4 ms·
So you want strong typing, but then are to lazy to properly type your function definitions?
by jsjohnst 1y ago
So you want strong typing, but then are to lazy to properly type your function definitions?
- toolslive 1y agono need to explicitly write the type if you have type inference: > # fun x -> x + 1;; > - : int -> int = <fun> >
- jsjohnst 1y ago1) the code you wrote isn’t Python. 2) inferring the type is int isn’t guaranteed to be correct in this case
- toolslive 1y agoI was merely giving an example that strong typing has nothing to do with having to write the types. (and, obviously, the inferred type (int -> int) is correct. )
- Izkata 1y agoOnly if reveal_type only accepts an int. Just because the default value of i is 0 doesn't mean anything about what could be passed in.
- toolslive 1y agonot my fault python is broken.
- jsjohnst 1y ago> and, obviously, the inferred type (int -> int) is correct. No it’s not. It’s Optional[int] -> int at minimum. There are other completely valid signatures beyond that too.
- int_19h 1y agoIt's guaranteed to be correct if you use different operators for ints and floats, which is what at least some ML dialects (notably, OCaml) do precisely so that types can be inferred from usage. That's the downside of operator overloading - since it relies on types to resolve, they need to be known and can't be inferred.
- hk__2 1y agoI want a typing system with a good inference that doesn’t require me to type each and every variable, just like in any good statically-typed language like OCaml or Typescript. Strong typing and explicit typing are two very different things.