[英]Python scoping mutable vs immutable
First I'm going to start like everyone else. 首先,我将像其他所有人一样开始。 I'm new to python.
我是python的新手。 My teacher gave me the problem:
我的老师给了我一个问题:
def f(a, b, c):
a = 1
c = b
c[0] = 2
a = 10
b = [11, 12, 13]
c = [13, 14, 15]
f(a, b, c)
print a, b, c
It prints: 它打印:
10 [2, 12, 13] [13, 14, 15]
I understand that a stays at 10 because integers are immutable, but I don't understand why b changes and c doesn't. 我知道a保持为10是因为整数是不可变的,但是我不明白为什么b改变而c不变。
c = b
c[0] = 2
Since you're setting c
to point to b
, You could just as easily do this: 由于将
c
设置为指向b
,因此您可以轻松地做到这一点:
def f(a, b, unused): # notice c isn't in the parameter list
a = 1
c = b # c is declared here
c[0] = 2 # c points to b, so c[0] is b[0]
Now it's obvious that c
is always the same as b
, so why not just remove it: 现在很明显,
c
始终与b
相同,所以为什么不删除它:
def f(a, b, unused):
a = 1
b[0] = 2
And now it's clear that you're changing the first element of b
and not doing anything to c
, and remember, this is functionally identical to the original. 现在很明显,您正在更改
b
的第一个元素,而不对c
进行任何操作,请记住,这在功能上与原始功能相同。
The key is to understand the variables as pointers under the hood: 关键是将变量理解为幕后的指针:
def f(a, b, c):
a = 1 # a is a single scalar value, so no pointing involved
c = b # point the local "c" pointer to point to "b"
c[0] = 2 # change the 2nd value in the list pointed to by "c" to 2
When you call f(a,b,c), only b actually gets changed. 当您调用f(a,b,c)时,实际上只有b被更改。 The "c" variable inside the function implementation is different from the "c" implementation outside the function.
函数实现内的“ c”变量不同于函数外的“ c”实现。
a
does not retain the value of 10 because it is immutable. a
不会保留10的值,因为它是不可变的。 It retains the value of 10 because when you call a = 1
in the local scope of f()
you create a new variable. 它保留10的值,因为在
f()
的局部范围内调用a = 1
,会创建一个新变量。
When you call c = b
inside f()
, the local c
becomes a local reference to the mutable object represented by b
. 当在
f()
调用c = b
,局部c
成为对b
表示的可变对象的局部引用。 When you re-assign the values in that mutable object the change is reflected in the original object. 当您在该可变对象中重新分配值时,更改将反映在原始对象中。
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