Dave Angel <da...@davea.name> writes: > On 05/06/2015 11:36 AM, Alain Ketterlin wrote: >> Yes, plus the time for memory allocation. Since the code uses "r *= >> ...", space is reallocated when the result doesn't fit. The new size is >> probably proportional to the current (insufficient) size. This means >> that overall, you'll need fewer reallocations, because allocations are >> made in bigger chunks. > > That sounds plausible, but a=5; a*=4 does not update in place. It > calculates and creates a new object. Updating lists can work as you > say, but an int is immutable.
Ah, okay. Even for big ints? If that is the case, my suggestion doesn't explain anything. Anyway, with so many allocations for so little arithmetic, the difference is probably due to the behavior of the allocator (which maybe always finds blocks big enough, since one was released after the previous multiplication, or something like that). The only way to know would be to profile the VM. > It's an optimization that might be applied if the code generator were > a lot smarter, (and if the ref count is exactly 1), but it would then > be confusing to anyone who used id(). "Abandon all hope, ye [optimizer] who enter here." Thanks for the clarification. -- Alain. -- https://mail.python.org/mailman/listinfo/python-list
