Entering a very large number
Steven D'Aprano
steve+comp.lang.python at pearwood.info
Mon Mar 26 05:34:22 EDT 2018
On Mon, 26 Mar 2018 02:37:44 +0100, bartc wrote:
> Calling a function that sets up C using 'C = 288714...' on one line, and
> that then calculates D=C+C, takes 0.12 seconds to call 1000000 times.
>
> To do D=C*C, takes 2.2 seconds (I've subtracted the function call
> overhead of 0.25 seconds; there might not be any function call).
Bart, this is a contrived example that really proves nothing. This is
literally the sort of premature optimization that Knuth etc so frequently
warn about. In *almost any* real program, you're not going to be
calculating C+C over and over and over and over again, millions of times
in a row, while doing *literally* nothing else.
If your point is that, *hypothetically*, there could be some amazingly
unlikely set of circumstances where that's exactly what I will need to
do, then fine, I'll deal with those circumstances when it happens and not
a moment before.
And I'll probably deal with it by calculating C+C in Python, then hard-
coding D= that number, reducing the runtime cost of the addition to zero.
> If I instead initialise C using 'C = int("288712...")', then timings
> increase as follows:
Given that the original number given had 397 digits and has a bit length
of 1318, I must admit to some curiosity as to how exactly you managed to
cast it to a C int (32 bits on most platforms).
It is too big for an int, a long (64 bits), a long-long (128 bits) or
even a long-long-long-long-long-long-long-long-long-long-long-long-long-
long-long-long (1024 bits), if such a thing even exists.
So what exactly did you do?
--
Steve
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