implement the logic in c++ using python, how?

Question

I want to implement below logic in c++ using python.

struct hash_string ///
{
    hash_string() {}

    uint32_t operator ()(const std::string &text) const
    {
        //std::cout << text << std::endl;
        static const uint32_t primes[16] =
        {
            0x01EE5DB9, 0x491408C3, 0x0465FB69, 0x421F0141,
            0x2E7D036B, 0x2D41C7B9, 0x58C0EF0D, 0x7B15A53B,
            0x7C9D3761, 0x5ABB9B0B, 0x24109367, 0x5A5B741F,
            0x6B9F12E9, 0x71BA7809, 0x081F69CD, 0x4D9B740B,
        };

        //std::cout << text.size() << std::endl;
        uint32_t sum = 0;
        for (size_t i = 0; i != text.size(); i ++) {
            sum += primes[i & 15] * (unsigned char)text[i];
            //std::cout << text[i] <<std::endl;
            // std::cout << (unsigned char)text[i] << std::endl;
        }
        return sum;
    }
};

python version is like this, which is not completed yet, since I haven't found a way to convert text to unsigned char. So, please help!

# -*- coding: utf-8 -*-

text = u'连衣裙女韩范'

primes = [0x01EE5DB9, 0x491408C3, 0x0465FB69, 0x421F0141,
                0x2E7D036B, 0x2D41C7B9, 0x58C0EF0D, 0x7B15A53B,
                0x7C9D3761, 0x5ABB9B0B, 0x24109367, 0x5A5B741F,
                0x6B9F12E9, 0x71BA7809, 0x081F69CD, 0x4D9B740B]

//*text[i] does not work (of course), but how to mimic the logic above
rand = [primes[i & 15]***text[i]** for i in range(len(text))]

print rand

sum_agg = sum(rand)

print sum_agg

Take text=u'连衣裙女韩范' for example, c++ version returns 18 for text.size() and sum is 2422173716, while, in python, I don't know how to make it 18.

The equality of text size is essential, as a start at least.

I think Python already has built-in good-enough hashing of strings. Have you checked that? — Cheers and hth. - Alf
– Cheers and hth. - Alf, Commented Nov 5, 2015 at 9:14
that's true, it is just that this logic is used universally in our application that I simply need to duplicate it to generate the same hash code. And my code is implemented in python all the way, so I just want to convert it to python for simplicity. — phoenix bai
– phoenix bai, Commented Nov 5, 2015 at 9:19
@SanderDeDycker I updated the post, ord doesnot solve it since text.size() in c++ returns 18, while using ord, I cannot get 18. — phoenix bai
– phoenix bai, Commented Nov 5, 2015 at 9:22

Dunes · Accepted Answer · 2015-11-05 11:59:45Z

2

Because you are using unicode, for an exact reproduction you will need to turn text in a series of bytes (chars in c++).

bytes_ = text.encode("utf8") 
# when iterated over this will yield ints (in python 3)
# or single character strings in python 2

You should use more pythonic idioms for iterating over a pair of sequences

pairs = zip(bytes_, primes)

What if bytes_ is longer than primes? Use itertools.cycle

from itertools import cycle
pairs = zip(bytes_, cycle(primes))

All together:

from itertools import cycle

text = u'连衣裙女韩范'

primes = [0x01EE5DB9, 0x491408C3, 0x0465FB69, 0x421F0141,
                0x2E7D036B, 0x2D41C7B9, 0x58C0EF0D, 0x7B15A53B,
                0x7C9D3761, 0x5ABB9B0B, 0x24109367, 0x5A5B741F,
                0x6B9F12E9, 0x71BA7809, 0x081F69CD, 0x4D9B740B]

# if python 3
rand = [byte * prime for byte, prime in zip(text.encode("utf8"), cycle(primes))]
# else if python 2 (use ord to convert single character string to int)
rand = [ord(byte) * prime for byte, prime in zip(text.encode("utf8"), cycle(primes))]
hash_ = sum(rand)

edited Nov 5, 2015 at 11:59

answered Nov 5, 2015 at 9:27

Dunes

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2 Comments

phoenix bai Over a year ago

it is throwing error and I print the bytes_, it is printing endlessly! could you help debug it?

Dunes Over a year ago

Sorry, the answer I posted was intended for python 3. I've made an edit to show what to do in python 2.

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