Why parameters are passed outside the parenthesis () in globals(), locals() and vars() in Python?

Question

I'm very new in Py but I know the syntax function(param1,.. paramN) and I can't understand this:

globals()['z'] = 8

Could not be more 'standar' ?

globals('z') = 8  
# or..
globals(param = 'z') = 8

EDIT: thanks to all for yours answers, I just have a couple of hours playing with Python

In retrospective it's "obvious" globals() returns a instance and I acceding to the index when I write globals('z') but the Dict is returned BY REFERENCE because I can change it just doing:

globals()['z'] = 8

or... just another member suggest:

x = globals(); 
x['z'] = 8

All the "magic" was just a pointer to the index :)

Answer 1

globals is actually a function that takes no parameters and returns a dict, which maps the names of identifiers to their values.

Trying typing just globals() in a python interpreter and examine the result.

>>> z = 8
>>> globals()
{'__builtins__': <module '__builtin__' (built-in)>, '__name__': '__main__', 'z': 8, '__doc__': None, '__package__': None}

Notice the 'z': 8 part.

But if you run this code you see actually it takes parameters and change that dict / list / whatever: z = 7 globals()['z'] = 8 print("z=", z) # z= 8

Yup, that's true. The dict returned by globals() is where python goes to get the values of global variables, in some sense. Modifying the dict modifies global values directly.

---

As an FYI, you generally shouldn't be doing stuff like globals()['z'] = 8. Just do z = 8 if you're in global scope already (ie, not inside of a function/class), and inside a function, do:

def foo():
    global z
    z = 8