Python——按值与按引用 [英] Python -- by value vs by reference

问题描述

直到最近，我还以为 Python 是按值传递参数的.例如，

def method1(n):n = 5返回 nn = 1m = 方法 1(n)>>>n == 5>>>错误的

但是如果我将一个列表传递给一个方法，比如

def method2(list):删除列表[0]返回列表list1 = [0,1,2,3,4,5]列表 2 = 方法 2(列表 1)>>>清单 1>>>[1,2,3,4,5]

它改变了列表.我又做了一个测试:

l1 = ['a','b','c','d']x = 方法 1(l1)>>>l1>>>['A B C D']

这里的列表没有改变.我的问题是

<块引用>

为什么会发生这些不同的情况?

解决方案

Python 总是 通过赋值传递 ，即使在第一种情况下，实际上传递的是对象引用.但是当您在函数内部进行赋值时，您会导致名称引用一个新对象.这不会改变用于将值传递给函数的名称/变量(在下面的示例中更好地解释).

示例 -

<预><代码>>>>定义函数(n):... n = [1,2,3,4]...返回 n...>>>l = [1,2,3]>>>功能(l)[1, 2, 3, 4]>>>升[1, 2, 3]

<小时>

在您的情况下，当您执行 del list[0] 时，您实际上是在适当地改变 list (从中删除第一个元素)，因此它反映了在调用函数的地方也是如此.

Until recently, I thought Python passed parameters by value. For example,

def method1(n):
   n = 5
   return n

n = 1
m = method1(n)
>>> n == 5
>>> False

But if I pass in a list to a method, like

def method2(list):
   del list[0]
   return list

list1 = [0,1,2,3,4,5]
list2 = method2(list1)
>>> list1
>>> [1,2,3,4,5]

it mutates the list. I did another test:

l1 = ['a','b','c','d']
x = method1(l1)
>>> l1
>>> ['a','b','c','d']

Here the list did not change. My question is

Why do these different cases happen?

解决方案

Python always passes by assignment , even in the first case, actually the object reference is passed. But when you do an assignment inside the function, you are causing the name to refer a new object. That would not mutate the name/variable that was used to pass the value to the function (Better explained in the below example).

Example -

>>> def func(n):
...     n = [1,2,3,4]
...     return n
...
>>> l = [1,2,3]
>>> func(l)
[1, 2, 3, 4]
>>> l
[1, 2, 3]

---

In your case, when you do del list[0] you are actually mutating the list in place (deleting the first element from it) , and hence it reflects in the place where the function was called from as well.

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