weird" implementations provide the same kind of consistency for pointer orderings as "normal" implementations that use a flat address spaces and have totally ordered <. If a < b is true for int pointers a and b, then less<int*>(a, b), less_equal<int*>(a, b), less<char*>(a, b), less<void*>(a, b), and greater<int*>(b, a) should all hold. I think this wording is sufficient to provide that.

Previous resolution [SUPERSEDED]:

This wording is relative to N4582.

  1. Alter [comparisons] to read:

    -14- For templates greater, less, greater_equal, and less_equal, the specializations for any pointer type yield athe same total order. That total order is consistent with the partial order imposed by, even if the built-in operators <, >, , and > do not. [Note: When a < b is well-defined for pointers a and b of type P, this implies (a < b) == less<P>(a, b), (a > b) == greater<P>(a, b), and so forth. — end note] For template specializations greater<void>, less<void>, greater_equal<void>, and less_equal<void>, if the call operator calls a built-in operator comparing pointers, the call operator yields a total order.

Date: 2016-05-21.14:26:43

N4567 [comparisons]/14 specifies that the comparison functors provide a total ordering for pointer types:

For templates greater, less, greater_equal, and less_equal, the specializations for any pointer type yield a total order, even if the built-in operators <, >, <=, >= do not.

It notably does not specify:

  • whether the specializations of all of the named templates for a given pointer type yield the same total order

  • whether the total order imposed respects the partial order imposed by the built-in operators

  • whether the total order imposed is consistent with the partition induced by ==

All of which are important for sane semantics and provided by common implementations, since the built-in operators provide a total order and the comparison functors yield that same order.

It would be extremely confusing — if not outright insane — for e.g.:

  • less<int*> and greater<int*> to yield different orders

  • less<int*> to disagree with < on the relative order of two pointers for which < is defined

  • less<int*> to order a before b when a == b, i.e., not preserve equality.

Consistent semantics for the various comparison functors and the built-in operators is so intuitive as to be assumed by most programmers.

Related issues: 2450, 2547.

Previous resolution [SUPERSEDED]:

This wording is relative to N4567.

  1. Alter [comparisons]/14 to read:

    For templates greater, less, greater_equal, and less_equal, the specializations for any pointer type yield athe same total order, even if the built-in operators <, >, <=, >= do not. The total order shall respect the partial order imposed by the built-in operators.

History
Date User Action Args
2017-07-30 20:15:43adminsetstatus: wp -> c++17
2016-11-14 03:59:28adminsetstatus: pending -> wp
2016-11-14 03:55:22adminsetstatus: ready -> pending
2016-08-05 03:58:49adminsetmessages: + msg8421
2016-08-05 03:58:49adminsetstatus: new -> ready
2016-08-04 21:34:55adminsetmessages: + msg8408
2016-05-21 14:26:43adminsetmessages: + msg8129
2015-12-14 22:03:28adminsetmessages: + msg7643
2015-11-18 00:00:00admincreate