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Thank you for writing the post. Newbie question about “nonlocal” from your example: def outer_function(): x = 11 def inner_function():

• ekimekim 5y ago
  Python assumes that all assignments assign to the current scope. So by default when you reach "x = 22", it would create a new variable called "x" in the inner_function() scope which overrides the variable "x" in the outer_function() scope. So when you print "Inner x" you would only be printing the inner_function() version of x, not the outer_function() version, which would remain at 11.
  • faho 5y ago
    This is a consequence of python not having explicit variable definition. Here it'll decide to define a new x instead of seeing the old one. And that's also usually what you want because otherwise a function would start altering variables in the enclosing scope if they happen to exist! E.g. foo = 42 def myfunc(bar): foo = bar + 1 print(foo) myfunc(6) print(foo) # would print "7" if the "foo =" above took the nonlocal foo automatically! So the trade-off is to require "nonlocal" if you ever need a variable from the enclosing scope.