Python variable scope and UnboundLocalError

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I don't understand the error in this example. Shouldn't print(b) print 6, by the Python LEGB rule? When the interpreter scans f(a) function in line 2, does it memorize that b = 33 and puts in the memory heap or just remembers the information that b is a local variable?

b = 6
def f(a):
    print(a)
    print(b)
    b = 33
f(20)
# UnboundLocalError: local variable 'b' referenced before assignment
global-variables python-3.x function python
1 Answers

When Python scans your functions it sees an assignment to b. This assignment creates specific bytecode (LOAD FAST) that instructs Python to load the name b as a local variable.

You can see this if you disassemble the bytecode (using the dis module) created for the function f:

from dis import dis
dis(f)
  2           0 LOAD_GLOBAL              0 (print)
              2 LOAD_FAST                0 (a)
              4 CALL_FUNCTION            1
              6 POP_TOP

  3           8 LOAD_GLOBAL              0 (print)
             10 LOAD_FAST                1 (b)
             12 CALL_FUNCTION            1
             14 POP_TOP

the relevant instruction here is:

LOAD_FAST                1 (b)

which loads the name b as a local variable.

When print(b) is encountered, it tries to load the name b but b hasn't been defined yet (the assignment has not been performed): you referenced it (print(b)) before an assignment to it has been made.

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