super, apply, or __init__ when subclassing?

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  • exhuma.twn

    super, apply, or __init__ when subclassing?

    This is something that keeps confusing me. If you read examples of
    code on the web, you keep on seeing these three calls (super, apply
    and __init__) to reference the super-class. This looks to me as it is
    somehow personal preference. But this would conflict with the "There
    one way to do it" mind-set.

    So, knowing that in python there is one thing to do something, these
    three different calls must *do* domething different. But what exactly
    *is* the difference?

    ------------ Exampel 1: -----------------------------

    class B(A):
    def __init__(self, *args):
    A.__init__(self , args)

    ------------ Exampel 2: -----------------------------

    class B(A):
    def __init__(self, *args):
    apply( A.__init__, (self,) + args)

    ------------ Exampel 3: -----------------------------

    class B(A):
    def __init__(self, *args):
    super(A,self)._ _init__(*args)

  • Ben Finney

    #2
    Re: super, apply, or __init__ when subclassing?

    "exhuma.twn " <exhuma@gmail.c omwrites:
    This is something that keeps confusing me. If you read examples of
    code on the web, you keep on seeing these three calls (super, apply
    and __init__) to reference the super-class. This looks to me as it is
    somehow personal preference. But this would conflict with the "There
    one way to do it" mind-set.
    >
    So, knowing that in python there is one thing to do something, these
    three different calls must *do* domething different. But what exactly
    *is* the difference?
    >
    ------------ Exampel 1: -----------------------------
    >
    class B(A):
    def __init__(self, *args):
    A.__init__(self , args)
    >
    ------------ Exampel 2: -----------------------------
    >
    class B(A):
    def __init__(self, *args):
    apply( A.__init__, (self,) + args)
    >
    ------------ Exampel 3: -----------------------------
    >
    class B(A):
    def __init__(self, *args):
    super(A,self)._ _init__(*args)
    Note that your examples 1 and 3 aren't different *calls*. They are
    different ways of *getting at* the same class: either name it
    explicitly (as in example 1) or use 'super' to get it for you.

    Also, example 3 should instead call 'super(B, self).__init__' . If
    you're going to go to the bother of actually *specifying* the class
    'A', it's silly to call 'super' to get at it *again*.


    The broader picture: I think you're right that this is ridiculously
    difficult in Python. The community elders would have us use 'super' to
    get at our inherited '__init__', but that doesn't work very well
    <URL:http://fuhm.org/super-harmful/so most people use your example
    1.

    --
    \ "The most merciful thing in the world... is the inability of |
    `\ the human mind to correlate all its contents." -- Howard |
    _o__) Philips Lovecraft |
    Ben Finney

    Comment

    • Duncan Booth

      #3
      Re: super, apply, or __init__ when subclassing?

      "exhuma.twn " <exhuma@gmail.c omwrote:
      So, knowing that in python there is one thing to do something, these
      three different calls must *do* domething different. But what exactly
      *is* the difference?
      >
      ------------ Exampel 1: -----------------------------
      >
      class B(A):
      def __init__(self, *args):
      A.__init__(self , args)
      >
      ------------ Exampel 2: -----------------------------
      >
      class B(A):
      def __init__(self, *args):
      apply( A.__init__, (self,) + args)
      >
      ------------ Exampel 3: -----------------------------
      >
      class B(A):
      def __init__(self, *args):
      super(A,self)._ _init__(*args)
      Yes, they are all different.

      The first one calls B's immediate base class but packs all of the
      arguments together into a single tuple. Probably not what you meant.

      The second one passes B's positional arguments to its immediate base
      class without messing them up but uses a deprecated function to do it.
      You should use "A.__init__(sel f, *args)" instead unless you are
      concerned about multiple inheritance.

      The third one skips over the __init__ method in the immediate base class
      and calls the __init__ method in whatever class follows A in the MRO
      instead. Probably not what you meant either.

      Comment

      • Bruno Desthuilliers

        #4
        Re: super, apply, or __init__ when subclassing?

        exhuma.twn a écrit :
        This is something that keeps confusing me. If you read examples of
        code on the web, you keep on seeing these three calls (super, apply
        and __init__) to reference the super-class. This looks to me as it is
        somehow personal preference. But this would conflict with the "There
        one way to do it" mind-set.
        apply is deprecated. Chances are that code using it is somewhat old.
        super came with the new object model in Python 2.2.1 (IIRC), and is only
        useful for some corner cases involving multiple inheritence . Directly
        calling the superclass's method (__init__ or whatever) is the canonical
        way in the most common cases.

        And BTW, the sentence is "there *should* be one - and *preferably* only
        one - *obvious* way to do it" (emphasis is mine). In this case, there's
        _at least_ one way do to do it, and only one (direct call) is really
        obvious IMHO !-)
        So, knowing that in python there is one thing to do something, these
        three different calls must *do* domething different.
        Indeed. But mostly because you managed to get 2 examples wrong !-)
        But what exactly
        *is* the difference?
        >
        ------------ Exampel 1: -----------------------------
        >
        class B(A):
        def __init__(self, *args):
        A.__init__(self , args)
        You want:

        class B(A):
        def __init__(self, *args):
        A.__init__(self , *args)

        ------------ Exampel 2: -----------------------------
        >
        class B(A):
        def __init__(self, *args):
        apply( A.__init__, (self,) + args)
        is the same as the previous, using the deprecated apply function.
        ------------ Exampel 3: -----------------------------
        >
        class B(A):
        def __init__(self, *args):
        super(A,self)._ _init__(*args)
        >
        You want:

        class B(A):
        def __init__(self, *args):
        super(B,self)._ _init__(*args)

        Comment

        • exhuma.twn

          #5
          Re: super, apply, or __init__ when subclassing?

          On Sep 18, 2:45 pm, Bruno Desthuilliers <bruno.
          42.desthuilli.. .@wtf.websitebu ro.oops.comwrot e:
          exhuma.twn a écrit :
          >
          This is something that keeps confusing me. If you read examples of
          code on the web, you keep on seeing these three calls (super, apply
          and __init__) to reference the super-class. This looks to me as it is
          somehow personal preference. But this would conflict with the "There
          one way to do it" mind-set.
          >
          apply is deprecated. Chances are that code using it is somewhat old.
          super came with the new object model in Python 2.2.1 (IIRC), and is only
          useful for some corner cases involving multiple inheritence . Directly
          calling the superclass's method (__init__ or whatever) is the canonical
          way in the most common cases.
          >
          And BTW, the sentence is "there *should* be one - and *preferably* only
          one - *obvious* way to do it" (emphasis is mine). In this case, there's
          _at least_ one way do to do it, and only one (direct call) is really
          obvious IMHO !-)
          >
          So, knowing that in python there is one thing to do something, these
          three different calls must *do* domething different.
          >
          Indeed. But mostly because you managed to get 2 examples wrong !-)
          >
          But what exactly
          *is* the difference?
          >
          ------------ Exampel 1: -----------------------------
          >
          class B(A):
          def __init__(self, *args):
          A.__init__(self , args)
          >
          You want:
          >
          class B(A):
          def __init__(self, *args):
          A.__init__(self , *args)
          Ah.. this was a typo in my original post. Oops ;)
          >
          ------------ Exampel 2: -----------------------------
          >
          class B(A):
          def __init__(self, *args):
          apply( A.__init__, (self,) + args)
          >
          is the same as the previous, using the deprecated apply function.
          >
          ------------ Exampel 3: -----------------------------
          >
          class B(A):
          def __init__(self, *args):
          super(A,self)._ _init__(*args)
          >
          You want:
          >
          class B(A):
          def __init__(self, *args):
          super(B,self)._ _init__(*args)
          Hmmm... and suddenly it all makes sense! Great!

          Thanks all for clarifying this to a Java-Convert ;)

          Comment

          • Gabriel Genellina

            #6
            Re: super, apply, or __init__ when subclassing?

            En Tue, 18 Sep 2007 04:33:11 -0300, exhuma.twn <exhuma@gmail.c omescribi�:
            This is something that keeps confusing me. If you read examples of
            code on the web, you keep on seeing these three calls (super, apply
            and __init__) to reference the super-class. This looks to me as it is
            somehow personal preference. But this would conflict with the "There
            one way to do it" mind-set.
            >
            So, knowing that in python there is one thing to do something, these
            three different calls must *do* domething different. But what exactly
            *is* the difference?
            There are a few typos in your examples. If you write them this way:
            ------------ Exampel 1: -----------------------------
            >
            class B(A):
            def __init__(self, *args):
            A.__init__(self , *args)
            >
            ------------ Exampel 2: -----------------------------
            >
            class B(A):
            def __init__(self, *args):
            apply( A.__init__, (self,) + args)
            >
            ------------ Exampel 3: -----------------------------
            >
            class B(A):
            def __init__(self, *args):
            super(B,self)._ _init__(*args)
            then 2 is exactly the same as 1 but using a deprecated function. And 3 is
            the same as 1 only when there is single inheritance involved (and you are
            using new-style classes). But see the thread "super() doesn't get
            superclass"

            --
            Gabriel Genellina

            Comment

            • exhuma.twn

              #7
              Re: super, apply, or __init__ when subclassing?

              On Sep 18, 2:50 pm, "Gabriel Genellina" <gagsl-...@yahoo.com.a r>
              wrote:
              En Tue, 18 Sep 2007 04:33:11 -0300, exhuma.twn <exh...@gmail.c omescribi?:
              >
              This is something that keeps confusing me. If you read examples of
              code on the web, you keep on seeing these three calls (super, apply
              and __init__) to reference the super-class. This looks to me as it is
              somehow personal preference. But this would conflict with the "There
              one way to do it" mind-set.
              >
              So, knowing that in python there is one thing to do something, these
              three different calls must *do* domething different. But what exactly
              *is* the difference?
              >
              There are a few typos in your examples. If you write them this way:
              >
              Example 3 was not really a typo on my side. While I was browsing
              around for python, I saw this in a code-fragment. So I thought this
              was the way to do it. Even though I thought "super(A... " does not make
              sense in a "semantic" way. Instead of just accepting this as a fact I
              probably should have trusted my gut-feeling and investigate.

              Nonetheless the posts (and explanations) here were really helpful for
              me to understand what's happening under the hood. But seeing these
              three variants of doing (nearly) the same was an itch I finally
              decided to scratch. So far I blindly used "Example 1" as it seemed to
              work for me.
              >
              >
              ------------ Exampel 1: -----------------------------
              >
              class B(A):
              def __init__(self, *args):
              A.__init__(self , *args)
              >
              ------------ Exampel 2: -----------------------------
              >
              class B(A):
              def __init__(self, *args):
              apply( A.__init__, (self,) + args)
              >
              ------------ Exampel 3: -----------------------------
              >
              class B(A):
              def __init__(self, *args):
              super(B,self)._ _init__(*args)
              >
              then 2 is exactly the same as 1 but using a deprecated function. And 3 is
              the same as 1 only when there is single inheritance involved (and you are
              using new-style classes). But see the thread "super() doesn't get
              superclass"
              >
              --
              Gabriel Genellina

              Comment

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