Problem in calculating checksum : casting int to signed int32

Question

I need to convert the following c code (to calculate checksum for a file) to python. I had written, the corresponding code in python but the result didn't match the c version. The problem was that python autmatically promotes int to long whenever overflow occurs and this results in wrong checksums.

Any idea how to overcome this problem ? or is there a python function that converts long to signed int32 ?

Thanks

int calcChecksum(const guchar *data, gsize len)
{ 

    const guchar *p = data;
    int checksum = 0, g, i = len;

    while(i--) {
            checksum = (checksum << 4) + *p++;

            if((g = (checksum & 0xf0000000)) != 0)
                    checksum ^= g >> 23;

            checksum &= ~g;
    }
    return checksum;
}

Solution:

Thanks for all the help. Here's the function that worked for me -

 def int32(x):
    x = 0xffffffff & x
    if x > 0x7fffffff :
        return - ( ~(x - 1) & 0xffffffff )
    else : return x

Isn't there something missing in the c code? data / p seems to be used in a very odd way... — mjv
– mjv, Commented Feb 20, 2010 at 2:11
Sorry, had accidently removed '+' during formatting. Corrections made . thanks — Code freak
– Code freak, Commented Feb 20, 2010 at 2:20

Ignacio Vazquez-Abrams · Accepted Answer · 2010-02-20 02:30:58Z

2

Use numpy.int32 or numpy.uint32 if you need to restrict the range. Or mod it by 1 << 32 after operations that could "overflow".

edited Feb 20, 2010 at 2:30

answered Feb 20, 2010 at 2:19

Ignacio Vazquez-Abrams

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2 Comments

ephemient Over a year ago

Or & 0xFFFFFFFF (& ((1<<32)-1)). I happen to like that more, but it does the exact same as % (1 << 32).

Code freak Over a year ago

Its not just about extracting the lower 32 bits, but the sign also needs to be considered - the way java handle the overflow. e.g. int i = 0xffffffff; system.out.println(i) would give -1.

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