gh-100239: use PyFloat_AS_DOUBLE and _PyLong_IsZero in the float / compactlong specializations - #144826
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Eclips4 merged 3 commits intoFeb 19, 2026
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gh-100239: use PyFloat_AS_DOUBLE and _PyLong_IsZero in the float / compactlong specializations#144826Eclips4 merged 3 commits into
Eclips4 merged 3 commits into
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know that we deal with exact floats here
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This microbenchmark import time
def bench():
t = time.perf_counter()
f = 1.0
for i in range(60000):
for j in range(1, 1000):
f = f + j
f = f - j
f = f / j
f = f * j
print(time.perf_counter() - t)
if __name__ == "__main__":
bench()shows even better speedup: old: 6.5 seconds new: 5.5 seconds |
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skirpichev
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Feb 15, 2026
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This is user-visible change. So, I believe it deserves a news entry. |
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chris-eibl
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Feb 17, 2026
| @@ -0,0 +1,2 @@ | |||
| Speedup ``BINARY_OP_EXTEND`` for exact floats and medium-size integers by up | |||
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Compact ints are an implementation detail and the term medium-size integers has been chosen in news entries in the past for them.
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Eclips4
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Feb 19, 2026
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Thank you Chris!
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brijkapadia
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…he float / compactlong specializations (python#144826)
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…he float / compactlong specializations (python#144826)
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Since we know that we deal with exact floats here.
This is just a small follow-up on #128722 and #128963.
I don't think it deserves a news entry, but I'd happily create it if deemed meaningful enough.
Microbenchmark from @iritkatriel 's PR:
Old:
>>> timeit("for i in range(10000):\n\tb = a+i", number=100000, setup="a = 1.0") 42.05178890001844 >>> timeit("for i in range(10000):\n\tb = i+a", number=100000, setup="a = 1.0") 41.31541720000678New:
>>> timeit("for i in range(10000):\n\tb = a+i", number=100000, setup="a = 1.0") 38.316913699993165 >>> timeit("for i in range(10000):\n\tb = i+a", number=100000, setup="a = 1.0") 38.37592560000485Fastmark results are within the noise 1, the three arithmetic benchmarks @Fidget-Spinner mentioned are a tad faster:
Footnotes
see my gist how I do them ↩