Question

>>> import sys
>>> print(sys.version)
2.4.4
>>> b = 11
>>> def foo2():
...    a = b
...    print a, b
...
>>> foo2()
11 11
>>> def foo3():
...    a = b 
...    b = 12
...    print a, b
...
>>> foo3()
Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
  File "<stdin>", line 2, in foo3
UnboundLocalError: local variable 'b' referenced before assignment
>>> def foo4():
...    global b
...    a = b
...    b = 12
...    print a, b
...
>>> foo4()
11 12

Question> In foo3, why you can access global variable without declaring it but you still cannot modify it.

Answer 1

Without a global declaration, the Python compiler scans the whole code of each function to see which variables are assigned to within the function code. In foo3(), you assign to both a and b so therefore they are both treated as local variables within the function.

When the method code executes, at the point where you do a = b, b does not have a value yet (because you have not assigned anything to it). Therefore, you get an UnboundLocalError.

This is done so that the use of a variable within a function always refers to the same location, even if nothing has been assigned to it yet.