新型类中的方法解析顺序（MRO）？

Question 1

In the book Python in a Nutshell (2nd Edition) there is an example which uses
old style classes to demonstrate how methods are resolved in classic resolution order and
how is it different with the new order.

I tried the same example by rewriting the example in new style but the result is no different than what was obtained with old style classes. The python version I am using to run the example is 2.5.2. Below is the example:

class Base1(object):  
    def amethod(self): print "Base1"  

class Base2(Base1):  
    pass

class Base3(object):  
    def amethod(self): print "Base3"

class Derived(Base2,Base3):  
    pass

instance = Derived()  
instance.amethod()  
print Derived.__mro__

The call instance.amethod() prints Base1, but as per my understanding of the MRO with new style of classes the output should have been Base3. The call Derived.__mro__ prints:

(<class '__main__.Derived'>, <class '__main__.Base2'>, <class '__main__.Base1'>, <class '__main__.Base3'>, <type 'object'>)

I am not sure if my understanding of MRO with new style classes is incorrect or that I am doing a silly mistake which I am not able to detect. Please help me in better understanding of MRO.

Question 2

The crucial difference between resolution order for legacy vs new-style classes comes when the same ancestor class occurs more than once in the “naive”, depth-first approach — e.g., consider a “diamond inheritance” case:

>>> class A: x = 'a'
... 
>>> class B(A): pass
... 
>>> class C(A): x = 'c'
... 
>>> class D(B, C): pass
... 
>>> D.x
'a'

here, legacy-style, the resolution order is D – B – A – C – A : so when looking up D.x, A is the first base in resolution order to solve it, thereby hiding the definition in C. While:

>>> class A(object): x = 'a'
... 
>>> class B(A): pass
... 
>>> class C(A): x = 'c'
... 
>>> class D(B, C): pass
... 
>>> D.x
'c'
>>>

here, new-style, the order is:

>>> D.__mro__
(<class '__main__.D'>, <class '__main__.B'>, <class '__main__.C'>, 
    <class '__main__.A'>, <type 'object'>)

with A forced to come in resolution order only once and after all of its subclasses, so that overrides (i.e., C’s override of member x) actually work sensibly.

It’s one of the reasons that old-style classes should be avoided: multiple inheritance with “diamond-like” patterns just doesn’t work sensibly with them, while it does with new-style.

Question 3

Python’s method resolution order is actually more complex than just understanding the diamond pattern. To really understand it, take a look at C3 linearization. I’ve found it really helps to use print statements when extending methods to track the order. For example, what do you think the output of this pattern would be? (Note: the ‘X’ is suppose to be two crossing edges, not a node and ^ signifies methods that call super())

class G():
    def m(self):
        print("G")

class F(G):
    def m(self):
        print("F")
        super().m()

class E(G):
    def m(self):
        print("E")
        super().m()

class D(G):
    def m(self):
        print("D")
        super().m()

class C(E):
    def m(self):
        print("C")
        super().m()

class B(D, E, F):
    def m(self):
        print("B")
        super().m()

class A(B, C):
    def m(self):
        print("A")
        super().m()


#      A^
#     / \
#    B^  C^
#   /| X
# D^ E^ F^
#  \ | /
#    G

Did you get A B D C E F G?

x = A()
x.m()

After a lot of trial an error, I came up with an informal graph theory interpretation of C3 linearization as follows: (Someone please let me know if this is wrong.)

Consider this example:

class I(G):
    def m(self):
        print("I")
        super().m()

class H():
    def m(self):
        print("H")

class G(H):
    def m(self):
        print("G")
        super().m()

class F(H):
    def m(self):
        print("F")
        super().m()

class E(H):
    def m(self):
        print("E")
        super().m()

class D(F):
    def m(self):
        print("D")
        super().m()

class C(E, F, G):
    def m(self):
        print("C")
        super().m()

class B():
    def m(self):
        print("B")
        super().m()

class A(B, C, D):
    def m(self):
        print("A")
        super().m()

# Algorithm:

# 1. Build an inheritance graph such that the children point at the parents (you'll have to imagine the arrows are there) and
#    keeping the correct left to right order. (I've marked methods that call super with ^)

#          A^
#       /  |  \
#     /    |    \
#   B^     C^    D^  I^
#        / | \  /   /
#       /  |  X    /   
#      /   |/  \  /     
#    E^    F^   G^
#     \    |    /
#       \  |  / 
#          H
# (In this example, A is a child of B, so imagine an edge going FROM A TO B)

# 2. Remove all classes that aren't eventually inherited by A

#          A^
#       /  |  \
#     /    |    \
#   B^     C^    D^
#        / | \  /  
#       /  |  X    
#      /   |/  \ 
#    E^    F^   G^
#     \    |    /
#       \  |  / 
#          H

# 3. For each level of the graph from bottom to top
#       For each node in the level from right to left
#           Remove all of the edges coming into the node except for the right-most one
#           Remove all of the edges going out of the node except for the left-most one

# Level {H}
#
#          A^
#       /  |  \
#     /    |    \
#   B^     C^    D^
#        / | \  /  
#       /  |  X    
#      /   |/  \ 
#    E^    F^   G^
#               |
#               |
#               H

# Level {G F E}
#
#         A^
#       / |  \
#     /   |    \
#   B^    C^   D^
#         | \ /  
#         |  X    
#         | | \
#         E^F^ G^
#              |
#              |
#              H

# Level {D C B}
#
#      A^
#     /| \
#    / |  \
#   B^ C^ D^
#      |  |  
#      |  |    
#      |  |  
#      E^ F^ G^
#            |
#            |
#            H

# Level {A}
#
#   A^
#   |
#   |
#   B^  C^  D^
#       |   |
#       |   |
#       |   |
#       E^  F^  G^
#               |
#               |
#               H

# The resolution order can now be determined by reading from top to bottom, left to right.  A B C E D F G H

x = A()
x.m()

Question 4

The result you get is correct. Try changing base class of Base3 to Base1 and compare with the same hierarchy for classic classes:

class Base1(object):
    def amethod(self): print "Base1"

class Base2(Base1):
    pass

class Base3(Base1):
    def amethod(self): print "Base3"

class Derived(Base2,Base3):
    pass

instance = Derived()
instance.amethod()


class Base1:
    def amethod(self): print "Base1"

class Base2(Base1):
    pass

class Base3(Base1):
    def amethod(self): print "Base3"

class Derived(Base2,Base3):
    pass

instance = Derived()
instance.amethod()

Now it outputs:

Base3
Base1

Read this explanation for more information.

Question 5

You’re seeing that behavior because method resolution is depth-first, not breadth-first. Dervied’s inheritance looks like

         Base2 -> Base1
        /
Derived - Base3

So instance.amethod()

Checks Base2, doesn’t find amethod.
Sees that Base2 has inherited from Base1, and checks Base1. Base1 has a amethod, so it gets called.

This is reflected in Derived.__mro__. Simply iterate over Derived.__mro__ and stop when you find the method being looked for.

新型类中的方法解析顺序（MRO）？

问题：新型类中的方法解析顺序（MRO）？

回答 0

回答 1

回答 2

回答 3

排行榜展示

Python 情人节超强技能导出微信聊天记录生成词云

你不得不知道的python超级文献批量搜索下载工具

Python 流程图 — 一键转化代码为流程图

7行代码 Python热力图可视化分析缺失数据处理

Python 优化—算出每条语句执行时间

你的10W块放哪里能赚最多钱？

文章展示

如何删除numpy.ndarray中包含非数字值的所有行

Python多重处理模块的.join（）方法到底在做什么？

在Python中向信号添加噪声

如何在广度优先搜索中跟踪路径？

我想exceptions处理“列表索引超出范围”。

生成文件以使用Django下载

新型类中的方法解析顺序（MRO）？

问题：新型类中的方法解析顺序（MRO）？

回答 0

回答 1

回答 2

回答 3

相关文章

排行榜展示

文章展示