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by Joker_vD 2055 days ago
So basically, "if you don't use newtype to implement an abstract data type, you don't get any type safety because the on-spot wrapping/unwrapping is still possible".

On one hand, yes. On another, no. I am working right now on a Go codebase with lots of newtypes (well, the Go's equivalent), and they're generally casted to/from underlying primitive types basically in two places: when they're serialized into JSON, and where they're deserialized from JSON. And in several places in the middle of the code where you need an explicit cast, well, the cast is explicit. You actually have to consider it.

Mind you, in Go it's almost impossible to hide the newtype's constructor unless you jump through some rather unintuitive hoops. Is Go fundamentally type-unsafe? I don't think it is, although I sometimes wish some structs were impossible to zero-initialize. Then again, that's generally amended by unexporting the struct and exporting its interface instead.

1 comments

> Is Go fundamentally type-unsafe?

Interface{} and nil pointers point to yes.

Neither of those is incompatible with Cardelli's definition of type safety. `interface{}` is not `void *`. Null pointer errors are not untrapped.
Cardelli's definitions are extremely odd; if you take them literally then Python is type safe but Java is not.

In everyday language a downcast, even a checked one, is not a type-safe operation, and so to the extent that Go's limited type system makes it impractical to write programs without downcasts I'd say that Go is fundamentally type-unsafe.

Well, it's a matter of taste, I guess. In my opinion any downcast that reliably distinguishes between valid and invalid values is type-safe. The behaviour of

    defer func() {
        if e := recover(); e != nil {
            fmt.Printf("not a string %v\n", e)
        }
    }()
    v := interface{}(5)
    u := v.(string)
    fmt.Println(u)
is well-defined: it will print "not a string", always.
By that logic there's no such thing as a non-type-safe language, because all programs have behaviour.

Normally one would say a checked downcast like that is not type-safe, because you can't reason locally about the type behaviour of the downcast based on knowing the type of v. You would have to know the value of v, which is a Turing-complete problem in the general case.