hex + Ascii display problem

thanks one addition i want on that is if it was possible to print out number of lines as well on left side before hex (in hex format)  starting from 0? Also what does memset do? I have raised 10 more points...

>>if it was possible to print out number of lines as well on left side before hex

Sure:

   while (f)
   {
        unsigned int nLines = 1;

        int n = 16;
        // array to store the line
        char acText [ 16 + 1]; // plus trailing '0'
        memset ( acText, 0, sizeof ( acText));

        cout << setbase ( ios_base::hex) << nLines << " ";
        for (int i =0; i<n; i++)
        {
             f.get(c); //get text from file
             acText [ i] = c;
             hexed(c); //convert to hex
                       
        }
        cout << acText << endl;

        nLines++;
   }


>>Also what does memset do?

'memset()' initializes the char array with bin. zeros.

BTW, 'setbase ( ios_base::hex)' might actually help you to get rid of the 'hexed()' function by using

cout << setbase ( ios_base::hex) << (int) c;

no i cannot use iso_base::hex or any such manipulators :( for conversion purposes

Well, then we could use

char acLine [ 10];
sprintf ( acLine, "0x%4.4x ", nLines); // change '4' to the witdh you actually want to use

cout << acLines;

if i try to output line numbers with the function you gave to output lines in hex (I made my original hex function to work with it to ouput lines in hex format) with numbers incrementing till eOf. It puts just 1 before each line and doesnt increment with lines..  

while (f)
           {
       unsigned int nLines = 1;

       int n = 16;
       // array to store the line
       char acText [ 16 + 1]; // plus trailing '0'
       memset ( acText, 0, sizeof ( acText));

       inthexed( nLines);
       for (int i =0; i<n; i++)
       {
            f.get(c); //get text from file
            acText [ i] = c;
            hexed(c); //convert to hex
                     
       }
       cout << acText << endl;

       nLines++;
  }

//my original char to hex converter that i used to convert number of lines from decimal to hex

void inthexed(unsigned int c) //char c to un..int c
{
     int LoNibble, HiNibble, LoChar, HiChar; //changed from char to int
     char HexTab[]= "0123456789ABCDEF";
     LoNibble = c & 0x0F;
     HiNibble = (c >>4) & 0x0F;
     LoChar = HexTab[ LoNibble ];
     HiChar = HexTab[ HiNibble ];
cout << HiChar << LoChar  << " " ;
}


what i want number output to be like is in 00000000 (for 0) and .........(000000A0) for 10 and so on format...

Just a little point.

When you want to display the text of the 16 bytes you should probably make sure that 'undisplayable' characters and characters that do things to cursor aren't simply printed out, make sure you replace those with a dot or space or something before you print the text.

This function does a hex display of 16 bytes at a time:

void hexline(unsigned int addr,
             const unsigned char * buf, int n,
             ostream & os)
{
   // addr is the address to display at start.
   // buf is a pointer to data, n is number of bytes in
   // data (usually 16 but can be less on the last
   // record.
   // os is the output stream.
   char abuf[16 + 1]; // buffer for the text.

   char tmp[4 + 1]; // address buffer 4 digits + null.

   sprintf( tmp, "%04x", addr);
   os << tmp << ':';

   static const char digits[17] = "0123456789abcdef";

   for (int i = 0; i < n; i++) {
       unsigned char uc = buf[i];

       if (i == 8)
          os << " -"; // output a dash in the middle.

       os << ' ' << digits[uc >> 4] << digits[uc & 0x0f];
       if (is_special(uc))
         uc = '.'; // replace uc with . if special.
       abuf[i] = static_cast<char>(uc);
   }
   abuf[n] = '\0';

   // if n < 16 we must add blanks to make the text buf
   // align with previous lines.
   while (n < 16) {
      if (n == 8)
         os << "  "; // use " -" if you want to.
      os << "   ";
      ++n;
   }
   os << ' ' << abuf << endl;
}

memset(dst,val,count); does the same as:

for (int i = 0; i < count; i++) {
   static_cast<char *>(dst)[i] = val;
}

only more efficient.

You can use memset to clear a buffer or set a portion of the buffer equal to blanks. Since I had my function do output to ostream I didn't need to use memset but if you want to output to a string instead you can use memset to blank out all the blanks as I did in that while loop.


The function written above can be called like this:

unsigned int addr = 0;
int n;
unsigned char buf[16];

while ((n = fread(buf,16,1,fp)) > 0) {
   hexline(addr,buf,n,cout);
   addr += n;
}

and the file will output in the 'usual' hexdump fashion.

<line> 0000 00 11 22 33 44 55 66 77 - 88 99 aa bb cc dd ee ff .."3DUfw........<line end>

The text may vary depending on your coding system for upper bytes in the range 0x80-0xff.

Alf

Thanks guys i will see if i could split up points and hand them to you both since what both of you did was of great help.

Thanks guys i will see if i could split up points and hand them to you both since what both of you did was of great help.

Dear MellowD0c

I've refunded 45 points to enable you to accept the comment for one expert and to post a
"Points for <expertname>" Q for the other expert in the same topic area.

Please:
1) Post the link to the original Q in the "Points for <expertname>" and
2) Add in the original Q a comment with the link to the "Points for <expertname>", thus the email notif will warn the expert.

modulo

Community Support Moderator
Experts Exchange

http://www.experts-exchange.com/Programming/Programming_Languages/Cplusplus/Q_20469017.html

There are your points Salte

Please can you help me convert to following code base_10_value to base_16_value.  I had been trying very hard bu I have been unsuccessful.  I would appreciated your help very much.  Thank you.  The program must add two very large hexadecima (unsigned) integers.  The two summads should be stored arrays. Starting at the least significant digits, the program should compute the sum by adding  the two integers digit by digit.  

My luck,

Please can you help me convert to following code base_10_value to base_16_value.  I had been trying very hard bu I have been unsuccessful.  I would appreciated your help very much.  Thank you.  The program must add two very large hexadecima (unsigned) integers.  The two summads should be stored arrays. Starting at the least significant digits, the program should compute the sum by adding  the two integers digit by digit.  

My luck, #include<iostream.h>



const char ch='A', 'B', 'C', 'D', 'E', 'F';


void main(){



      char first_summand[256];
      char second_summand[256];
      char sum[258];
      char ch=isum;
      char ch=icarry;
      int isum, icarry;
      char ch * first, * second, * csum;
      

      //Input the first summand
      cout << "Input the first summand:(1-9,A-F(UPPERCASE LETTERS)) ";
      first = first_summand;
      *first++ = '\0';
      while ( (*first++ = cin.get() )!= '\n' )
            ;

      *--first = '\0';
      first--;

      //Input the second summand
      cout << "Input the second summand:(1-9,A-F(UPPERCASE LETTERS))";
      second = second_summand;
      *second++ = '\0';
      while ( (*second++ = cin.get()) != '\n')
            ;

      *--second = '\0';
      second--;


      //Calculate the sum
      csum = sum + 256;
      *csum-- = '\0';
      icarry = 0;

      while ( *first || *second) {
            if ( *first && *second) {
                  isum = *first-- + *second-- - 2 * int('0') + icarry;
                  icarry = isum / 16;
                  isum = isum % 16;
                  *csum-- = isum + '0';

            }

            if ( *first && (! *second) ) {
                  isum = *first-- - int ('0') + icarry;
                  icarry = isum / 16;
                  isum = isum % 16;
                  *csum-- = isum + '0';

            }
            if ( (! *first) && *second) {
                  isum = *second-- - int('0') + icarry;
                  icarry = isum / 16;
                  isum = isum % 16;
                  *csum-- = isum + '0';
            }


      }
      char ch = *csum;

      if ( icarry ) {
            *csum-- = icarry + '0';
      }
      csum++; // adjust csum to point to the start of the output string
      cout << (csum) << '\n';
      }