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Are you saying `o` is passed by value? I think this behavior is due to the return from `id()` being interned, or not. `id(o) == id(o)` will be true in all cases

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I mean that the `id` function returns by value. It's not interning since that explicitly refers to something allocated, which isn't the case here.

This is incorrect. The returned integer is a regular Python object, not some "unboxed" integer value.

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