当我尝试这段代码:
a, b, c = (1, 2, 3)
def test():
print(a)
print(b)
print(c)
c += 1
test()
我从打印(c)行得到一个错误，它说:
UnboundLocalError: local variable 'c' referenced before assignment
在Python的新版本中，或者
UnboundLocalError: 'c' not assigned
在一些老版本中。
如果注释掉c += 1，两次打印都成功。
我不明白:如果c不行，为什么打印a和b可以?c += 1是如何导致print(c)失败的，即使它出现在代码的后面?
赋值c += 1似乎创建了一个局部变量c，它优先于全局变量c。但是一个变量如何在它存在之前“窃取”作用域呢?为什么c是局部的?
请参见在函数中使用全局变量，了解如何从函数中重新分配全局变量的问题，以及是否可以在python中修改位于外部(封闭)但不是全局范围的变量?用于从封闭函数(闭包)重新赋值。
参见为什么不需要'global'关键字来访问全局变量?对于OP预期错误但没有得到错误的情况，从简单地访问一个没有global关键字的全局变量。
参见如何在Python中“解除绑定”名称?什么代码可以导致“UnboundLocalError”?对于OP期望变量是本地的，但在每种情况下都有阻止赋值的逻辑错误的情况。
看一看这个分解程序可能会清楚发生了什么:
>>> def f():
... print a
... print b
... a = 1
>>> import dis
>>> dis.dis(f)
2 0 LOAD_FAST 0 (a)
3 PRINT_ITEM
4 PRINT_NEWLINE
3 5 LOAD_GLOBAL 0 (b)
8 PRINT_ITEM
9 PRINT_NEWLINE
4 10 LOAD_CONST 1 (1)
13 STORE_FAST 0 (a)
16 LOAD_CONST 0 (None)
19 RETURN_VALUE
As you can see, the bytecode for accessing a is LOAD_FAST, and for b, LOAD_GLOBAL. This is because the compiler has identified that a is assigned to within the function, and classified it as a local variable. The access mechanism for locals is fundamentally different for globals - they are statically assigned an offset in the frame's variables table, meaning lookup is a quick index, rather than the more expensive dict lookup as for globals. Because of this, Python is reading the print a line as "get the value of local variable 'a' held in slot 0, and print it", and when it detects that this variable is still uninitialised, raises an exception.
这里有两个链接可能会有所帮助
1: docs.python.org/3.1/faq/programming.html?highlight=nonlocal#why-am-i-getting-an-unboundlocalerror-when-the-variable-has- a-value
2: docs.python.org/3.1/faq/programming.html?highlight=nonlocal#how-do-i-write-a-function-with-output-parameters-call-by-ref erence
link 1描述了错误UnboundLocalError。链接二可以帮助您重新编写测试函数。基于连杆二，原问题可以改写为:
>>> a, b, c = (1, 2, 3)
>>> print (a, b, c)
(1, 2, 3)
>>> def test (a, b, c):
... print (a)
... print (b)
... print (c)
... c += 1
... return a, b, c
...
>>> a, b, c = test (a, b, c)
1
2
3
>>> print (a, b ,c)
(1, 2, 4)
