How to do C language calculation in Python

Question

I want to do some simulation of C language calculation in Python. For example, unsigned short, single precision float ...

ushort(0xffff) + 1 -> 0
0.1f * 0.1f -> ...

Are there some library to do this in Python?

I can use ctypes to create unsigned short, single float, but they cann't do math operation:

a = c_uint16(0xffff)
b = c_uint16(0x01)
a+b -> TypeError

Or, I can use numpy:

>>> np.uint16(0xffff) + np.uint16(0x01)
Warning: overflow encountered in ushort_scalars
0

but it's very slow comparing to Python's normal calculation:

>>> timeit.timeit("a+b", "import numpy as np;a=np.uint16(0xfffe);b=np.uint16(0x01)")
0.35577465681618037
>>> timeit.timeit("0xfffe+0x01")
0.022638104432360251
>>> timeit.timeit("np.uint16(0xfffe) + np.uint16(0x01)", "import numpy as np")
5.904765399236851

Edit:

>>> timeit.timeit("a+b", "a=0xfffe;b=0x01")
0.040062221014295574

how about: def add(a,b): return (a+b) % 65536 ?

JBernardo
– JBernardo

2011-06-16 02:21:43 +00:00
Commented Jun 16, 2011 at 2:21 — JBernardo
– JBernardo, Commented Jun 16, 2011 at 2:21

Sven Marnach · Accepted Answer · 2011-06-16 02:02:54Z

6

When compiling 0xfffe+0x01, this will be folded into the constant 65535. You aren't timing how long the addition takes -- you are just measuring the time of loading the constant:

>>> dis.dis(compile("0xfffe+0x01", "", "eval"))
  1           0 LOAD_CONST               2 (65535)
              3 RETURN_VALUE

The addition of NumPy scalars is slower than adding built-in integers nevertheless, but it won't get better than that in pure Python. Consider using Cython -- it will allow you to declare types and execute the computations in C speed. Alternatively, try to vectorise your code in NumPy (that is, if speed really matters).

edited Jun 16, 2011 at 2:02

answered Jun 16, 2011 at 1:04

Sven Marnach

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2 Comments

senderle Over a year ago

yes, the gerund vs. infinitive thing doesn't make a lot of sense does it? I was thinking about it after I saw this. For example "try to use cython" and "try using cython" are both idiomatic English, but they mean subtly different things; the former means something like "as a rule, use cython if you can," while the latter is closer to "consider using cython." (And for what it's worth, it should probably be "to learn better English.") All that said, I did mean it, about your posts being spotless, generally. Now that I know you don't mind a direct edit, perhaps I'll do that next time.

Sven Marnach Over a year ago

@senderle: Thanks again -- I'll clarify the message in my profile!

JBernardo · Accepted Answer · 2011-06-16 02:39:25Z

1

You can make a function for each operation using modulo % with 2**sizeof (in your case, 2**16 or 65536)

def add(a, b, mod=2**16):
    return (a+b) % mod

def sub(a, b, mod=2**16):
    return (a-b) % mod

and any other function you need.

>>> add(0xffff, 1)
0
>>> sub(10, 20)
65526

Note this will work only for unsigned types. For signed ones, you can use half the value used to mod (i.e. 2**15) and will have to validate the result before applying modulo

edited Jun 16, 2011 at 2:39

answered Jun 16, 2011 at 2:34

JBernardo

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1 Comment

HYRY Over a year ago

Thanks, your method is useful for integers. For single floats, I am considering using ctypes call some DLL.

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