Why this different behavior? Am I right?

Question

I ran the following code

global_var1 = 1
global_var2 = 2

def func():
    print(global_var1)
    print(global_var2)
    global_var2 = 3

func()

It threw error

Traceback (most recent call last):
  File "/home/harish/PycharmProjects/try/scope.py", line 10, in <module>
    func()
  File "/home/harish/PycharmProjects/try/scope.py", line 7, in func
    print(global_var2)
UnboundLocalError: local variable 'global_var2' referenced before assignment

But the following code worked

__author__ = 'harish'
global_var1 = 1
global_var2 = 2

def func():
    print(global_var1)
    print(global_var2)
func()

My expectation was when the funcion func was called. It would look for global_var1 and it was not it local so it looked in global and print it. then similarly it would look for global_var1 and it was not it local so it looked in global and print it. then it would create a variable global_var2 in localscope and assign 3 .

May be it was not done like above because then within the same functions scope the same variable global_var2 had different meaning . till line 2 it was referring to global then there after local .. Is my guess correct ?

score 1 · Accepted Answer · answered Jun 27 '15 at 11:37

global declarations should be put inside the function:

def func():
    global global_var2
    print(global_var1)
    print(global_var2)
    global_var2 = 3

They tell python that inside the function the name global_var2 refers to the global scope instead of local. Putting a global declaration at the top level is useless.

Python doesn't allow for the same name to refer to two different scopes inside a single scope so the assignment global_var2 = 3 imposes that the variable is local. Thus if you try to print it before it raises an error. But if you say to python that the name is actually the global name then python knows that he should take the value from that scope and assign to the global scope.

In particular:

a = 1

def f():
    print(a)
    a = 2

Here the a in print(a) and the a in a = 2 always refer to the same variable. In the above code the inside f is local because no declaration is provided and there is an assignment. So print(a) tries to access a local variable that doesn't have a value.

If you add global:

a = 1
def f():
    global a
    print(a)
    a = 2

Now all the as in f refer to the global a, so print(a) prints 1 because it's the value of the global variable and after executing f the value will become 2.

I understand that function def will be executed even before it being called (for eg this helps during default parameters setting).. will the entire function body be scanned ? how does it work ? how did python know about this local variable global_var2 untill it got executed ? — Harish Kayarohanam, Jun 27 '15 at 11:41
@HarishKayarohanam I don't get what you are saying. In order to create a function object python has to compile all its body. When compiling all its body it looks for all assignments and so discovers the names of all local variables. The fact that the assignment is after its first use does **not** mean that python doesn't already know that there *is* an assignment somewhere in the function. The simple fact that such an assignment exist forces the variable to be local in *all* its uses inside the function. Period. — Bakuriu, Jun 27 '15 at 11:45
@HarishKayarohanam Think about it in this way: every local variable is *implicitly* declared at the beginning of the function. So it's like if `f` were: `def f(): local a; print(a); a = 2`. Simply the `local a` is implicit so you don't write it. If you put a `global a` or `nonlocal a` declaration you override it. — Bakuriu, Jun 27 '15 at 11:47

score 1 · Answer 2 · answered Jun 27 '15 at 11:37

1

Python looks for local variables in functions at compile time. There is only one scope per function. So global_var2 is local, because it is defined as local variable in the last line of the function.

answered Jun 27 '15 at 11:37

Daniel

42,087
4
55
81

I asked @bakaru . but this doubt can be asked to you too ,,,,I understand that function def will be executed even before it being called (for eg this helps during default parameters setting).. will the entire function body be scanned ? how does it work ? how did python know about this local variable global_var2 untill it got executed ? – Harish Kayarohanam Jun 27 '15 at 11:44
1

@HarishKayarohanam: you know, what a compiler is? The code is parsed and analyzed, before execution. This way, the executer can reserve the needed space on the stack for local variables. – Daniel Jun 27 '15 at 11:55

Why this different behavior? Am I right?

2 Answers2