到底为什么 Python 允许更改而不是函数中的全局声明列表？
重新更新
numbers = []
num = 4
def add(n, thisnum=None):
# changing global list without global declaration!
numbers.append(n)
if thisnum:
num = thisnum
print 'num assigned', thisnum
##numbers = ('one', 'two', 'three')
## adding this line makes error:
"""Traceback (most recent call last):
File "J:\test\glob_vals.py", line 13, in <module>
add(i)
File "J:\test\glob_vals.py", line 6, in add
numbers.append(n)
UnboundLocalError: local variable 'numbers' referenced before assignment
"""
for i in (1,2,3,564,234,23):
add(i)
print numbers
add(10, thisnum= 19)
# no error
print num
# let the fun begin
num = [4]
add(10, num)
print num
# prints:
"""[1, 2, 3, 56, 234, 23]
num assigned 19
4
num assigned [4]
[4]
"""
如果我将赋值给具有相同名称的变量，那么该行之前的操作将变为错误，而不是添加的行(我猜是字节码编译器发现了它)。
请您参考如下方法:
您不是在为全局变量赋值，而是在其上调用一个方法来更改其内容。这是允许的。
没有 global 关键字你不能做的是:
def add(n):
#global numbers
numbers = numbers + [n]
结果:
Traceback (most recent call last):
File "C:\Users\Mark\Desktop\stackoverflow\python\test.py", line 8, in
add(i)
File "C:\Users\Mark\Desktop\stackoverflow\python\test.py", line 5, in add
numbers = numbers + [n]
UnboundLocalError: local variable 'numbers' referenced before assignment
The difference is that here I am not mutating the existing list - I'm trying to create a new list and reassign back to the global scope. But this cannot be done without the global keyword.
Regarding your update:
The following line is OK because it creates a new local name num in the scope of the function. This does not affect the value of the variable at the global scope.
num = thisnum
