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weberc2 | 5 years ago

This doesn't add anything. You get the same protection in Go with pointers. For example, a `*T` can't be passed to a function taking a `T` without explicitly dereferencing it. The problem of course is that the type system doesn't guarantee that the pointer isn't nil when you go to dereference it, similarly your `Option<T>` doesn't guarantee that the option.Value is set correctly. You need sum types to provide this substantial guarantee.

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TheDong|5 years ago

Pointers in go aren't a signal for optionality, they're a signal for stack or heap or optionality. They also don't play nice with interfaces.

Let's say I have a hashmap that returns a `&T` and I have a `func foo(s Stringer)`. Because 'Stringer' is an interface, it's possible for it to take both `T` and `&T` without a compile-time complaint.

In addition, I may wish to have my function `bar(t &T)` always take a pointer because I want the object to exist on the heap, or to be able to mutate its value for the caller.

Since pointers mean so many other things, they're not a good way to have a compile-error indicate optionality.

On the other hand, if my caller does `hm.Get` and gets an `Option<s: Stringer>`, it's clear what to do to pass it to foo, and if it's an `Option<&T>`, it's clear both that I want a pointer, and that it should be checked / unwrapped.

I agree that `T` / `&T` would be just as powerful as option types in go (without sum types) if pointers didn't already have other substantial meaning, and if they could sensibly interact with interfaces.

As it stands, I think you're off the mark though.

(note: All the asterisks are & because I dunno how to escape stuff on hn)

bradfitz|5 years ago

> Pointers in go aren't a signal for optionality, they're a signal for stack or heap

No, a pointer in Go doesn't mean that it's on the heap. The compiler keeps it on the stack if it's safe to do so, regardless of whether you're using pointers.

You can even write code like `t := new(T); t.Foo()` that very much looks like you're allocating on the heap, but it can stay on the stack, yet t is then a pointer to the stack.

Unlike C, you don't need to worry about the heap-vs-stack in Go. It's never even mentioned in the language spec as a concept people need to be concerned with. It's an implementation detail.

alkonaut|5 years ago

It guarantees that the value is set if the flag is saying it is set (that's the invariant of the type). It screams "check flag before accessing the value".

To compare with references which are also 0-or-1-thing effectively: A reference where you as a developer know it's never null but always a ref to exactly one thing is denoted "* T" and a reference to "one thing or null" is denoted "* T"! There is no difference in the types! so you can accidentally send one that is "0-or-1-things" to a method accepting a * T that MUST be a thing. Type system didn't help you document which case it was.

Options, apart from the annotation benefit it also helps making the syntax nicer in many cases, with e.g. "or()" fallbacks etc.

    let data = get_cached().or(load_from_disk()).or_panic();

weberc2|5 years ago

I agree that there are semantic issues with pointers, but my point was to illustrate that you need sum types in Go to get any real benefit out of an Option type. If you need an option type and you aren't content with a pointer, you can use `(T, bool)`, but this is still a far cry from a real Option type.