Question

我试图在运行init之前更改args但它没有改变并保持为主要给出的第一个args，我如何更改 new 中的args？

class A(object):
            def __init__(self,ip,st):
                    print 'A arrived to init '+st
                    self.ip=ip

        def __new__(cls,ip,st):
                print "A arrived to new"
                if ip>10:
                        return object.__new__(cls,ip,"A")
while True:                
        s=input("input?")
        a=A(s,"a")
        print type(a)

输出：

input?88
A arrived to new
A arrived to init a
<class '__main__.A'>
input?44
A arrived to new
A arrived to init a
<class '__main__.A'>
input?22
A arrived to new
A arrived to init a
<class '__main__.A'>
input?12
A arrived to new
A arrived to init a
<class '__main__.A'>

Answer 1

这是元类的__call__()方法，每次调用它都会调用YourClass.__new__()和YourClass.__init__()。因此，如果您希望在参数到达YourClass.__init__()之前更改它们，您有两个解决方案：装饰__init__()或使用自定义元类覆盖type.__call__()。

（Q＆amp; D）装饰者版本：

def changeargs(func):
    # fixme : make this a well-behaved decorator
    def wrapper(self, *args, **kw):
        print("changearg.wrapper(%s, %s)" % (args, kw))
        args = (1, 2)
        kw = {"hacked": True}
        return func(self, *args, **kw)
    return wrapper

class Bar(object):
    @changeargs
    def __init__(self, *args, **kw):
        self.args = args
        self.kw = kw

    def __repr__(self):
        return "<Bar(%s, %s)>" % (self.args, self.kw)

（Q＆amp; D）元类版本（py 2.7.x）：

class FooType(type):
    def __call__(self, *args, **kw):
        print("FooType.__call__(%s, %s)" % (args, kw))
        args = (1, 2)
        kw = {"hacked": True}
        # fixme : make this collaborative super() call
        return type.__call__(self, *args, **kw)

class Foo(object):
    __metaclass__ = FooType

    def __init__(self, *args, **kw):
        self.args = args
        self.kw = kw

    def __repr__(self):
        return "<Foo(%s, %s)>" % (self.args, self.kw)

但正如Rawing在评论中正确提到的那样，你可以直接在你的班级__init__方法中直接这样做。

python从new转移到init

1 个答案:

python从__new__转移到__init__

1 个答案:

python从new转移到init