2 ms·

...it wouldn't be as big an issue if Python's closures and scoping weren't problematic as well.

• eru 17y ago
  As far as I know Python has true lexical scoping for functions, if not for blocks. (They used to have very limited scoping ages ago.) What do you mean?
  • silentbicycle 17y ago
    In Python 2.6: def acc(x): foo = x def closures_are_broken(): foo += 1 return foo return closures_are_broken def acc2(x): foo = [x] def closure_with_workaround(): foo[0] += 1 return foo[0] return closure_with_workaround The former is an error ("UnboundLocalError: local variable 'foo' referenced before assignment"), the latter is a workaround by manually boxing the variable in a list. I'm not sure why that works, but it's rather ugly. The same, in Lua: function acc(x) local foo = x return function() foo = foo + 1 return foo end end
    • eru 17y ago
      Oh, yes. You are mutating variables. You shouldn't do that anyway in a functional setting. Python's syntax forces the language to guess whether you want to make a new variable or use the variable of the same name from the outer scope. The heuristic they have, says that if you assign something to the variable, you get a new one by default. You can declare global x before to assign to the variable in the outerscope.