Porting c++ to python code

Question

I want to convert this C++ code to Python v2:

static unsigned char asConvCode[3] = {0xFC, 0xCF, 0xAB};

void asConv(char* str, int size)
{
    int i = 0;

    for (i = 0; n< size; n++)
    {
        str[i] ^= asConvCode[n % 3];
    }
}

tried to make like that:

def asConv(self, data):
    asConvCode= [0xFC, 0xCF, 0xAB]

    for i in range(len(data)):
        data[i] ^= asConvCode[i % 3] # Error: Unsupported operand type(s) for ^=: ...

        return data

I will be happy for any hint

@David: The OP already posted the error message, as a comment in the code. — C. K. Young, Mar 11 '11 at 22:50
But the important detail (*which* types are unsupported) is missing... — , Mar 11 '11 at 22:52
@delnan: It's actually pretty obvious if you notice that `data` is a string (as that's the intent of the C code). — C. K. Young, Mar 11 '11 at 22:54
This might help http://stackoverflow.com/questions/432842/how-do-you-get-the-logical-xor-of-two-variables-in-python — Endophage, Mar 11 '11 at 22:55
@Endophage: That post actually won't help in this instance. The OP is not doing "logical xor". — C. K. Young, Mar 11 '11 at 22:57
@Chris Jester-Young ok... I'm not particularly fluent with C++ but the info floating around indicates the `^=` does an XOR... I'd love to be corrected if I'm wrong. — Endophage, Mar 12 '11 at 20:28

score 2 · Accepted Answer · answered Mar 11 '11 at 22:52

In Python, characters in strings are simply strings of length 1, not integers. So you must use this:

data[i] = chr(ord(data[i]) ^ asConvCode[i % 3])

Also, as I wrote in a comment, your return data is at the wrong indentation level, and will cause your function to return after processing the first character.

Porting c++ to python code

1 Answers1