Semantics of ==

[Erik Max Francis]

Axel Boldt wrote:

> Still trying to understand "=="... It appears as if two equal objects
> can become unequal if you perform the same operations on them:
> 
>   >>> l=[1]
>   >>> s=l
>   >>> l.append(s)
>   >>> w=[1]
>   >>> r=[1,w]
>   >>> w.append(r)
>   >>> s
>   [1, [...]]
>   >>> w
>   [1, [1, [...]]]
>   >>> s==w
>   True

You're creating recursive lists (which is extremely rare in the real
world; I've never seen anyone create a self-referencing list in
real-life code.  Tree structures and even recursive data structures are
one thing, but you're considering a very weird corner case here.

> Note that they're equal, yet are printed differently.

That's fine.  1 and 1.0 are printed differently (and even are of
different types) but they're equal.  If you try running through the
element-wise algorithm that Python uses to determine equality of lists,
you'll see that they are in fact equal at this point.  The first element
is 1, the second element is a list whose first element is 1, the second
element of that list is a list whose first element is 1, ad infinitum.

>   >>> s[0]=2
>   >>> w[0]=2
>   >>> s==w
>   False
> 
> All of a sudden they have become unequal.

Because now they really are not equal.  Show them:

>>> s
[2, [...]]
>>> w
[2, [1, [...]]]

The [...] doesn't indicate which sublist is recursively contained, so
you have to dig deeper:

>>> s[1]
[2, [...]]
>>> w[1][1]
[2, [1, [...]]]

So s is

	[2, [2, [2, [2, ...]]]]

but w is

	[2, [1, [2, [1, ...]]]]

Despite both recursing, it's clear that these lists are not element-wise
equal, and so Python is completely right in saying they're not.

Again, though, real-world creation of self-referencing lists is
extremely rare at best.

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