- Title
- Common and CommonReference have a common defect
- Status
- c++20
- Section
- [concept.common]
- Submitter
- Casey Carter

Created on **2018-08-10.00:00:00**
last changed **1 week ago**

Date: 2018-11-12.04:39:29

**Proposed resolution:**

This wording is relative to N4762.

Modify [concept.commonref] p2 as follows:

-2- Let

`C`be`common_reference_t<T, U>`. Let`t`be a function whose return type is`t1`and`t2`be equality-preserving expressions ([concepts.equality]) such that`decltype((t1))`and`decltype((t2))`are each`T`, and let`u`be a function whose return type is`u1`and`u2`be equality-preserving expressions such that`decltype((u1))`and`decltype((u2))`are each`U`.`T`and`U`model`CommonReference<T, U>`~~is satisfied~~only if:(2.1) —

`C(t1`equals~~()~~)`C(t2`if and only if~~()~~)`t1`equals~~()~~`t2`, and~~is an equality-preserving expression ([concepts.equality]).~~(2.2) —

`C(u1`equals~~()~~)`C(u2`if and only if~~()~~)`u1`equals~~()~~`u2`~~is an equality-preserving expression~~.Modify [concept.common] p2 similarly:

-2- Let

`C`be`common_type_t<T, U>`. Let`t`be a function whose return type is`t1`and`t2`be equality-preserving expressions ([concepts.equality]) such that`decltype((t1))`and`decltype((t2))`are each`T`, and let`u`be a function whose return type is`u1`and`u2`be equality-preserving expressions such that`decltype((u1))`and`decltype((u2))`are each`U`.`T`and`U`model`Common<T, U>`~~is satisfied~~only if:(2.1) —

`C(t1`equals~~()~~)`C(t2`if and only if~~()~~)`t1`equals~~()~~`t2`, and~~is an equality-preserving expression ([concepts.equality]).~~(2.2) —

`C(u1`equals~~()~~)`C(u2`if and only if~~()~~)`u1`equals~~()~~`u2`~~is an equality-preserving expression ([concepts.equality])~~.

Date: 2018-11-12.04:39:29

*[ 2018-11, Adopted in San Diego ]*

Date: 2018-08-22.12:55:05

*[ 2018-08 Batavia Monday issue prioritization ]*

Priority set to 0, status to 'Tentatively Ready'

Date: 2018-08-10.00:00:00

The semantic requirements of both `Common`
([concept.common]):

-2- Let

Cbecommon_type_t<T, U>. Lettbe a function whose return type isT, and letube a function whose return type isU.Common<T, U>is satisfied only if:(2.1) —

C(t())equalsC(t())if and only ift()is an equality-preserving expression ([concepts.equality]).(2.2) —

C(u())equalsC(u())if and only ifu()is an equality-preserving expression ([concepts.equality]).

and similarly `CommonReference` ([concept.commonreference]):

-2- Let

Cbecommon_reference_t<T, U>. Lettbe a function whose return type isT, and letube a function whose return type isU.CommonReference<T, U>is satisfied only if:(2.1) —

C(t())equalsC(t())if and only ift()is an equality-preserving expression ([concepts.equality]).(2.2) —

C(u())equalsC(u())if and only ifu()is an equality-preserving expression.

don't properly reflect the intended design that conversions to the common type /
common reference type are identity-preserving: in other words, that converting
two values to the common type produces equal results if and only if the values
were initially equal. The phrasing "`C(E)` equals `C(E)` if and only if `E` is an equality-preserving expression" is also clearly
defective regardless of the intended design: the assertion "`E` is not
equality-preserving" does not imply that every evaluation of `E` produces
different results.

History | |||
---|---|---|---|

Date | User | Action | Args |

2021-02-25 10:48:01 | admin | set | status: wp -> c++20 |

2018-11-12 04:39:29 | admin | set | messages: + msg10208 |

2018-11-12 04:39:29 | admin | set | status: voting -> wp |

2018-10-08 05:13:59 | admin | set | status: ready -> voting |

2018-08-22 12:55:05 | admin | set | messages: + msg10101 |

2018-08-22 12:55:05 | admin | set | status: new -> ready |

2018-08-13 22:37:52 | admin | set | messages: + msg10065 |

2018-08-10 00:00:00 | admin | create |