在python中，实现构造函数可能失败的类的正确方法是什么?

For example, in the constructor I need to grab some resources which could fail, in that case the class shouldn't be instantiated.

例如，在构造函数中，我需要获取一些可能失败的资源，在这种情况下，类不应该被实例化。

Should I throw exception in the constructor?

我应该在构造函数中抛出异常吗?

class Foo(object):
    def __init__(self):
        self.rsc = get_resource():
        assert self.rsc

And try to catch it in the instantiation code?

试着在实例化代码中捕获它?

try:
    a = Foo()
except:
    a = None

Then later code checks if a is None.

然后，代码检查a是否为None。

2 个解决方案

#1

Expanding my earlier comment into an answer:

把我先前的评论扩展成一个答案:

It's better to add the resource as a constructor parameter, like this:

最好将资源添加为构造函数参数，如下所示:

class Foo(object):
    def __init__(self, rsc):
        self.rsc = rsc

This approach has a number of benefits:

这种方法有很多好处:

The constructor can no longer fail
构造函数不能再失败。
You no longer have an expensive constructor
您不再拥有一个昂贵的构造函数。
You have looser coupling between Foo and its resource type
在Foo和它的资源类型之间有更松散的耦合。

And if you like the interface of implicitly acquiring a resource, it is easy to wrap this. A simple function like make_foo attempts to acquire the resource automatically, returning None (or raising an error, if you prefer) when it cannot:

如果您喜欢隐式获取资源的接口，那么很容易将其包装起来。像make_foo这样的简单函数试图自动获取资源，如果不能，则返回None(或引发错误):

def make_foo():
    rsc = get_resource()
    if rsc:
        return Foo(rsc)
    else:
        return None

#2

-3

Not really a problem, just add try...except in the init function and have it assign a specific value like None in case of failure, or add additional features

不是真的有问题，只是增加尝试……除了在init函数中，并在失败时指定一个特定的值，或者添加其他特性。

class Foo(object):
    def __init__(self):
        try:
            self.rsc = get_resource()
            self.success = True
        except:
            self.rsc = None
            self.success = False

a = Foo()
if a.success:
    print(a.rsc)
else:
    do_something_else()

#1