Python内存分配

我对Python内存分配的工作原理非常感兴趣

我已经写了两段代码：

一,

二,

当我编译第一个代码时，我得到结果[[1,2,3]]和[3,2,1]

然而，当我编译第二个代码时，得到的结果是[1,50,70]

在第一种情况下，我创建了一个对象“a”和一个数组“list”。当我将对象“a”附加到数组中时，数组实际上指向它。然后，当我指定“a”新数组[3,2,1]时，对象[1,2,3]保存在数组中，“a”指向新数组[3,2,1]

在第二种情况下，我还创建了一个对象“a”和一个数组“list”。我在数组中附加了“a”，还有一个指针，指向“a”。然后我调用数组“a”上的一个方法。在调用函数并更改元素的值之后，数组“list”仍然指向对象“a”，而不创建新实例

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有人能解释一下它的工作原理吗？

在Python中创建的每个变量都是对对象的引用。列表是包含对不同对象的多个引用的对象

python中几乎所有的操作都修改这些引用，而不是对象。因此，当您执行

list[1]=50

时，您正在修改列表项包含的第二个引用

我发现一个有用的可视化工具是

第一个例子是

list = [] # you create a reference from `list` to the the new list object you have created
a = [1,2,3] # you create a reference from `a` to the new list you have created which itself contains 3 references to three objects, `1`, `2` and `3`.
list.append(a) # this adds another reference to `list` to the object referenced by `a` which is the object `[1, 2, 3]`
a = [3,2,1] # create a reference from `a` to the new list you have created which itself contains 3 references to three objects, `3`, `2` and `1`.
print(list) # print the object referenced by `list`
print(a) # print the object referenced by `a`

第二个例子

def change(list): # create a reference from `change` to a function object
    list[1] = 50 #change the second reference in the object referenced by `list` to `50`
    list[2] = 70 #change the third reference in the object referenced by `list` to `70`

list = [] # create a reference from `list` to a new list object
a = [1,2,3] # create a reference from `a` to a new list object which itself contains three references to three objects, `1`, `2` and `3`.
list.append(a) # add a reference to list to the object pointed to by `a`
change(a) # call the function referenced by change
print(list) # print the object referenced by `list`
print(a) # print the object referenced by `a`

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此时，内存分配确实不是您应该考虑的问题。先读一读。内存分配是一个实现细节。你能澄清你到底不明白什么吗？如果我理解正确，您已经描述了它的工作原理。这些不是数组，而是

list

对象。但无论如何，请阅读以下内容：Python导师值得几票！很好的工具，真的：-）

list = [] # you create a reference from `list` to the the new list object you have created
a = [1,2,3] # you create a reference from `a` to the new list you have created which itself contains 3 references to three objects, `1`, `2` and `3`.
list.append(a) # this adds another reference to `list` to the object referenced by `a` which is the object `[1, 2, 3]`
a = [3,2,1] # create a reference from `a` to the new list you have created which itself contains 3 references to three objects, `3`, `2` and `1`.
print(list) # print the object referenced by `list`
print(a) # print the object referenced by `a`

def change(list): # create a reference from `change` to a function object
    list[1] = 50 #change the second reference in the object referenced by `list` to `50`
    list[2] = 70 #change the third reference in the object referenced by `list` to `70`

list = [] # create a reference from `list` to a new list object
a = [1,2,3] # create a reference from `a` to a new list object which itself contains three references to three objects, `1`, `2` and `3`.
list.append(a) # add a reference to list to the object pointed to by `a`
change(a) # call the function referenced by change
print(list) # print the object referenced by `list`
print(a) # print the object referenced by `a`