Note also that prod(range(1,n)) does *NOT* just multiple 1 times 2
times 3 in order. Instead, it uses a tree approach so that the
multiplications involve objects with more balanced sizes, which is
much faster, e.g., for large integers multiplyling n*m with n and m
having similar sizes can overall be massively better than n * m with n
large and m tiny...
On Thu, Sep 22, 2022 at 2:54 PM kcrisman <kcris...@gmail.com> wrote:
>
> I'm not sure what you were expecting. It does return 0, but it does so after
> all the print statements are done. Remember, your "test" function just
> returns the input, so the map function just returns a list (well, a map
> object, but ...) and then we prod everything in the list. But that happens
> AFTER you have done the map on test, and that requires printing out all of
> them. The prod can't finish up until the entire list has been processed via
> map and the function "test". I could imagine an alternate implementation of
> prod which checked each time whether zero had been produced in an
> intermediate step, but that doesn't appear to be in the code in "prod??".
>
> On Thursday, September 22, 2022 at 10:39:03 AM UTC-4 axio...@yahoo.de wrote:
>>
>> sage: def test(n):
>> ....: print("n:", n)
>> ....: return n
>> ....:
>> sage: l = [2,3,5,0,7,11,17,19]
>> sage: prod(map(test, l))
>> n: 2
>> n: 3
>> n: 5
>> n: 0
>> n: 7
>> n: 11
>> n: 17
>> n: 19
>> 0
>> I expected that it would return 0 once we multiply with 0.
>>
>> Martin
>
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