How to do C language calculation in Python

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")
5.9047653992开发者_运维知识库36851  

Edit:

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

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).

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