Please, help me to convert a Python code to C++

[Bokka]

Hello Forumers!

I am new to C++, but I have to write a simple program to calculate the CRC16 of a Modbus string.
I have only to write that, I have not to do anything else.

The program writed in Python is the following:

[coode]
def crc16(data, bits=8):
crc = 0xFFFF
for op, code in zip(data[0::2], data[1::2]):
crc = crc ^ int(op+code, 16)
for bit in range(0, bits):
if (crc&0x0001) == 0x0001:
crc = ((crc >> 1) ^ 0xA001)
else:
crc = crc >> 1
msb = crc >> 8
lsb = crc & 0x00FF
return '{:X}{:X}'.format(lsb, msb)
[/code]


For example, with the string "11010003000C" the output of the conversion must be "CE9F".
I need only that function for my first and I think only program in C++.

Someone could help me please?

Thanks a lot in advance!

[keskiverto]

If you already have functional code, then why should you rewrite it in another language, like C++?

Are you fluent in Python, i.e. know the existing algorithm's logic intimately?

Have you done a websearch with "CRC16 of a Modbus string"?

[Bokka]

Thanks for the reply! But I need the code in C++ because I have to insert it in a QT project, and I can't use Python.

I'm not fluent in python, but I'm studying it, but for now I need only that function in C++ for a small project, then I will use only Python probabily

[Computergeek01]

Let's try a different approach. This python function looks a little sparse for something like redundancy checking so we're going to drop it on the floor and look at a C\C++ solution instead. Check here: https://github.com/d-bahr/CRCpp .Why is it only 16-bit? Is that a requirement?

[icy1]

Bokka, i call BS, that's not Python. Python is all about indentation, and your post is completely left-aligned. Can't even tell when the nested(?) for loops end.

The zip() with extended slices of 2 just assigns "op" to every even index element of data, and "code" to every odd index of data. The zip seems unnecessary to me since it's zipping from the same array. Then again, I come from Ruby background, not python.

The range() is just up to but not including bits. So a normal indexed loop.

Rest is math and a formatted string at the end.

[icy1]

Well, just guess+check I found out the indentation

def crc16(data, bits=8):
    crc = 0xFFFF
    for op, code in zip(data[0::2], data[1::2]):
        crc = crc ^ int(op+code, 16)
        for bit in range(0, bits):
            if (crc&0x0001) == 0x0001:
                crc = ((crc >> 1) ^ 0xA001)
            else:
                crc = crc >> 1
    msb = crc >> 8
    lsb = crc & 0x00FF
    return '{:X}{:X}'.format(lsb, msb)

print(crc16("11010003000C"))

Running at https://repl.it/repls/ComposedSevereApache

CE9F

[Bokka]

The python code is not indented because the forum won't allow me to do it.

Btw I repeat it, I need only that part of code for my first and probabily last C++ project.

For computerGeek01: yes, I need only a 16 bit check, I don't need another things

[keskiverto]

Google offers for me now: https://stackoverflow.com/questions/19347685/calculating-modbus-rtu-crc-16

Is that the same algorithm?

[icy1]

Upon closer inspection, "op" and "code" are basically unused. They just wanted the substring of every two characters. std::hex returns lowercase hex by default, so that's why std::uppercase is employed here as well.

#include <iostream>
#include <string>
#include <sstream>

using namespace std;

string Crc16(const string& data, unsigned short bits=8)
{
    unsigned short crc = 0xFFFF;
    for (int i=0; i<data.size(); i+=2)
    {
        crc = crc ^ stoi(data.substr(i,2), nullptr, 16);
        for (int x=0; x<bits; ++x)
        {
            if (crc&0x0001 == 0x0001)
                crc =  (crc>>1) ^ 0xA001;
            else
                crc = crc >> 1;
        }
    }
    ostringstream oss;
    oss << uppercase << hex << (crc&0x00FF) << (crc>>8);
    return oss.str();
}

int main()
{
    cout << Crc16("11010003000C") << endl;
    return 0;
}

Running at https://repl.it/repls/AltruisticEthicalZettabyte

CE9F

[Bokka]

EXACTLY what I want! YOU are my HERO! Thanks a lot to everybody!!!