Initializing object in init parameter list for default value returning same instance

Question

Given the sample code below:

class Foo:
    def __init__(self):
        return

class Bar:
    def __init__(self, foo = Foo()):
        print(foo)

bar1 = Bar()
bar2 = Bar()

print(bar1)
print(bar2)

Outputs the following:

<__main__.Foo object at 0x000002A989A2E9B0>
<__main__.Foo object at 0x000002A989A2E9B0>
<__main__.Bar object at 0x000002A989A2EA20>
<__main__.Bar object at 0x000002A989A2EA90>

It was unexpected that both instances of Foo were identical. I was hoping that each subsequent creation of a Bar() object would have it’s own unique version of a Foo object.

How can I modify my example so that each call to Bar() will create its own unique version of Foo? Do I need to use __new__ somehow?

Asked By: roundtheworld

||

Source

Answer 1

class Foo:
    pass

class Bar:
    def __init__(self, foo = None):
        if foo is None:
            foo = Foo()
        print(foo)

bar1 = Bar()
bar2 = Bar()

print(bar1)
print(bar2)

or

import attr

class Foo:
    pass

@attr.s
class Bar(object):
    foo = attr.ib(default=attr.Factory(Foo))

bar1 = Bar()
bar2 = Bar()

print(bar1.foo)
print(bar2.foo)

Answered By: Jean-Paul Calderone

Initializing object in init parameter list for default value returning same instance

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Initializing object in __init__ parameter list for default value returning same instance

Question:

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Initializing object in init parameter list for default value returning same instance