4 ms·

lol. in python: (2 ^ 32)*upper + lower edit: dbl star goes blank .. so in almost python

tptacek 17y ago
You're XORing 32 into 2, multiplying it by the top word, and adding the bottom word? Also, you think exponentiation and multiplication is faster than a shift and an OR? I think you're the target audience for this article, and I revise my opinion (thankfully unstated) about whether this is material every developer "must" know.
- notaddicted 17y ago
  Okay okay ... I should have written: import math; math.pow(2,32) I tried to use the double star operator, but that is markup for italics in this software, so it was edited out.
- comatose_kid 17y ago
  Surprisingly (I just benchmarked it myself), the times are near the same. Here's my benchmark (in case I made a mistake): import time def combine(upper,lower): return (2**32)*upper + lower def combine2(upper,lower): return (upper << 32) | lower def test_combine(): start = time.clock() for i in range(10000000): combine(i,i) end = time.clock() output = 'The time taken for combine() is ' + str(end - start) print output def test_combine2(): start = time.clock() for i in range(10000000): combine2(i,i) end = time.clock() output = 'The time taken for combine() is ' + str(end - start) print output Also see http://wiki.python.org/moin/PythonSpeed/PerformanceTips#PythonisnotC http://wiki.python.org/moin/PythonSpeed/PerformanceTips#Pyth...
  - tptacek 17y ago
    That's surprising; it's more than twice as slow in Ruby (yeah, yeah). It's also (for obvious reasons) orders of magnitude faster in C --- if you were crazy enough to actually write code that way.
  - ramchip 17y ago
    I don't know much about Python so forgive me if this is stupid - but could 2^32 have been optimized as a constant, so that the result is just a multiplication and an addition?
jf 17y ago
I considered that solution. And in thinking about why I chose to do bit-shifting, I remembered why: I actually needed to combine two _signed_ 32 bit integers in MySQL, and since I wasn't sure how signed integers are stored in MySQL, I figured that I could just use bit-shifting and move on.

You're XORing 32 into 2, multiplying it by the top word, and adding the bottom word? Also, you think exponentiation and multiplication is faster than a shift and an OR? I think you're the target audience for this article, and I revise my opinion (thankfully unstated) about whether this is material every developer "must" know.

notaddicted 17y ago
Okay okay ... I should have written: import math; math.pow(2,32) I tried to use the double star operator, but that is markup for italics in this software, so it was edited out.
comatose_kid 17y ago
Surprisingly (I just benchmarked it myself), the times are near the same. Here's my benchmark (in case I made a mistake): import time def combine(upper,lower): return (2**32)*upper + lower def combine2(upper,lower): return (upper << 32) | lower def test_combine(): start = time.clock() for i in range(10000000): combine(i,i) end = time.clock() output = 'The time taken for combine() is ' + str(end - start) print output def test_combine2(): start = time.clock() for i in range(10000000): combine2(i,i) end = time.clock() output = 'The time taken for combine() is ' + str(end - start) print output Also see http://wiki.python.org/moin/PythonSpeed/PerformanceTips#PythonisnotC http://wiki.python.org/moin/PythonSpeed/PerformanceTips#Pyth...
- tptacek 17y ago
  That's surprising; it's more than twice as slow in Ruby (yeah, yeah). It's also (for obvious reasons) orders of magnitude faster in C --- if you were crazy enough to actually write code that way.
- ramchip 17y ago
  I don't know much about Python so forgive me if this is stupid - but could 2^32 have been optimized as a constant, so that the result is just a multiplication and an addition?

jf 17y ago

I considered that solution. And in thinking about why I chose to do bit-shifting, I remembered why: I actually needed to combine two _signed_ 32 bit integers in MySQL, and since I wasn't sure how signed integers are stored in MySQL, I figured that I could just use bit-shifting and move on.