UnboundLocalError when manipulating variables yields inconsistent behavior

Question

In Python, the following code works:

a = 1
b = 2

def test():
    print a, b

test()

And the following code works:

a = 1
b = 2
         


        

Answer 1

When you modify a, it becomes a local variable. When you're simply referencing it, it is a global. You haven't defined a in the local scope, so you can't modify it.

If you want to modify a global, you need to call it global in your local scope.

Take a look at the bytecode for the following

import dis

a = 9 # Global

def foo():
    print a # Still global

def bar():
    a += 1 # This "a" is local


dis.dis(foo)

Output:

  2           0 LOAD_GLOBAL              0 (a)
              3 PRINT_ITEM
              4 PRINT_NEWLINE
              5 LOAD_CONST               0 (None)
              8 RETURN_VALUE

For the second function:

dis.dis(bar)

Output:

  2           0 LOAD_FAST                0 (a)
              3 LOAD_CONST               1 (1)
              6 INPLACE_ADD
              7 STORE_FAST               0 (a)
             10 LOAD_CONST               0 (None)
             13 RETURN_VALUE

The first function's bytecode loads the global a (LOAD_GLOBAL) because it is only being referenced. The second function's bytecode (LOAD_FAST) tries to load a local a but one hasn't been defined.

The only reason your second function works is because a is equal to 1. If a was anything but 1, the local assignment to b wouldn't happen and you'd receive the same error.