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Comment by omnicognate | Hacker News Reader

The distinct types are the whole point. You wouldn't want a std::tuple<> to be implicitly convertible to a std::optional<T> (for arbitrary T), and std::nullptr_t exists to be the type of nullptr, which captures the conversion behaviours appropriate for null pointer literals and has nothing to do with the variant use case std::monostate exists to serve.

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If there was a std::unit_t and it was implicitly convertible to optional, tuple and pointer, I don't think that would be worse in terms of usability at all (maybe worse in readability for people who haven't heard of a 'unit' type).

As for the std::variant use case, using std::monostate is only a matter of convention there. You could use any of the other unit types just the same.

std::monostate is explicitly provided for use with std::variant. It's in the <variant> header. Sometimes people use it for other things, but that's really an abuse, especially given defining your own type suitable for such cases is typically as simple as `struct mytype{};`.

Using one type to represent empty literals for optional, tuple and pointer types, implicitly convertible to all of them, would make the compiler accept many obviously accidental constructs. In a world where the maintainers of C++ are trying their hardest to make the language safer what conceivable benefit would there be?

Then you're basically back to "anything can convert back and forth with void " -- the point is to avoid* that.

Why wouldn't you want std::tuple<> to be the same as std::monostate, though? In many languages with a proper unit type such as Haskell and Rust, the zero tuple is the unit type.

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