我定义了以下类:
class A(object):
def __init__(self, a, b):
self.a = a
self.b = b
self.c = ""
def mA1(self):
print "Method 1 in A"
def mA2(self):
print "Method 2 in A"
def mA3(self):
print "Method 3 in A"
class B(object):
def __init__(self, a, b):
self.a = a
self.b = b
self.x = 0
def mB1(self):
print "Method 1 in B"
class C(object):
def __init__(self, a, b):
self.a = a
self.b = b
self.c = ""
self.y = 1
def mC1(self):
print "Method 1 in C"
def mC2(self):
print "Method 2 in C"
可以看出,它们在构造函数中采用相同的输入参数。现在,我想创建一个类D
,该类继承自A
,B
和C
的全部,以便可以将参数直接传递给D
'的构造函数如下:
clsT = D(1, 2)
因此,如here所述,我尝试了以下修订的定义:
class A(object):
def __init__(self, a, b):
self.a = a
self.b = b
self.c = ""
def mA1(self):
print "Method 1 in A"
def mA2(self):
print "Method 2 in A"
def mA3(self):
print "Method 3 in A"
class B(A):
def __init__(self, **kw):
super(B, self).__init__(**kw)
self.x = 0
def mB1(self):
print "Method 1 in B"
class C(A):
def __init__(self, **kw):
super(C, self).__init__(**kw)
self.c = ""
self.y = 1
def mC1(self):
print "Method 1 in C"
def mC2(self):
print "Method 2 in C"
class D(A, B, C):
def __init__(self, a, b):
super(D, self).__init__(a=a, b=b)
ci = D(1, 2)
print "a = ", ci.a
print "b = ", ci.b
print "c = ", ci.c
print "x = ", ci.x
print "y = ", ci.y
以上内容似乎无效,并给我以下错误:
class D(A, B, C):
TypeError: Error when calling the metaclass bases
Cannot create a consistent method resolution
order (MRO) for bases B, C, A
可能出了什么问题?确实需要B
和C
必须包含super()
并在中级基类上创建线性继承吗?除了我的解释,还有其他选择吗?如果没有,我该如何调整才能使它正常工作?
答案 0 :(得分:2)
我如下更改了D
的类定义,并且它起作用了
class D(B, C, A):
def __init__(self, a, b):
super(D, self).__init__(a=a, b=b)
所以整个类看起来像(我删除了使代码简短的方法)
class A(object):
def __init__(self, a, b):
self.a = a
self.b = b
self.c = ""
class B(A):
def __init__(self, **kw):
super(B, self).__init__(**kw)
self.x = 0
class C(A):
def __init__(self, **kw):
super(C, self).__init__(**kw)
self.c = ""
self.y = 1
class D(B, C, A):
def __init__(self, a, b):
super(D, self).__init__(a=a, b=b)
c = D(1, 2)
print(c.a, c.b, c.x, c.c, c.y)
输出将为
1 2 0 1
这是由于MRO算法中的一条规则所致(此处的更多详细信息,answer,但要旨是
一个类总是出现在其祖先之前(“单调性”)
因此B和C必须出现在A之前,因为A是B和C的祖先
或者换句话说:
D从A,B和C继承。由于B和C已经从A继承,因此python现在无法确定首先要查找哪种类的方法。如果使用旧的定义顺序,则为A或B和C。 D(A, B, C)