如果在类的构造函数中不满足某些条件,是否可以不创建对象?
E.g:
class ABC:
def __init__(self, a):
if a > 5:
self.a = a
else:
return None
a = ABC(3)
print(a)
这应该打印None
(因为它不应该创建一个Object但在这种情况下返回None
)但是当前打印了Object ...
答案 0 :(得分:3)
您可以使用classmethod
作为替代构造函数并返回您想要的内容:
class ABC:
def __init__(self, a):
self.a = a
@classmethod
def with_validation(cls, a):
if a > 5:
return cls(a)
return None
a = ABC.with_validation(10)
a
<__main__.ABC at 0x10ceec288>
a = ABC.with_validation(4)
a
type(a)
NoneType
答案 1 :(得分:2)
此代码似乎表明__init__()
中引发的异常会为您提供所需的效果:
class Obj:
def __init__(self):
raise Exception("invalid condition")
class E:
def __call__(self):
raise Exception("raise")
def create(aType):
return aType()
def catchEx():
e = E()
funcs=[Obj, int, e]
for func in funcs:
try:
func()
print('No exception:', func)
except Exception as e:
print(e)
catchEx()
输出:
invalid condition
No exception: <class 'int'>
raise
答案 2 :(得分:2)
我认为这表明了原则。请注意,返回 IE.Visible = true;
System.Threading.Thread.Sleep(500);
并不返回新对象,因为None
是Python中的单例,但当然它仍然是一个对象。另请注意,None
不会被调用,因为__init__
不是None
类对象。
A
输出:
class A():
def __new__(cls, condition):
if condition:
obj = super().__new__(cls)
return obj
a = A(True)
print(a)
a1 = A(False)
print(a1)
答案 3 :(得分:1)
在@progmatico的帮助下,我尝试了一点点尝试和错误:
class ABC:
def __new__(cls, *args, **kwargs):
if len(args) > 0:
arg = args[0]
else:
arg = kwargs['a']
if arg <= 5:
return None
return object.__new__(cls)
def __init__(self, a):
self.a = a
def __str__(self):
return str(self.a)
a = ABC(a=3)
print(a)
b = ABC(a=7)
print(b)