[SOLVED] Function behave differently with similar input (C++)

wowy · 05-12-2015, 03:58 AM

I am trying to work with the modbus protocol and right now i am calculating the LRC of the messages. I made a function which worked with no issue whatever i was putting into and then i noticed that id did not worked with one input and i can't find a logical explanation on why this don't work.

The function is :

Code:

void LRCstring(std::string example)
{

std::stringstream ss;
std::string hex =example.substr(1, example.length()-5);
std::vector<unsigned char> hexCh;
unsigned int buffer;
int offset = 0;
while (offset < hex.length()) {
   ss.clear();
   ss << std::hex << hex.substr(offset, 2);
   ss >> buffer;
   hexCh.push_back(static_cast<unsigned char>(buffer));
   offset += 2;
}

unsigned char LRC=0x00;
int i;
for (i=0;i<hexCh.size();i++)
{
    LRC=LRC+hexCh[i];
}

LRC = 0xFF-LRC; // 1 complement
LRC = LRC+1; // 2 complement

//std::string s = std::to_string(LRC);

//int deci = atoi(s.c_str());
int deci = LRC;
int reste=deci  16;
std::string temp;
int partiehexa=(deci-reste)/16;
std::string temp2;

std::cout << "deci : " << deci << std::endl;
std::cout << "reste : " << reste << std::endl;
std::cout << "partiehexa : " << partiehexa << std::endl;

std::stringstream ss2;
ss2 << reste;
ss2 >> temp;
ss2 << partiehexa;
ss2 >> temp2;

if (partiehexa<10) {LRCascii+=temp2;}
if (partiehexa==10) {LRCascii+='A';}
if (partiehexa==11) {LRCascii+='B';}
if (partiehexa==12) {LRCascii+='C';}
if (partiehexa==13) {LRCascii+='D';}
if (partiehexa==14) {LRCascii+='E';}
if (partiehexa==15) {LRCascii+='F';}
if (reste<10) {LRCascii+=temp;}
if (reste==10) {LRCascii+='A';}
if (reste==11) {LRCascii+='B';}
if (reste==12) {LRCascii+='C';}
if (reste==13) {LRCascii+='D';}
if (reste==14) {LRCascii+='E';}
if (reste==15) {LRCascii+='F';}

std::cout << "LRC : " << LRCascii << std::endl;

return;
}

Examples on what is the input and the result when it is working :

input > ":040100130013??\r\n"

The cout display "LRC : D5"

input > ":0401CD6B05??\r\n"

The cout display "LRC : BE"

D5 and BE are the right results.

I tried with other inputs and had no problem until this :

input > ":0403006B0003??\r\n"

The cout display "LRC : B"

input > ":040306022B00000064??\r\n"

The cout display "LRC : 2"

It should be 8B and not simply B and it should be 62 and not simply 2.

We can see that the last part of the LRC is good but the other part is ignored. What is even stranger is that in this case the cout of "partiehexa" is showing "8" and "6", it is not like this int was empty. I fail to understand why this is happening in this case.

rtmistler · 05-12-2015, 07:04 AM

Can't yet compile this in a test program to provide any assistance. I run into a problem understanding what hexCh is. What headers do you use when you compile this program?

Note that my strategy here would be to run in GDB and step through it to see the construction of LRCacsii on a case by case basis. I.e. my main simply calls this function four or five times using each of your test input strings.

Have you tried to run under GDB and stepped through this program?

Have you just put debug output for LRCascii as it gets constructed?

pan64 · 05-12-2015, 07:54 AM

you need to print partiehexa and reste just to check their contents, looks like one of them is out of range and there is no error check in your code.

wowy · 05-12-2015, 08:56 AM

Quote:

Originally Posted by rtmistler

Can't yet compile this in a test program to provide any assistance. I run into a problem understanding what hexCh is. What headers do you use when you compile this program?

Note that my strategy here would be to run in GDB and step through it to see the construction of LRCacsii on a case by case basis. I.e. my main simply calls this function four or five times using each of your test input strings.

Have you tried to run under GDB and stepped through this program?

Have you just put debug output for LRCascii as it gets constructed?

Sadly i use a kit of qt with a faulty debugger and no rights on this session to modify it right now.

The header i use are (some of them useless here for this function):
include <iostream>
#include <stdint.h>
#include <stdio.h>
#include <sstream>
#include <string>
#include <vector>
#include <fstream>

It seems that if i replace :

Code:

if (partiehexa<10) {LRCascii+=temp2;}

by :

Code:

if (partiehexa<10) 
{ 
    std::stringstream ss3;
    ss3 << partiehexa;
    ss3 >> temp2;
    LRCascii += temp2;
}

The problem is solved. So it works but for me this is not logical, since the modification is really close to what it was before.

johnsfine · 05-12-2015, 10:57 AM

First, this is an amazingly ugly way to convert a number into two digits in ascii hex. That ought to be a trivial task (many different trivial ways are possible) yet you made it both difficult and wrong. Was there a reason?

Second, you have shown something is wrong with a specific small bit of code you posted:

Code:

std::stringstream ss2;
ss2 << reste;
ss2 >> temp;
ss2 << partiehexa;
ss2 >> temp2;

I don't use stringstream often enough myself to recognize what is wrong. I've never reversed direction repeatedly like that and don't even know if that is supported.

If I were a bit more motivated, I might look it up and/or test it. But I'm not that motivated. You could test it yourself.

reste is a one or two digit int. partiahexa is a one digit int. temp and temp2 are std::string.

You have indicated the above code puts the "wrong" thing into temp2. You want it to put (std::string)( (char)('0'+partiahexa) ) into temp2. It should be easy to test what it actually puts into temp2. But why you ever even needed temp2 is questionable.