There’s one thing wrong with Python

Finally! I found something that is wrong with Python! With Python2, that is – Python3 saves itself from being wrong by being stupid.

And it’s really wrong, it’s of “absolutely broken” wrongness. On a scale from 0 (right) to 12 (wrong) this would score a straight twenty-seven.

It’s about scope in nested functions.

Normally, scope in python is pretty straight-forward. Okay, to access the global scope from a non-global scope, one has to use the global keyword:

foo = 0
def bar():
    global foo # no access to foo without this
    print foo

That’s stupid but it doesn’t really qualify as being wrong. Now enter nested functions:

def foo():
    state = 1
    def bar():
        pass # "pass" is a no-op
    bar()
    return state
print foo()

That’s the (artifical) example I’ll be using. It doesn’t do something useful and the nested function is not needed here, but that’s not the point. I “discovered” the behaviour I’m going to describe while writing a lexer. Which did do something useful. Or at least it would have – if python had let it.

So, the program above is perfectly valid python and prints 1, as you would expect.

This, on the other hand, is not valid python:

def foo():
    state = 1
    def bar():
        state += 1
    bar()
    return state
print foo() # should print 2, but doesn't

It gives an UnboundLocalError: “local variable ‘state’ referenced before assignment”.

The unhelpful error message notwithstanding, that seems to be the same case as with global variables. Seems like we always have to be explicit to access higher-than-local scope.

Well, yes and no.

Yes, that might have been the thought behind it.

No, global doesn’t work in this case – and there is no complement for use in functions. I’m told that python 3 has nonlocal which solves this issue – but again: I think that requiring that explicitness is just stupid.

But wait, there’s more! We were wrong about the need to be explicit: in fact, bar() does have access to state – by default! How come?

Well, local functions can access all variables local to their parent function. They just can’t rebind them.

To see how utterly, utterly wrong this is, take a look at the following code. Which does exactly what the last snipped intended to do, but is absolutely legal python. Because this time, we cheat our ways around that idiotic “no rebind” policy:

def foo():
    state = [1]
    def bar():
        state[0] += 1
    bar()
    return state[0]
print foo() # prints 2

Really, are you kidding me or what? How can “access, but not rebind” ever have seemed to be a good idea? I sincerly hope that this behaviour is only due to some implementation detail and fell out for free.

Because I just cannot imagine anyone being stupid enough to implement this deliberately.

Ironically, this is also legal:

def foo():
    global state
    state = 1
    def bar():
        global state
        state += 1
    bar()
    return state
print foo() # prints 2

Great, your scope-access control bullshit allows introducing global variables that nobody knows about because they are not declared anywhere in the global scope – but it doesn’t allow nested functions to access perfectly encapsulated local variables of their parents. Gnah!

But let’s explore this some more: solely accessing state in bar() is allowed – which, for instance, makes the following legal python:

def foo():
    state = 1
    def bar():
        print state # prints 1
    bar()
    return state
print foo() # prints 1

That was expected. What was unexpected: this is also legal:

def foo():
    state = 1
    def bar():
        state = 2
    bar()
    return state
print foo() # prints 1 [sic!]

That’s because in this case the state that bar() sets to 2 is a different state than the one in foo(). No Error, everything is fine. How… why… I don’t even… ARG!

But I’m not mad. There’s one thing horribly wrong with Python.

That’s pretty good for a programming language, I’d say.

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