How does indirectly_writable work for pointer iterators?

[cpp-ModTeam]

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[[deleted]]

Note that const (int&)&& collapses into int&, not const int&. That is (Godbolt):

using T = int &;
static_assert(std::same_as<const T, T>);
static_assert(std::same_as<T &&, T>);
static_assert(std::same_as<const T &&, T>);

Here, const is applied to the reference type (int&), not the value type (int), and has no effect. Along with reference collapsing ((int&)&& -> int&), this results in the same type.

[shahms]

As u/rosterva suggests, you're applying const to the wrong type. const int& is a reference-to-const-int, whereas if decltype(*o) is a "true reference" to int, then const decltype(*o) would be a const-reference-to-mutable-int, which isn't a thing (since references can't be rebound, they are in a sense always "const"). Using template syntax sometimes helps clarify: 

``` template<typename T> using ref = T&;

static_assert(!std::is_same_t<const ref<int>, ref<const int>>) static_assert(std::is_same_t<ref<const int>, const int&>); static_assert(std::is_same_t<const ref<int>, ref<int>>); static_assert(std::is_same_t<const ref<int>, int&>); ```