why does dead code costs time?
Neil Cerutti
neilc at norwich.edu
Wed Dec 5 11:15:59 EST 2012
On 2012-12-05, Bruno Dupuis <python.ml.bruno.dupuis at lisael.org> wrote:
> Hi,
>
> I'm interested in compilers optimizations, so I study python
> compilation process
>
> I ran that script:
>
> import timeit
>
> def f(x):
> return None
>
> def g(x):
> return None
> print(x)
>
> number = 10000
>
> print(timeit.timeit('f(1)',setup="from __main__ import f", number=number))
> print(timeit.timeit('g(1)',setup="from __main__ import g", number=number))
>
> print(dis.dis(f))
> print(dis.dis(g))
>
> It gives this output:
>
> 0.003460251959040761
> 0.004164454061537981
> 17 0 LOAD_CONST 0 (None)
> 3 RETURN_VALUE
> None
> 20 0 LOAD_GLOBAL 0 (None)
> 3 RETURN_VALUE
>
> 21 4 LOAD_GLOBAL 1 (print)
> 7 LOAD_FAST 0 (x)
> 10 CALL_FUNCTION 1 (1 positional, 0 keyword pair)
> 13 POP_TOP
> None
>
> I do not understand why the dead code `print(x)` takes time (~20% in
> that case). As we see in the opcode, a call to g(1) returns immediately, so
> there should be no delay at all. Where am i wrong?
>
> mmhh... it comes to me now that the gap must be in function loading time...
> I'll check ceval.c
>
> However, isn't there a room for a slight optim here? (in this case, the
> dead code is obvious, but it may be hidden by complex loops and
> conditions)
Maybe it's the difference between LOAD_CONST and LOAD_GLOBAL. We
can wonder why g uses the latter.
--
Neil Cerutti
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