Sorry, I'm not good at English enough to explain my mental model.
I meant no skip, no ignorance, no throw away.
In case of 1+2=3, both of 1 and 2 are not skipped, ignored or thrown away.
On the other hand, in case of {a:1, b:2}+{a:2}={a:2, b:2}, I feel {a:1} is
skipped, ignored, or thrown away. I used "lost" to explain it.
And I used "lossless" for "there is no lost". Not for reversible.
If it isn't understandable to you, please ignore me.
I think Rémi’s comment is very similar to my thought. Merging mapping is
more complex than concatenate sequence and it seems hard to call it "sum".
Regards,
2019年3月1日(金) 23:19 Ivan Levkivskyi <levkivs...@gmail.com>:
> On Fri, 1 Mar 2019 at 13:48, INADA Naoki <songofaca...@gmail.com> wrote:
>
>> >
>> >
>> > Is that an invariant you expect to apply to other uses of the +
>> > operator?
>> >
>> > py> x = -1
>> > py> x <= (x + x)
>> > False
>> >
>> > py> [999] <= ([1, 2, 3] + [999])
>> > False
>> >
>>
>> Please calm down. I meant each type implements "sum"
>> in semantics of the type, in lossless way.
>> What "lossless" means is changed by the semantics of the type.
>>
>> -1 + -1 = -2 is sum in numerical semantics. There are no loss.
>>
>
> TBH I don't understand what is lossless about numeric addition. What is
> the definition of lossless?
> Clearly some information is lost, since you can't uniquely restore two
> numbers you add from the result.
>
> Unless you define what lossless means, there will be just more
> misunderstandings.
>
> --
> Ivan
>
>
>
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