`COMMON-DOMAIN` should test for convertibility to `default_domain`

[ericniebler]

P3826R4 adds the following paragraph to [exec.snd.expos]:

6. For a pack of subexpressions domains, COMMON-DOMAIN(domains...) is expression-equivalent to common_type_t<decltype(auto(domains))...>() if that expression is well-formed, and indeterminate_domain<Ds...>() otherwise, where Ds is the pack of types consisting of decltype(auto(domains))... with duplicate types removed.

my intention with this paragraph was that two domains such as below would have a COMMON-DOMAIN of default_domain:

struct my_thread_pool_domain : std::execution::default_domain
{};

struct my_parallel_runtime_domain : std::execution::default_domain
{};

using domain-t = decltype(COMMON-DOMAIN(my_thread_pool_domain{}, my_parallel_runtime_domain{}));
static_assert(std::same_as<domain-t, default_domain>); // FAILS

but that's not how common_type_t works. even though both domains derive from default_domain, they do not have a common type as determined by common_type_t.

we can patch this up easily enough: if common_type_t<Domains...> is ill-formed, we should check if common_type_t<default_domain, Domains...> is well-formed. this will correctly handle the case when all the domains are convertible to default_domain.

Proposed Resolution

Change [exec.snd.expos]/p6 as follows:

 6. For a pack of subexpressions domains, COMMON-DOMAIN(domains...) is
    expression-equivalent to common_type_t<decltype(auto(domains))...>()
-   if that expression is well-formed,
+   if that expression is well-formed; otherwise,
+   common_type_t<default_domain(), decltype(auto(domains))...>() if that
+   expression is well-formed;
    and indeterminate_domain<Ds...>() otherwise, where Ds is the pack of
    types consisting of decltype(auto(domains))... with duplicate types
    removed.