4 ms·

Doesn't all of Python miss that, having (close to) no compile time?

• amelius 1y ago
  Python does some checks before it runs code. E.g.: print("hello") def f(): nonlocal foo gives: SyntaxError: no binding for nonlocal 'foo' found before printing hello, and note that f() wasn't even called.
  • nomel 1y ago
    I think it's just giving an error because a valid AST can't be made, which means valid bytecode can't be made. "<word> <word>" is only valid syntax if one is a reserved word. `nonlocal(foo)` is just fine, of course.
    • pansa2 1y ago
      > "<word> <word>" is only valid syntax if one is a reserved word. `nonlocal` is a keyword
    • zahlman 1y ago
      No, it gives an error because `nonlocal foo` requests that the name `foo` be looked up in a closure, but `f` doesn't have such a closure (the `foo` defined outside the function is global instead). `nonlocal` is the same sort of keyword as `global` but for enclosing functions instead of the global namespace; see also https://stackoverflow.com/questions/1261875 https://stackoverflow.com/questions/1261875 .
      • nomel 1y ago
        Here's the statement checking code, which I believe is pre-AST [1]. I would have to dig more to see if that check is there to prevent invalid AST or to just "help the user" (would depend on how they reference the original variable I suppose). But wow, that's the first time I've seen "nonlocal". In the ~100 packages I have installed, I see 0 usages! [1] https://github.com/python/cpython/blob/a6a3dbb7db0516a72c5ef149223d904f4e6a31ae/Python/ast.c#L714 https://github.com/python/cpython/blob/a6a3dbb7db0516a72c5ef...
        • zahlman 1y ago
          Well, yes, not a lot of people write closures except perhaps when they implement decorators. So there's ordinarily no non-local scope to worry about. People tend to write classes instead, because that's what's familiar.