考慮下面的代碼示例 import abc
class ABCtest(abc.ABC):
@abc.abstractmethod
def foo(self):
raise RuntimeError("Abstract method was called, this should be impossible")
class ABCtest_B(ABCtest):
可以說我有， class A(object):
def __init__(self, var):
print("The parent class is A")
def methodA(self):
print("This method should only be accessed by a child of class A")
class B(ob
爲現有第三方（非Django）Python類添加Django模型的好方法是什麼？我試圖多重繼承，像這樣： class ThirdPartyClass(object):
pass
class ModelForThirdPartyClass(models.Model, ThirdPartyClass):
uid = models.ForeignKey(settings.AUTH
我遇到了一個問題，我想使用collections.abc.MutableSequence中的mixin方法，但我也必須從其他方面繼承。 class Thing(urwid.Pile, collections.abc.MutableSequence):
...
我結束了 TypeError: metaclass conflict: the metaclass of a derived
這編譯並基於Visual C++ 2015年更新3 RC運行正常發現： class A
{
template <class T> void f() {}
};
class B : A {};
class C : A {};
class D : B, C {};
int main()
{
D d;
d.f<int>();
}
有兩個問題與此代碼