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### 3431. `<=>` for containers should require `three_way_comparable<T>` instead of `<=>`

Section: 24.2.2.4 [container.opt.reqmts] Status: New Submitter: Jonathan Wakely Opened: 2020-04-17 Last modified: 2022-04-24

Priority: 2

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Discussion:

The precondition for `<=>` on containers is:

"Either `<=>` is defined for values of type (possibly `const`) `T`, or `<` is defined for values of type (possibly `const`) `T` and `<` is a total ordering relationship."

I don't think `<=>` is sufficient, because `synth-three-way` won't use `<=>` unless `three_way_comparable<T>` is satisfied, which requires `weakly-equality-comparable-with<T, T>` as well as `<=>`.

So to use `<=>` I think the type also requires `==`, or more precisely, it must satisfy `three_way_comparable`.

The problem becomes clearer with the following example:

>     #include <compare>
>     #include <vector>

>     struct X
>     {
>       friend std::strong_ordering operator<=>(X, X) { return std::strong_ordering::equal; }
>     };

>     std::vector<X> v(1);
>     std::strong_ordering c = v <=> v;

This doesn't compile, because despite `X` meeting the preconditions for `<=>` in [tab:container.opt], `synth-three-way` will return `std::weak_ordering`.

Here is another example:

>     #include <compare>
>     #include <vector>

>     struct X
>     {
>       friend bool operator<(X, X) { return true; } // The return value is intentional, see below
>       friend std::strong_ordering operator<=>(X, X) { return std::strong_ordering::equal; }
>     };

>     std::vector<X> v(1);
>     std::weak_ordering c = v <=> v;

This meets the precondition because it defines `<=>`, but the result of `<=>` on `vector<X>` will be nonsense, because `synth-three-way` will use `operator<` not `operator<=>` and that defines a broken ordering.

So we're stating a precondition which implies "if you do this, you don't get garbage results" and then we give garbage results anyway.

The proposed resolution is one way to fix that, by tightening the precondition so that it matches what `synth-three-way` actually does.

[2020-04-25 Issue Prioritization]

Priority to 2 after reflector discussion.

Previous resolution [SUPERSEDED]:

> This wording is relative to N4861.

> 1. Modify 24.2.2.1 [container.requirements.general], Table [tab:container.opt], as indicated:

>    > Table 75: Optional container operations [tab:container.opt]
>    > Expression | Return type | Operational semantics | Assertion/note pre-/post-condition | Complexity
>    > --- | --- | --- | --- | ---
>    > `…`
>    > `a <=> b` | `synth-three-way-result<value_type>` | `lexicographical_compare_three_way( a.begin(), a.end(), b.begin(), b.end(), synth-three-way)` | Preconditions: Either `<=>` is defined for values of type (possibly `const`) `T` satisfies `three_way_comparable`, or `<` is defined for values of type (possibly `const`) `T` and `<` is a total ordering relationship. | linear
>    > `…`

[2022-04-24; Daniel rebases wording on N4910]

Proposed resolution:

This wording is relative to N4910.

1. Modify 24.2.2.4 [container.opt.reqmts] as indicated:

   >     a <=> b

   > > -2- Result: `synth-three-way-result<X::value_type>`.

   > > -3- Preconditions: Either `<=>` is defined for values of type (possibly `const`) `T` satisfies `three_way_comparable`, or `<` is defined for values of type (possibly `const`) `T` and `<` is a total ordering relationship.

   > > -4- Returns: `lexicographical_compare_three_way(a.begin(), a.end(), b.begin(), b.end(), synth-three-way)`

   > > [Note 1: The algorithm `lexicographical_compare_three_way` is defined in Clause 27. — end note]

   > > -5- Complexity: Linear.