Python local vs global variables

Question

I understand the concept of local and global variables in Python, but I just have a question about why the error comes out the way it is in the following code. Python execut

Answer 1

Setup and Testing

To analyze your question, let's create two separate test functions that replicate your issue:

a=0

def test1():
    print(a)

test1()

prints 0. So, calling this function is not problematic, but on the next function:

def test2():
    print(a)  # Error : local variable 'a' referenced before assignment
    a=0  

test2()

We get an error:

Traceback (most recent call last):
  File "", line 1, in 
  File "", line 2, in test2
UnboundLocalError: local variable 'a' referenced before assignment

Disassembly

We can disassemble the two functions (first Python 2):

>>> import dis
>>> dis.dis(test1)
  2           0 LOAD_GLOBAL              0 (a)
              3 PRINT_ITEM          
              4 PRINT_NEWLINE       
              5 LOAD_CONST               0 (None)
              8 RETURN_VALUE        

And we see that the first function automatically loads the global a, while the second function:

>>> dis.dis(test2)
  2           0 LOAD_FAST                0 (a)
              3 PRINT_ITEM          
              4 PRINT_NEWLINE       

  3           5 LOAD_CONST               1 (0)
              8 STORE_FAST               0 (a)
             11 LOAD_CONST               0 (None)
             14 RETURN_VALUE      

seeing that a is assigned inside it, attempts to LOAD_FAST from the locals (as a matter of optimization, as functions are pre-compiled into byte-code before running.)

If we run this in Python 3, we see nearly the same effect:

>>> test2()
Traceback (most recent call last):
  File "", line 1, in 
  File "", line 2, in test2
UnboundLocalError: local variable 'a' referenced before assignment
>>> 
>>> import dis
>>> dis.dis(test1)
  2           0 LOAD_GLOBAL              0 (print) 
              3 LOAD_GLOBAL              1 (a) 
              6 CALL_FUNCTION            1 (1 positional, 0 keyword pair) 
              9 POP_TOP              
             10 LOAD_CONST               0 (None) 
             13 RETURN_VALUE     

>>> dis.dis() # disassembles the last stack trace
  2           0 LOAD_GLOBAL              0 (print) 
    -->       3 LOAD_FAST                0 (a) 
              6 CALL_FUNCTION            1 (1 positional, 0 keyword pair) 
              9 POP_TOP              

  3          10 LOAD_CONST               1 (0) 
             13 STORE_FAST               0 (a) 
             16 LOAD_CONST               0 (None) 
             19 RETURN_VALUE        

We see our error is on the LOAD_FAST again.