为了分析您的问题，让我们创建两个独立的测试函数来复制您的问题：
a=0
def test1():
print(a)
test1()
def test2():
print(a) # Error : local variable 'a' referenced before assignment
a=0
test2()
Traceback (most recent call last):
File "<stdin>", line 1, in <module>
File "<stdin>", line 2, in test2
UnboundLocalError: local variable 'a' referenced before assignment
>>> 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
>>> 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
看到a分配给它的内容，尝试从本地进行LOAD_FAST（作为优化的问题，因为函数在运行前已预先编译为字节码）。
如果我们在 运行此命令， 看到几乎相同的效果：
>>> test2()
Traceback (most recent call last):
File "<stdin>", line 1, in <module>
File "<stdin>", 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
我们看到错误再次出现在LOAD_FAST上。
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