Am 26.03.18 um 12:31 schrieb bartc:
On 26/03/2018 10:34, Steven D'Aprano wrote:
So what exactly did you do?
I did this:
def fn():
C = int(
"28871482380507712126714295971303939919776094592797"
"22700926516024197432303799152733116328983144639225"
"94197780311092934965557841894944174093380561511397"
"99994215424169339729054237110027510420801349667317"
"55152859226962916775325475044445856101949404200039"
"90443211677661994962953925045269871932907037356403"
"22737012784538991261203092448414947289768854060249"
"76768122077071687938121709811322297802059565867")
# C = 2887148238050771212671429... [truncated for this post]
D=C+C
for i in range(1000000):
fn()
The purpose was to establish how such int("...") conversions compare in
overheads with actual arithmetic with the resulting numbers.
I still claim that it is not necessary to perform the conversion on each
function call. For instance, the following change speeds it up by a
factor of 7 on my machine:
=======================================================
def fn(C = int(
"28871482380507712126714295971303939919776094592797"
"22700926516024197432303799152733116328983144639225"
"94197780311092934965557841894944174093380561511397"
"99994215424169339729054237110027510420801349667317"
"55152859226962916775325475044445856101949404200039"
"90443211677661994962953925045269871932907037356403"
"22737012784538991261203092448414947289768854060249"
"76768122077071687938121709811322297802059565867")):
# C = 2887148238050771212671429... [truncated for this post]
D=C+C
return D
for i in range(1000000):
fn()
#print(fn())
=======================================================
Out of pure laziness, I haven't compared it to the version with the long
string. Maybe tha gurus here find yet another elegant way to assign the
constant only once; my trial with "fn.C = foobar" was not successful.
Christian
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