It'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.

I 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. )

Only 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.

not my fault python is broken.

> 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.