help on debugging

hello, i dont know if its righ to say sth like this here , so im sorry if i shouldnt have posted this !
anyway ,i need to debug my program , (current project) , but the catch here is i dont know what those debugging stuffs are ! i start debuging some windows poping up and say sth , but i dont understand it , if there is anyone who can help me understand where the problem is and what it is , i will be very thankful ,

here are some screen shots and also some text file that i dont know how to work with them ! i dont understand them !
( if the source code can help you , i can give you that, (its not complete, and is some how very messed up! it needs more work, but some member functions are complete and should work , i just dont understand why i get a "send error, dont send " windows!!)
when i entered some instruction , after that the program did its work, it crashes!

Code:

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http://master.huricane.googlepages.com/fulllength.GIF

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while debugging! stuff i dont understand!

Code:

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http://master.huricane.googlepages.com/screenshot.GIF

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callstack.txt

debugassembely.txt
and this is the projects exe file so far . if you want to see it error! here it is:
claexe.rar
you should start with ORG a number (e.g 10 or 88)
then enter your instructions .. you can find some instruction here .: #4

again , i would be very thankful if any one tell me what those texts mean !
--------------------------------
by the way , gotta tell you sth , i fyou disable the loop in main.cpp, the program runs once, so far so good, and it wont generate any error!!(i mean dont send error! ) . but you know i=its not good either m cause i need to process all of the strings ,
(fetch is giving one string at each loop, and the other members work with this string , the loops continue until the fetch reaches "exit". exit must not be processed !this is a brief summery of what i was doing there! i know its realllly messed up , meany redeclarations !but believe me , i had no ther choices, i didnt know debugging , so that my way of finding error! sorry for being that messed up.!!)

Actual code would help.

Quote:

---

Originally Posted by Salem

Actual code would help.

---

tanx for your attention
REVISEDCLAnewsearch.rar
REVISEDCLAnewsearch.zip

by the way , gotta tell you sth , i fyou disable the loop in main.cpp, the program runs once, so far so good, and it wont generate any error!!(i mean dont send error! ) . but you know i=its not good either m cause i need to process all of the strings ,
(fetch is giving one string at each loop, and the other members work with this string , the loops continue until the fetch reaches "exit". exit must not be processed !this is a brief summery of what i was doing there! i know its realllly messed up , meany redeclarations !but believe me , i had no ther choices, i didnt know debugging , so that my way of finding error! sorry for being that messed up.!!)
------------------------
new problem i found , when you enter the value of a variable in declaration e.g num, 65 or 12 or 15 it crashes!!buf if you enter 99 or any thing else and plus if you enter one digit numbers such as 5 or 7 or 8 , it wont generate error !! im reallly confused. i checked the number conversion function works! it convers 86 or stuff . but i dont know why it generates error and crashes on 65! or 10 or 12(these are decimal so they must be converted to octal! )!!
sample input !:
(ORG 10
CLA
RED num
num: 5
exit

Runs fine. to the end , and processes all of the strings, but crashes at the end.
------------------

ORG 10
CLA
RED num
num: 65
exit
Generates Error! before reaching the end and process all of the strings it crashes
--------------------------
ORG 10
CLA
RED num
num: 99
exit
conversion occured! and rest of the info. at the end program crashes.
Runs fine again!
)

Code:

---

      while (CLAEX.Fetch()!=0)
        {
                //اکه فچ بره آخر و حلقه رو هم غیر فعال کنیم .ارروی نمیده و خط او
ل چک میشه .اما اگه بره اخر حتی اونخطر اخر هم چک نمیشه.
                CLAEX.Fetch();
                CLAEX.AnalizeString();
                CLAEX.Tokenizer();
                CLAEX.machinLanguageConvertor();
                CLAEX.MemoryDumper();
              //CLAEX.Fetch();
        }
        CLAEX.ExecutionEngine();

---

If you look at the code for CLAEX.Fetch(), you'll see that it modifies the state of the class. I didn't investigate in too much detail, but I think that perhaps you should only call Fetch once, i.e.

Code:

---

      while (CLAEX.Fetch()!=0)
        {
                //اکه فچ بره آخر و حلقه رو هم غیر فعال کنیم .ارروی نمیده و خط او
ل چک میشه .اما اگه بره اخر حتی اونخطر اخر هم چک نمیشه.
                //CLAEX.Fetch(); do not call Fetch() again
                CLAEX.AnalizeString();
                CLAEX.Tokenizer();
                CLAEX.machinLanguageConvertor();
                CLAEX.MemoryDumper();
              //CLAEX.Fetch();
        }
        CLAEX.ExecutionEngine();

---

Quote:

---

Originally Posted by dwks

Code:

---

      while (CLAEX.Fetch()!=0)
        {
               
              //CLAEX.Fetch();
                CLAEX.AnalizeString();
                CLAEX.Tokenizer();
                CLAEX.machinLanguageConvertor();
                CLAEX.MemoryDumper();
              //CLAEX.Fetch();
        }
        CLAEX.ExecutionEngine();

---

If you look at the code for CLAEX.Fetch(), you'll see that it modifies the state of the class. I didn't investigate in too much detail, but I think that perhaps you should only call Fetch once, i.e.

Code:

---

      while (CLAEX.Fetch()!=0)
        {
             
                //CLAEX.Fetch(); do not call Fetch() again
                CLAEX.AnalizeString();
                CLAEX.Tokenizer();
                CLAEX.machinLanguageConvertor();
                CLAEX.MemoryDumper();
              //CLAEX.Fetch();
        }
        CLAEX.ExecutionEngine();

---

---

tanx dear dwk, i commented that part too, but again at the end it crashes!(dont send error window appears,)
( by the way , why is it that after 3 or 4 times of compiling and running the program, at the fifth try, it doesnt even scan these commands (upon reaching REM num e.g, it crashes) and generates errors! is it because of the fault the program is suffering?, cause when i log off and on again , for 3 times again , its works fine, but if you do try more , it crashes more and more even before program reaches the end of the map!)

anyway thanks so far dears.im waiting for your comments.

Well if it's a zip, I might download it. But I'm not interested enough to download and install whatever I need to unpack a rar file just to help you.
Don't make it hard for people to help you.

Who does not have rar/7z these days?
On a school, or a workplace, I might believe you, but on a home computer?

again im really thank you for your attention , while in a deep desperation and confusion ,i extremely thankyou for sharing your precious time with me for solving this problem .

ive uploaded the zip file too, hope you help me find the problem.

ive faced another error . ( i mean an error that kept happening before and after a while after some changes, it disappeared i thought . now it just poped up. and i found what was the cause.
when i enter number 65 or 10 or 12 as the variables value in a declaration, the program crashes immediately by typing exit,(and thus executing the rest of the functions!)

EDITED : using stringstream , the error is gone , now they are all right ! i used itoa() improperly and the samples below showed up! , but upon using string stream , they are gone .! thanks all

for example :
//declaring 1 digit number is harmless!! to the program, it process all of the strings, and at the end it crashes.

ORG 10
CLA
RED num
num: 5
exit

Runs fine. to the end , and processes all of the strings, but crashes at the end.
------------------

ORG 10
CLA
RED num
num: 65
exit
Generates Error! before reaching the end and process all of the strings it crashes
--------------------------
ORG 10
CLA
RED num
num: 99
exit
conversion occured! and rest of the info. at the end program crashes.
Runs fine again!
)

its really really confusing me, because i dont see any difference between 65 and 99 or 86 .
99 and 86 are fine but 65 not!!
you know what. i checked it . the program converts the number 65 to octal but after that there is an error ( it doesnt even show the conversion occured! message! as it does in 99 and 86s !) some time loging off an on will make it right!! and some time no!! just by chance i think!!

edited! double post

I can't even get this to compile under Linux due to itoa; around 310 lines of warnings though.

Here's something, in SearchEngine.h:

Code:

---

else if( message != "no" )
{
    //if it was octal . well do the routin .copy it to offset,convert it to binary
    char * oo;
    char * o;
    o = itoa(temp,oo,8);    //itoa(number,a buffer to hold the string,the base you want the number to be saved in buffer)
    offsetptr = o;//itoa(temp,ptr,8);    //itoa(number,a buffer to hold the string,the base you want the number to be saved in buffer)
    //cout<<"\nThis is in Search,"<<offsetptr;
}
...
return offsetptr;

---

You need to allocate storage for the buffer passed to itoa.
You're returning a pointer to a local variable.

Quote:

---

Originally Posted by rags_to_riches

I can't even get this to compile under Linux due to itoa; around 310 lines of warnings though.

Here's something, in SearchEngine.h:

Code:

---

else if( message != "no" )
{
    //if it was octal . well do the routin .copy it to offset,convert it to binary
    char * oo;
    char * o;
    o = itoa(temp,oo,8);    //itoa(number,a buffer to hold the string,the base you want the number to be saved in buffer)
    offsetptr = o;//itoa(temp,ptr,8);    //itoa(number,a buffer to hold the string,the base you want the number to be saved in buffer)
    //cout<<"\nThis is in Search,"<<offsetptr;
}
...
return offsetptr;

---

You need to allocate storage for the buffer passed to itoa.
You're returning a pointer to a local variable.

---

im using gcc under windows, it should compile nder linux too!! and by the way for me it doesnt generate any warning concerning itoa() !! and most of the warnings are because of comaprison betwean size_t found and -1!

after all , how to allocate that! i dont understand! what should i do ?

You need to make sure that the variable oo that you pass to itoa represents some real memory - oo currently points at some random location, which is highly likely not a good place to store the string produced by itoa.

--
Mats

how to do that!? im not familiare with char * string initialization! will
int i;
char * oo =& i ; ! work? !!

Quote:

---

Originally Posted by Masterx

how to do that!? im not familiare with char * string initialization! will
int i;
char * oo =& i ; ! work? !!

---

In theory, that would work. I'd probably use char oo[13] instead [13 is long enough for a 32-bit integer].

However, you return 'o', which is a pointer to the oo variable (yes, itoa() returns the string it got as buffer, which is oo), so you will still return a pointer to a local variable, which is bad. You either should pass a character array to the function, or call malloc() to get a permanent piece of memory (just make sure you free it later).

--
Mats

Quote:

---

Originally Posted by Masterx

how to do that!? im not familiare with char * string initialization! will
int i;
char * oo =& i ; ! work? !!

---

You know, this code is full of char * (C-style strings) usage. Sorry, but if you don't know how to use them, you're destined to fail. This looks like a complicated project to undertake if you don't understand such a rudimentary part of C & C++ programming. If you're used to using the C++ string class and its relatives, then I would use those instead.

Quote:

---

Originally Posted by matsp

In theory, that would work. I'd probably use char oo[13] instead [13 is long enough for a 32-bit integer].

However, you return 'o', which is a pointer to the oo variable (yes, itoa() returns the string it got as buffer, which is oo), so you will still return a pointer to a local variable, which is bad. You either should pass a character array to the function, or call malloc() to get a permanent piece of memory (just make sure you free it later).

--
Mats

---

tanx dear Mats, but when i did

Code:

---

              char  oo[13];
                        char * o;
                        o = (char*) malloc (13);
                        if (o==NULL) exit (1);

                        o = itoa(temp,oo,8);    //itoa(number,a buffer to hold the string,the base you want the number to be saved in buffer)
                        offsetptr = o;//itoa(temp,ptr,8);    //itoa(number,a buffer to hold the string,the base you want the number to be saved in buffer)
                        free (o);

---

it didnt work. it just even corrupts the output! i used to get before changing this !

Quote:

---

Originally Posted by rags_to_riches

You know, this code is full of char * (C-style strings) usage. Sorry, but if you don't know how to use them, you're destined to fail. This looks like a complicated project to undertake if you don't understand such a rudimentary part of C & C++ programming. If you're used to using the C++ string class and its relatives, then I would use those instead.

---

yeah , that why i asked for your help! :(

Why are you refusing to use std::string and standard containers? Read Stroustrup's answer to the FAQ: How do I deal with memory leaks?

Oh, and what's with o and oo? Read Stroustrup's answer to the FAQ: How do you name variables?

Quote:

---

Originally Posted by laserlight

Why are you refusing to use std::string and standard containers? Read Stroustrup's answer to the FAQ: How do I deal with memory leaks?

Oh, and what's with o and oo? Read Stroustrup's answer to the FAQ: How do you name variables?

---

tanx, i am using standard container!! and im using strings! too, i cant use std::string because of the conversions i have to deal with . atoi() only accepts C style strings , or and also itoa() . thats my worst luck ever!

and about o and oo, i used them temporarily just to make that itoa() function to work, i was just testing , and just it happend and it worked, after that , i didnt change them !! ok , ill change their name,

and after all , do i get these errors because i do have a memory leak or sth in my program !! ?
by the way , when i used

Code:

---

cstr = new char [str.size()+1];
  strcpy (cstr, str.c_str());

---

in my program , ( more than 20 times)
i didnt write "delete[] cstr;" after using them , is it importatn to do so ? i mean could it be the cause to crash ? !!?

and why do i get that error in malloc? what did i do wrong?

1. Is there a reason to use itoa and atoi? If you're using strings, why not use stringstreams?

2. If you're not seeing the warnings, why not use -Wall -pedantic flags when you compile?

Quote:

---

Originally Posted by Masterx

tanx, i am using standard container!! and im using strings! too, i cant use std::string because of the conversions i have to deal with . atoi() only accepts C style strings , or and also itoa() . thats my worst luck ever!

---

Look at it from another perspective: you can't use atoi() because it only accepts C style strings (but then you can get an unmodifiable C style string from a std::string, so that is not entirely true), and you can't use itoa() because it produces a C style string (aside from the fact that it is non-standard).

I believe we have mentioned this before, but the solution is to use a stringstream, or if you are willing to use a non-standard component, then you can use boost::lexical_cast.

Quote:

---

Originally Posted by Masterx

and after all , do i get these errors because i do have a memory leak or sth in my program !! ?

---

They are probably more likely due to an attempt to access memory that you do not own.

Quote:

---

Originally Posted by Masterx

i didnt write "delete[] cstr;" after using them , is it importatn to do so ? i mean could it be the cause to crash ? !!?

---

It is good practice to do so, but it is even better practice to use a class that manages memory for you.

im going to use string stream instead of itoa(). wish me luck .
.please sb tell me if im wrong .

Code:

---

#include <iostream>
#include <sstream>

int main() {
  int number = 123;

  std::stringstream ss;
  ss << number;

  std::cout << ss.str() << endl;
}

---

we have an integer number. and we want to convert it to a string ! so we use ss<<number; to copy the number value to ss , . now whats the ss.str()? can i simply write
std::offset = std::ss now?
will it work?

Quote:

---

Originally Posted by Masterx

tanx, i am using standard container!! and im using strings! too, i cant use std::string because of the conversions i have to deal with . atoi() only accepts C style strings , or and also itoa() . thats my worst luck ever!

---

atoi DOES work with std::string:
But it accepts C-style strings, so use the c_str() member function to get a C-style string from a std::string object.
itoa is non-standard, so it should not be used anyway.
But better yet, get rid of both atoi and itoa and use boost::lexical_cast. It will serve the purpose of both - and more!

Quote:

---

and after all , do i get these errors because i do have a memory leak or sth in my program !! ?
by the way , when i used

Code:

---

cstr = new char [str.size()+1];
  strcpy (cstr, str.c_str());

---

in my program , ( more than 20 times)
i didnt write "delete[] cstr;" after using them , is it importatn to do so ? i mean could it be the cause to crash ? !!?

---

Of course you must! Otherwise there will be a memory leak!
But you should not do it in the first place (using dynamic memory with new).

Quote:

---

and why do i get that error in malloc? what did i do wrong?

---

malloc should not be used in C++, either. If you need to use dynamic memory, use std::vector. In rare cases where it will not do, use new.

Quote:

---

Originally Posted by Elysia

atoi DOES work with std::string:
But it accepts C-style strings, so use the c_str() member function to get a C-style string from a std::string object.
itoa is non-standard, so it should not be used anyway.
But better yet, get rid of both atoi and itoa and use boost::lexical_cast. It will serve the purpose of both - and more!


Of course you must! Otherwise there will be a memory leak!
But you should not do it in the first place (using dynamic memory with new).


malloc should not be used in C++, either. If you need to use dynamic memory, use std::vector. In rare cases where it will not do, use new.

---

first of all tanx . but second .im really dieing !!!im owerhelmed ! i learn one thing , another just pops up!
well, when trying to use string stream , i encountered a problem !

Code:

---

|80|error: ambiguous overload for 'operator=' in 'offsetptr = ss'|

---

edited : i used offsetptr = ss.str(); and voila it works ! i successfully took a step ahead .! thank GOD. and thank you guys too, !
when i tried to std::offsetptr = ss;!
any reason!

and third : boost::lexical_cast, how should i know if i have it ! if i dont where can i get it ! ?

and at last ,im trying to remove any C style string that i used, let me see if i can do such a thing at my current stage !! :(

You use >> to get something out of a stringstream.

Boost.org
whatever_type_I_want var = boost::lexical_cast<whatever_type_I_want>(source);
Just remember that it throws.

Quote:

---

Originally Posted by tabstop

You use >> to get something out of a stringstream.

---

tanx .dear tabstop. and Oh , GOD i cant believe it , it just wiped out one of the most recent annoying errors i used to get!! i mean this one( #8) is solved now .
im going to change all of the C based stuff now. to see if the errors will be gone !

Code:

---

              char  oo[13];
                        char * o;
                        o = (char*) malloc (13);
                        if (o==NULL) exit (1);

                        o = itoa(temp,oo,8);    //itoa(number,a buffer to hold the string,the base you want the number to be saved in buffer)
                        offsetptr = o;//itoa(temp,ptr,8);    //itoa(number,a buffer to hold the string,the base you want the number to be saved in buffer)
                        free (o);

---

There are so many things wrong with this code that I hardly know where to start.
First of all, itoa() returns a pointer to the same address as the input buffer (oo). So, you must not malloc() o, and then assign it with the return from itoa() - it's the same as assigning o = oo after malloc. Then you set offsetptr to o, followed by free(o) - so what memory [assuming the other stuff wasn't all wrong] is o pointing to? Who owns it?

Yes, you must free memory allocated by malloc, but NOT UNTIL YOU HAVE FINISHED USING IT!

Now, it looks like you are planning to use C++ style strings, so it's a moot point anyways, but you it really seems like you need to learn a bit more about the basics of memory and how you deal with it before you start on a complex programming project.

--
Mats

tanx dear mats, in this case i would appreciate a proper and yet sample demosntrating how to use malloc() just for learning purposes, im not gona use it anymore.

and the rest :
ive disabled all of the C string variables so far , and used their equivalent C++ counter parts ,(str::string) .
i even managed to replace the atoi() with the new string stream based Atoi() function i modified :

Code:

---

#include <cstdlib>
#include <iostream>
#include <sstream>
using namespace std;

int Atoi(string input)
{
        int temp;
            stringstream ss( input);
            ss>>temp;
            return temp;
}

---

it was primarily accepting a char * as an argumnet , but i changed it into string ! i dont know whether this makes an error or not !, so please tell me about this too.

and now, im having a slight big problem using stringstream in converting octal to decimal. the problem is i dont know , how im supposed to make the string stream understand that , it should treat the number a octal not decimal as default!
well conversion from decimal to octal is fine , because it treats the number as decimal at first ( as the number itself is decimal indeed!)
but when i have an octal number, how can i do this ? i mean simply doing :

Code:

---

#include <cstdlib>
#include <iostream>
#include <sstream>
using namespace std;

int main(int argc, char *argv[])
{
   
    int num = 12;


    std::stringstream stream;
    stream <<dec<<num;
    std::cout << stream.str();
   
    system("PAUSE");
    return EXIT_SUCCESS;
}

---

wont give me a correct answer though!
but this program is just fine with stream<<oct<<num, meaning the conversion from decimal to octal.! so im stuck in it !

and by the way , is not a function for converting octal to binary just like the above ! ?
im asking these, because im revising my whole project ! so i do change the functions , and im trying to use standard stuff at first.

( i cant download boost library, im using Dialup now, so , please help me with these stuff. tanx .
-------------------------
logs:
replaced itoa() with a better equivalent by string stream
replace atoi() with a stringstream based function . (Atoi(). i)
replaced all of C style string variables with their std::string counterparts.
DectoOctal() is removed and instead the new function :

Code:

---

string DectoOct(int dec)
{
    std::stringstream stream;
    stream <<std::oct<<dec;
    return stream.str();

}

---

is used.
meanwhile im seeking a better function for Octal to bin
the current function i have for cotal to binary conversion is :

Code:

---

string OcttoBin(int n)
{


        int a[6],i=0,t=0,num;
        string str,str1,str2,message;

        char buffer[9];

        string str_array[9];
        const size_t str_array_size = sizeof(str_array) / sizeof(str_array[0]);

        while(n!=0)
        {
                a[i]=n%10;
                n=n/10;
                if(a[i]>7)
                t=1;
                i++;
        }
        i--;
        if(t==0)
          for(;i>=0;i--)
          {
                  switch(a[i])
                  {
                          case 0:
                          if (!str.empty())
                          str.append("000");
                          else
                          str="000";
                          break;

                          case 1:
                          if (!str.empty())
                          str.append("001");
                          else
                          str="001";
                          break;

                          case 2:
                          if (!str.empty())
                          str.append("010");
                          else
                          str="010";
                          break;

                          case 3:
                          if (!str.empty())
                          str.append("011");
                          else
                          str="011";
                          break;

                          case 4:
                          if (!str.empty())
                          str.append("100");
                          else
                          str="100";
                          break;

                          case 5:
                          if (!str.empty())
                          str.append("101");
                          else
                          str="101";
                          break;

                          case 6:
                          if (!str.empty())
                          str.append("110");
                          else
                          str="110";
                          break;

                          case 7:
                          if (!str.empty())
                          str.append("111");
                          else
                          str="111";
                          break;
                  }
          }
          str1="000000";
          str2="000";
              if (str.length()==3)
              {
                        str1.append(str);
                        str=str1;
              }

              if (str.length()==6)
              {
                        str2.append(str);
                        str=str2;
              }



        if(t==1)
          {

                  message="no";
                // cout<<"Not a Octal number\n";
                return s;
          }

return str;
}

---

and the last one is to find a way to get stringstream to work properly as i said above.

your problem not in the stringstream

int num = 12;

12 is decimal constant - so you already have a decimal 12 in your var when starting using the stringstream and it is late to recognize 12 as oct or hex

change that line to
int num = 0x12;
for example to represent 12 as hex number

Quote:

---

Originally Posted by vart

your problem not in the stringstream

int num = 12;

12 is decimal constant - so you already have a decimal 12 in your var when starting using the stringstream and it is late to recognize 12 as oct or hex

change that line to
int num = 0x12;
for example to represent 12 as hex number

---

well , tanx, but how am i supposed to do such a thing , when i have a variable that gets its value form some other calculations! and its not just that simple!
e.g suppose " num " has an octal value inside it(e.g 12 is octal here and it = 10 decimal!)! , so in this case , im asking for help!because i cant just add O before the number to imply its octal! so how to convert it now ! how to let the function understand its dealing with an octal integer!?

and any suggestion about octal to binary conversion?

Well, you either have a BINARY representation of the number, in an int or some such. Or you have a string representation, in which case you need to know what base the number is (by prefix or implicitly by context or some such).

--
Mats

Quote:

---

Originally Posted by matsp

Well, you either have a BINARY representation of the number, in an int or some such. Or you have a string representation, in which case you need to know what base the number is (by prefix or implicitly by context or some such).

--
Mats

---

Quote:

---

Well, you either have a BINARY representation of the number, in an int or some such. Or you have a string representation,

---

i have an int which would be converted to a string (containing the numbers binary representation )

Quote:

---

in which case you need to know what base the number is (by prefix or implicitly by context or some such).

---

r by context !
( if i got you right!)

So, let's try to explain in a different way.

Right, so if you have an integer, and you set it x = 12, then x has the value 12 decimal. It is also 14 octal, 0x0C in hex, 1100 in binary and 10 in base 12.

If you have a string with the number '1', '2' in it, it would depend on what the base is, what the value of the number is.
Here's a list of base vs. decimal:

Code:

---

base value
10  12
3    5
8    10
16  18

---

--
Mats

Quote:

---

Originally Posted by matsp

So, let's try to explain in a different way.

Right, so if you have an integer, and you set it x = 12, then x has the value 12 decimal. It is also 14 octal, 0x0C in hex, 1100 in binary and 10 in base 12.

If you have a string with the number '1', '2' in it, it would depend on what the base is, what the value of the number is.
Here's a list of base vs. decimal:

Code:

---

base value
10  12
3    5
8    10
16  18

---

--
Mats

---

tanx, right here !

Quote:

---

if you have an integer, and you set it x = 12, then x has the value 12 decimal. It is also 14 octal

---

yeah this is my problem , i know about strings , and i do understand what you mean .when there is a string containing "123" or stuff if it can be in any base!
this is straight forward!
but when i have an integer that simultaneously can be interpreted as"decimal" and again as Octal, how can i just make the difference and make it understandable for the stringstream function which has the duty to convert between octal and decimals ! could i convey what i wanted to say ?

Quote:

---

to convert between octal and decimals

---

you cannot conver integer between that and this...

integer is always stored as binary...

what you could do - to take integer that contains say 12 and represent it in string as
"12" or "014" or "0xC"

That is what you are trying to do?

If you have an integer, then the value is stored in binary. Most often, however, we present that binary number as a decimal, hex or octal number because humans tend to get confused with very large numbers - and a relatively small number of 146 is still an 8-digit number in binary, so it's not that easy to read even for such a limited number.

You can make streams use octal for input or output (e.g. if your stringstream contains the letter '1', '2', you can read that into an integer and get it to have the binary value 1010, or which would be 10 decimal).

Code:

---

strinstream ss;
int n;
ss << "12";
ss >> oct >> n;
cout << "n = " << n;

---

I may have misunderstood completely what you want to do, but I hope I'm actually helping ;)

--
Mats

Quote:

---

Originally Posted by vart

you cannot conver integer between that and this...

integer is always stored as binary...

what you could do - to take integer that contains say 12 and represent it in string as
"12" or "014" or "0xC"

That is what you are trying to do?

---

well , im trying to make a better counter part for this function with the help of stringstream or any other standard function (except boost , cause i dont have access to it)

Code:

---

int OcttoDec(int oct)
{
        int n,r,s=0,i;
        n=oct;
        double S=0;
        for(i=0;n!=0;i++)
        {
                r=n%10;
                s=s + r * ((pow(8.0,i)));
                n=n/10;
        }
        return static_cast<int>(s);
}

---

as you see, im seeking a way to convert an integer Octal value to an integer decimal value ! . or convert an integer octal value to a decimal string!

just like what i do above or below:

Code:

---

string DectoOct(int dec)
{
    std::stringstream stream;
    stream <<std::oct<<dec;
    return stream.str();

}

---

( if i have it in string , big deal i can convert it to integer using my state of the art Atoi() function . so it doesnt make any differecen though .! but i prefer to have sth like the second function for conversion , , did i convey it completley now?

Quote:

---

Originally Posted by vart

you cannot conver integer between that and this...

integer is always stored as binary...

what you could do - to take integer that contains say 12 and represent it in string as
"12" or "014" or "0xC"

That is what you are trying to do?

---

10 i would say (octal to decimal! )

Quote:

---

Originally Posted by matsp

If you have an integer, then the value is stored in binary. Most often, however, we present that binary number as a decimal, hex or octal number because humans tend to get confused with very large numbers - and a relatively small number of 146 is still an 8-digit number in binary, so it's not that easy to read even for such a limited number.

You can make streams use octal for input or output (e.g. if your stringstream contains the letter '1', '2', you can read that into an integer and get it to have the binary value 1010, or which would be 10 decimal).

Code:

---

strinstream ss;
int n;
ss << "12";
ss >> oct >> n;
cout << "n = " << n;

---

I may have misunderstood completely what you want to do, but I hope I'm actually helping ;)

--
Mats

---

many tanx dear Mats, this would be very useful when i have the number in decimal, but the reverse is not true , meaning i fthe number stored in integer is 12, i must be treated as octal! so when it is octal , and i use sush a thing , this would just simply act as if it was decimal! ( this is exactly my problem , just this! the reverse of the example wont work.) meaning

Code:

---

strinstream ss;
int n;
ss << "12";
ss >> dec >> n;
cout << "n = " << n;

---

wont work simply !

you mean like this?

Code:

---

int OcttoDec(int y)
{
    std::stringstream test;
    test << y;

    int x;
    test >> std::oct >> x;
    return x;
}

---

Quote:

---

Originally Posted by vart

you mean like this?

Code:

---

int OcttoDec(int y)
{
    std::stringstream test;
    test << y;

    int x;
    test >> std::oct >> x;
    return x;
}

---

---

yeah ,exactly , i wrote instead :

Code:

---

    stringstream ss;
    ss << num; //get the number ,( if you think this is a decimal number, ),
    ss>>oct>>num; //save it as octal
   
    ss<<dec<<num;//then retrieve it as decimal
    ss<< num;//palce it in num again
    cout<<ss.str();

---

just doesnt work!from octal to decimal !!
by the way your code is working just fine ! great, but mine ! no, why ?

Quote:

---

Originally Posted by Masterx

by the way your code is working just fine ! great, but mine ! no, why ?

---

Just compare line by line

Quote:

---

ss<< num;//palce it in num again

---

comment specifies opposit action to the made by code

Quote:

---

Originally Posted by vart

Just compare line by line


comment specifies opposit action to the made by code

---

tanx, about the comment, its just a mistake , ( i changed the code , but didnt remove the comment) .

so comparing the lines with each other, i just did the reverse huh ! ?

now, how an i convert an octal to binary using stringstream? is it even possible ?
ive posted my octal to binary function already (here it is again) , is there any better function doing the job?

Code:

---

string OcttoBin(int n)
{


        int a[6],i=0,t=0,num;
        string str,str1,str2,message;

        char buffer[9];

        string str_array[9];
        const size_t str_array_size = sizeof(str_array) / sizeof(str_array[0]);

        while(n!=0)
        {
                a[i]=n%10;
                n=n/10;
                if(a[i]>7)
                t=1;
                i++;
        }
        i--;
        if(t==0)
          for(;i>=0;i--)
          {
                  switch(a[i])
                  {
                          case 0:
                          if (!str.empty())
                          str.append("000");
                          else
                          str="000";
                          break;

                          case 1:
                          if (!str.empty())
                          str.append("001");
                          else
                          str="001";
                          break;

                          case 2:
                          if (!str.empty())
                          str.append("010");
                          else
                          str="010";
                          break;

                          case 3:
                          if (!str.empty())
                          str.append("011");
                          else
                          str="011";
                          break;

                          case 4:
                          if (!str.empty())
                          str.append("100");
                          else
                          str="100";
                          break;

                          case 5:
                          if (!str.empty())
                          str.append("101");
                          else
                          str="101";
                          break;

                          case 6:
                          if (!str.empty())
                          str.append("110");
                          else
                          str="110";
                          break;

                          case 7:
                          if (!str.empty())
                          str.append("111");
                          else
                          str="111";
                          break;
                  }
          }
          str1="000000";
          str2="000";
              if (str.length()==3)
              {
                        str1.append(str);
                        str=str1;
              }

              if (str.length()==6)
              {
                        str2.append(str);
                        str=str2;
              }



        if(t==1)
          {

                  message="no";
                // cout<<"Not a Octal number\n";
                return message;
          }

return str;
}

---

I'm even more confused. If you have a integer, it is stored in binary in the computer. If it prints as 12, 14, 0x0c or something else is presentation. If you have a number you want to present in octal, then use the oct. You can not "convert an integer in decimal to octal", because integers are stored internally in the computer as binary. If you want to display an integer as binary, then do this:

Code:

---

std::string result;
int x = 12;  // Or 012 if you want it to be octal. 0x12 for hex.
const int numdigits = 10;  // Number of bits you want to display.
for(i = 0; i < numdigits; i++)
{
    result += ('0' + (x & 1)) + result;
    x >>= 1;
}

---

--
Mats

Something like

Code:

---

std::string OcttoBin2(int n)
{
        std::bitset<CHAR_BIT * sizeof(int)> bs(OcttoDec(n));

        std::stringstream test;
        test << bs;
        return test.str();
}

---

Quote:

---

Originally Posted by matsp

I'm even more confused. If you have a integer, it is stored in binary in the computer. If it prints as 12, 14, 0x0c or something else is presentation. If you have a number you want to present in octal, then use the oct. You can not "convert an integer in decimal to octal", because integers are stored internally in the computer as binary. If you want to display an integer as binary, then do this:

Code:

---

std::string result;
int x = 12;  // Or 012 if you want it to be octal. 0x12 for hex.
const int numdigits = 10;  // Number of bits you want to display.
for(i = 0; i < numdigits; i++)
{
    result += ('0' + (x & 1)) + result;
    x >>= 1;
}

---

--
Mats

---

well i got mixed up !!! wait a minute , if its not conversion from decimal to octal and vice versa, how does the oct do the magic? it just gives me a number that is not the previous one any more, i mean , when i enter 10 as decimal , ( here i am the only one who knows the base , ) and when i use oct. teh output number will be 12 , which is to me an octal number in integer! so then it would be stored in memory just like "001 010" representing an octal value( i think). am i right?! so what do we call this process, ? we do arithmatics on these figures i mean scintific stuff works in different bases, so if this is not conversion , what is it?, !i mean , you know , well from what you ve mentioned, i gather, to the system teh binary matters, ! no octal or decimal or stuff. , when we are explicitly working with octal and stff, system works with binary anyway, but it represents as if it is octal! is it right?!!? ( exteremly confused! sorry for being stupid! i just dont understand!)

Quote:

---

Originally Posted by vart

Something like

Code:

---

std::string OcttoBin2(int n)
{
    std::bitset<CHAR_BIT * sizeof(int)> bs(OcttoDec(n));

    std::stringstream test;
    test << bs;
    return test.str();
}

---

---

it calls octal to dec and then use the output to get the its binary form?!

If you enter 10 (decimal) to a input operator, it will (unless you have specified otherwise) take it as a decimal number. It will be stored as 000...01010 internally in the computer. If you then print that using oct, it will come out in octal. The octal representation of 00...001010 is 12. So, the number stored in the integer is exactly the same whether you entered 0x0a, 012 or 10 [assuming the machine understands that 0x.. is hex, and 0... is octal, and not starting with 0 is decimal].

Integers, no matter how they got into the computer, are stored as binary. You can choose, when you output them and when you input them, given the right input and output methods, to DISPLAY the number in different bases. But it doesn't really change the internal value of the number.

To explain it a different way: We have a piece of wood, that happens to be two meters long. Now, we can express that as 2000 mm, 200 cm or 2m. Or as 6'6" or 78". But it's still the same piece of wood, whichever unit we describe it's length with. The same applies to displaying numbers in computers - we can use "any unit", but the actual number is the same either way.

--
Mats

Quote:

---

Originally Posted by matsp

If you enter 10 (decimal) to a input operator, it will (unless you have specified otherwise) take it as a decimal number. It will be stored as 000...01010 internally in the computer. If you then print that using oct, it will come out in octal. The octal representation of 00...001010 is 12. So, the number stored in the integer is exactly the same whether you entered 0x0a, 012 or 10 [assuming the machine understands that 0x.. is hex, and 0... is octal, and not starting with 0 is decimal].

Integers, no matter how they got into the computer, are stored as binary. You can choose, when you output them and when you input them, given the right input and output methods, to DISPLAY the number in different bases. But it doesn't really change the internal value of the number.

To explain it a different way: We have a piece of wood, that happens to be two meters long. Now, we can express that as 2000 mm, 200 cm or 2m. Or as 6'6" or 78". But it's still the same piece of wood, whichever unit we describe it's length with. The same applies to displaying numbers in computers - we can use "any unit", but the actual number is the same either way.

--
Mats

---

thanks thanks thanks , thanks ! i just got it dude, im really amazed! so thats how it is done down there !
iwont forget this . really tanx ..Dear Mats. you taught me couple of new things!

hello again , ive almost completely revised the project. removed almost all of the C style string variables, but still i get that error at the end of the program! !
when i pressed debug, the program terminated, and then it pointed at the line: !

Code:

---

  OurString = (*usrstr_it).second;

---

in this function.:

Code:

---

        int CLA::Fetch()
        {
                if (cntt == 0)
                usrstr_it = StringCollector.begin();

                while ( usrstr_it != StringCollector.end() && check2)
                {
                        if (cntt != 0) //for first element.after a successful loop,
//this condition becomes true and next string will be fetched.
                        usrstr_it++;

                        OurString = (*usrstr_it).second; //get the string.
                        if (OurString == "exit" ||OurString == "Exit" ||OurString =="EXIT")
                        {
                              EXIT = true;
                                break;
                        }

                        cntt++;

                        check = 1;      //this  indicates fetch is running and should be running.
                        check2 = false; //geting away from infinite loop.(gets out of the loop when we have he string.


                }
                if ( usrstr_it == StringCollector.end() || EXIT )
                            check = 0; //end of loop . end of stringcollector, the end of string processing
                if (check ==0 )
                condition = false;


                return check;
           
           

        }//end of Fetch();

---

and again it indicates that the loop in main.cpp has a probelm :
ive taken a screenshot , so that you can see the actual line !!

well i dont know what to do now, heres the new source code !:
newCLA.zip

pic2
pic1
pic3
hope you can help me with it.

any help! :(

Code:

---

                        usrstr_it++;

                        OurString = (*usrstr_it).second; //get the string.

---

After incrementing the . . . well, I'm assuming it's an iterator . . . how do you know it's still valid? You should probably check to make sure it's still in range after you increment it.

[edit] That's pretty funny that after some 12 hours, we both post at exactly the same time (I didn't see your post) . . . [/edit]

Quote:

---

Originally Posted by dwks

Code:

---

                        usrstr_it++;

                        OurString = (*usrstr_it).second; //get the string.

---

After incrementing the . . . well, I'm assuming it's an iterator . . . how do you know it's still valid? You should probably check to make sure it's still in range after you increment it.

[edit] That's pretty funny that after some 12 hours, we both post at exactly the same time (I didn't see your post) . . . [/edit]

---

thanks dear dwks,
( about post :yeah i didnt even notice that)

----------------------
edited: Got it , thanks im checking it now!.
===============================
edited. it works like a charm now.

OHHH GOD , Oh GOD , DWKS , YOU ARE AN ANGLE MAN, its Gone , i cant BEleive it , it just happened ! it worked .!! i realllly dont know how to thank you .to thank you alll Dears , my treasures ! ( deeply touched !)

A big big big enormous Thank to you all specially dear DWKS!

But if userit is just before end userit++ will put it AT end. That is not a valid place to look at the content.

--
Mats

Quote:

---

Originally Posted by matsp

But if userit is just before end userit++ will put it AT end. That is not a valid place to look at the content.

--
Mats

---

yeah, so what should i do ? nothing crosses my mind now to fix this!!
at the moment , ill try putting it lets say at the end!. i ll you what happens.(
but if is not right , ( as it is not right! ) how come the program just works fine?!)

changed to :

Code:

---

        int CLA::Fetch()
        {
             
                if (cntt == 0)
                usrstr_it = StringCollector.begin();

                while ( usrstr_it != StringCollector.end() && check2)
                {
                        if (cntt != 0) /*for first element.after a successful loop,
                                            this condition becomes true and next string will be fetched.*/

                        if (usrstr_it == StringCollector.end())
                        {
                                break;
                        }
                        OurString = (*usrstr_it).second; //get the string.
                        if (OurString == "exit" ||OurString == "Exit" ||OurString =="EXIT")
                        {
                              EXIT = true;
                                break;
                        }

                        cntt++;

                        check = 1;      //this  indicates fetch is runing and should be running.
                        check2 = false; //geting away from infinit loop.(gets out of the loop when we have he string.

                        usrstr_it++;
                }
                if ( usrstr_it == StringCollector.end() || EXIT )
                            check = 0; //end of loop . end of stringcollector, the end of string processing
                if (check ==0 )
                condition = false;


                return check;

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and progarm still works fine ! ( i think its better and more secure than before ! many thanks to you Dear Mats.)