为什么对象的id会根据python shell中的行而改变 [英] Why the id of an object would change depending on the line in the python shell

问题描述

这个问题只是出于好奇.

This questions is just out of curiosity.

虽然我在阅读 python的对象模型文档，我决定尝试使用类方法的id进行实验，并发现以下行为:

While I was reading the python's object model documentation, I decided to experiment a little with the id of a class method and found this behavior:

 Python 3.2.2 (default, Sep  4 2011, 09:07:29) [MSC v.1500 64 bit (AMD64)] on win32
 Type "copyright", "credits" or "license()" for more information.
 >>> class A():
    def a(self):
        pass


 >>> id(A().a)
 54107080
 >>> id(A().a)
 54108104
 >>> id(A().a)
 54107080
 >>> id(A().a)
 54108104
 >>> 
 >>> id(A().a)
 54108104
 >>> 
 >>> id(A().a)
 54108104
 >>> id(A().a)
 54107080
 >>> 

该方法的ID随行的奇偶校验而改变！

The id of the method changes with the parity of the line!

我实际上想创建几个相同类的实例，看看它们是否具有相同的方法对象，并且我希望它们是完全相同的，或者每次都更改，我没想到的是方法ID与解释器行是否均匀相关！有什么想法吗?

I actually wanted to create a couple of instances of the same class and see if they had the same method object, and I expected that they would be the exact same one, or change every time, what I did not expect was that the method id would be related with the interpreter line being even or not! Any ideas?

注意:我知道文档和解释器的版本不匹配，只是我在Windows上并且只安装了3.2版本

Note: I know that there is a mismatch of version from the docs and the interpreter, it just happens that I'm on windows and I have only 3.2 installed

推荐答案

我将解释id(A().a)行的作用:

A() # creates a new object I call a

然后

A().a # creates a function f bound to a
A.a.__get__(A(), A) # same as above

>>> A.a.__get__(A(), A)
<bound method A.a of <__main__.A object at 0x02D85550>>
>>> A().a
<bound method A.a of <__main__.A object at 0x02D29410>>

此绑定函数始终是另一个，因为它在__self__

This bound function is always another because it has another object in __self__

>>> a = A()
>>> assert a.a.__self__ is a

__self__将作为第一个参数self传递给函数A.a

__self__ will be passed as first argument self to the function A.a

这是它的样子:

Python 3.3.0 (v3.3.0:bd8afb90ebf2, Sep 29 2012, 10:55:48) [MSC v.1600 32 bit (Intel)] on win32
Type "copyright", "credits" or "license()" for more information.
>>> class A:
    def a(self):
        pass


>>> id(A().a)
43476312
>>> id(A().a)
49018760

此处的ID像abab一样重复

Here the id repeats like abab

Python 3.2.2 (default, Sep  4 2011, 09:51:08) [MSC v.1500 32 bit (Intel)] on win32
Type "copyright", "credits" or "license()" for more information.
>>> class A:
    def a(self):
        pas
        s


>>> id(A().a)
50195512
>>> id(A().a)
50195832
>>> id(A().a)
50195512
>>> id(A().a)
50195832

对于linux或不是我的机器，什么都不知道

For linux or what is not my machine and whatsoever I do not know

id(A().a)

将始终给出相同的结果，除非将其存储到变量中. 我不知道为什么，但是我认为这是因为性能优化. 对于堆栈上的对象，您无需在每次调用函数时为该对象分配新的空间.

will always give the same result except if you store this to a variable. I do not know why but I would think that it is because of performance optimization. For objects on the stack you do not need to allocate new space for an object every time you call a function.

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