Python:构造函数初始化__

Question

我试图理解，即使在注释中有解释，此代码中错误的原因是什么：

    """

    This program is part of an exercise in
    Think Python: An Introduction to Software Design
    Allen B. Downey

    WARNING: this program contains a NASTY bug.  I put
    it there on purpose as a debugging exercise, but
    you DO NOT want to emulate this example!

    """

    class Kangaroo(object):
        """a Kangaroo is a marsupial"""

        def __init__(self, contents=[]):
            """initialize the pouch contents; the default value is
            an empty list"""
            self.pouch_contents = contents

        def __str__(self):
            """return a string representaion of this Kangaroo and
            the contents of the pouch, with one item per line"""
            t = [ object.__str__(self) + ' with pouch contents:' ]
            for obj in self.pouch_contents:
                s = '    ' + object.__str__(obj)
                t.append(s)
            return '\n'.join(t)

        def put_in_pouch(self, item):
            """add a new item to the pouch contents"""
            self.pouch_contents.append(item)

    kanga = Kangaroo()
    roo = Kangaroo()
    kanga.put_in_pouch('wallet')
    kanga.put_in_pouch('car keys')
    kanga.put_in_pouch(roo)

    print kanga

    # If you run this program as is, it seems to work.
    # To see the problem, trying printing roo.

解释报告如下：

^{pr2}$

但我，因为我是Python（和通用）编程新手，还不了解它。在

谢谢。在

Answer 1

问题是，如果将数组作为默认值放在成员定义中，则对于没有传入数组的类的所有实例，它都是相同的数组。因此，不是每个实例都获得一个新的空数组，而是在所有实例之间共享内容。在

Answer 2

问题是在kangaroo类的init部分使用了一个可变对象-list。这导致问题的原因是可变对象为您提供了对对象的引用，有时当您认为您正在制作对象的新副本时，您只是在对同一对象进行新的引用。解决方案会移动定义，以便每次类初始化时都将其计算为新列表，而不是仅当函数init被定义为kangaroo类的一部分时。在