Python 局部变量与全局变量 [英] Python local vs global variables
问题描述
我了解 Python 中局部变量和全局变量的概念，但我只是有一个问题，为什么错误会在以下代码中以这种方式出现.Python 一行一行地执行代码，所以它直到读取第 5 行才知道 a 是局部变量.Python 在尝试执行第 5 行后是否会返回一行并将其标记为错误?
a=0定义测试():打印#line 4，错误:赋值前引用了局部变量a"a=0 #第5行测试()设置和测试
为了分析您的问题，让我们创建两个单独的测试函数来复制您的问题:
a=0def test1():打印(一)测试1()打印 0.所以，调用这个函数没有问题，但是在下一个函数上:
def test2():print(a) # 错误:赋值前引用了局部变量a"a=0测试2()我们得到一个错误:
回溯(最近一次调用最后一次):文件<stdin>"，第 1 行，在 <module> 中文件<stdin>"，第 2 行，在 test2 中UnboundLocalError:赋值前引用了局部变量a"拆解
我们可以反汇编这两个函数(首先是Python 2):
<预><代码>>>>导入文件>>>dis.dis(test1)2 0 LOAD_GLOBAL 0 (a)3 PRINT_ITEM4 PRINT_NEWLINE5 LOAD_CONST 0(无)8 RETURN_VALUE并且我们看到第一个函数自动加载全局a，而第二个函数:
看到 a 被分配在其中，尝试从本地加载 LOAD_FAST(作为优化问题，因为函数在运行之前被预编译为字节码.)
如果我们在 Python 3 中运行它，我们会看到几乎相同的效果:
<预><代码>>>>测试2()回溯(最近一次调用最后一次):文件<stdin>"，第 1 行，在 <module> 中文件<stdin>"，第 2 行，在 test2 中UnboundLocalError:赋值前引用了局部变量a">>>>>>导入文件>>>dis.dis(test1)2 0 LOAD_GLOBAL 0(打印)3 LOAD_GLOBAL 1 (a)6 CALL_FUNCTION 1(1 个位置，0 个关键字对)9 POP_TOP10 LOAD_CONST 0 (无)13 RETURN_VALUE>>>dis.dis() # 反汇编最后一个堆栈跟踪2 0 LOAD_GLOBAL 0(打印)-->3 LOAD_FAST 0 (a)6 CALL_FUNCTION 1(1 个位置，0 个关键字对)9 POP_TOP3 10 LOAD_CONST 1 (0)13 STORE_FAST 0 (a)16 LOAD_CONST 0 (无)19 RETURN_VALUE我们看到我们的错误再次出现在 LOAD_FAST 上.
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 execute the codes line by line, so it does not know that a is a local variable until it reads line 5. Does Python go back one line and tag it as an error after it tries to execute line 5?
a=0
def test():
print a #line 4, Error : local variable 'a' referenced before assignment
a=0 #line 5
test()
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 "<stdin>", line 1, in <module>
File "<stdin>", 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 "<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
We see our error is on the LOAD_FAST again.
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