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Feb 25th, 2008

EOF problem

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Hello i am using AES to encrypt/decrypt data/ pixel values.. saved in a file myfile.txt and the AES encrypts tha data and writes the ciphertext in a file ciphertext.txt
When decrypting the data i am getting some of the pixel values decrypted correctly followed by garbage..

Code to open myfile.txt and save ciphertext to Ciphertext.txt file

c++ Syntax (Toggle Plain Text)
char d;
	ifstream myfile1 ("myfile.txt");
	ofstream myfile3 ("Ciphertext.txt");
 
	myfile1.get(d);  //priming read
 
  while(!myfile1.eof())
{
	  for ( j=0; j<BC; j++)
	  for ( i=0; i < 4; i++)        
	  {
		  a[i][j] = d; // plaintext
		  //cout<<a[i][j]; // to chk how many plaintext is copied
		  myfile1.get(d);  //reads first element of next block
	  } 
 
		  Encrypt(a, rk);
		  counter1();
 
	  if (myfile3.is_open())
	  {
		  for(j=0; j< BC; j ++)
			  for ( i=0; i<4; i++)
				  myfile3<<a[i][j];
	  }   
	  else        
		cout << "Unable to open file";
  }	 
   	myfile3.close();
	myfile1.close();
char d;
	ifstream myfile1 ("myfile.txt");
	ofstream myfile3 ("Ciphertext.txt");
	
	myfile1.get(d);  //priming read

  while(!myfile1.eof())
{
	  for ( j=0; j<BC; j++)
	  for ( i=0; i < 4; i++)        
	  {
		  a[i][j] = d; // plaintext
		  //cout<<a[i][j]; // to chk how many plaintext is copied
		  myfile1.get(d);  //reads first element of next block
	  } 

		  Encrypt(a, rk);
		  counter1();
		
	  if (myfile3.is_open())
	  {
		  for(j=0; j< BC; j ++)
			  for ( i=0; i<4; i++)
				  myfile3<<a[i][j];
	  }   
	  else        
		cout << "Unable to open file";
  }	 
   	myfile3.close();
	myfile1.close();

Code to open Ciphertext.txt and save decrypted data to text.txt file

c++ Syntax (Toggle Plain Text)
char buffer[4][4];
	char dxc;
	ifstream myfile4 ("Ciphertext.txt");
	ofstream myfile5 ("Text.txt");
 
	myfile4.get(dxc);  //priming read
 
  while(!myfile4.eof())
  {
	  for ( j=0; j<BC; j++)
	  for ( i=0; i < 4; i++)        
	  {
		  buffer[i][j]=dxc;
		  //a[i][j] = dxc; // plaintext
		  //cout<<a[i][j]; // to chk how many plaintext is copied
		  myfile4.get(dxc);  //reads first element of next block
	  }
 
	  for ( j=0; j<BC; j++)
	  for ( i=0; i < 4; i++)        
	  {
		  a[i][j] = buffer[i][j];
 
	  }
		  Decrypt(a, rk);
		  counter2();
 
		if (myfile5.is_open())
		{
		    for(j=0; j<BC; j ++)
			for ( i=0; i<4; i++)
			myfile5<<a[i][j];
		}   
		  else        
			  cout << "Unable to open file";
  }			 
  	myfile5.close();
	myfile4.close();
char buffer[4][4];
	char dxc;
	ifstream myfile4 ("Ciphertext.txt");
	ofstream myfile5 ("Text.txt");

	myfile4.get(dxc);  //priming read

  while(!myfile4.eof())
  {
	  for ( j=0; j<BC; j++)
	  for ( i=0; i < 4; i++)        
	  {
		  buffer[i][j]=dxc;
		  //a[i][j] = dxc; // plaintext
		  //cout<<a[i][j]; // to chk how many plaintext is copied
		  myfile4.get(dxc);  //reads first element of next block
	  }

	  for ( j=0; j<BC; j++)
	  for ( i=0; i < 4; i++)        
	  {
		  a[i][j] = buffer[i][j];

	  }
		  Decrypt(a, rk);
		  counter2();

		if (myfile5.is_open())
		{
		    for(j=0; j<BC; j ++)
			for ( i=0; i<4; i++)
			myfile5<<a[i][j];
		}   
		  else        
			  cout << "Unable to open file";
  }			 
  	myfile5.close();
	myfile4.close();

E.g:- pixel values im myfile.txt
-13224404 -14211046 -14605042 -13682410 -12298975 -11245015 -10322644 -9532112 -8609993 -7490750 -6240432 -5516455 -4990625 -4332695 -3675534 -3412107 -3082376 -2753413 -2687618 -2884997 -3082374 -3082628 -3345799 -3740555 -3675016 -3675013 -3675267 -3609474 -3412093 -3214714 -3017333 -2885749 -2557300 -2425716 -2622326 -2885500 -3082625 -3345797 -4069263 -4727191 -5319066 -6108833 -6766499 -7424678 -8609203 -9333176 -9925051 -10714820 -11240648 -11503818 -11635406 -11437774 -11437776 -11634903 -11503318 -11239892 -11503066 -11503066 -11503064 -11503318 -11503317 -11503317 -11503315 -11503315 -11109329 -11109329 -10978000 -10978000 -10912461 -10912461 -10715850 -10715850 -10847949 -10847949 -10782667 -10848460 -10914252 -10914252 -10849229 -10849227 -10782918 -10782916 -10848709 -10914502 -11046088 -11111881 -11177674 -11177674 -11506639 -11506639 -11506639 -11506639 -11440846 -11440846 -11440846 -11440846
-13158611 -14276582 -14670578 -13682410 -12364768 -11245015

Pixel values in text.txt
-13224404 -14211046 -14605042 -13682410 -12298975 -11245015 -10322644 -9532112 \Ó¶¬6Ewƒ›SÆû53yè

However, some of the encrypted text is the same as EOF and thus i cant read the text
that after it. What should i do?
How to remove the EOF character from the encrypted text...

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kartouss

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Feb 25th, 2008

Re: EOF problem

Use binary mode to open files which deal with encrypted text.

That should solve the problem with the EOF char

P.S Refer to my reply to your mail.

If you get it working post the working program here

FireNet

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Feb 25th, 2008

Re: EOF problem

As I'd said privately, you don't want to change the EOF char in your ciphertext - that will corrupt the data.

In order to read in the file, without the EOF character halting the read, do so in a binary fashion. Grab 16 bytes, then place them in your 4x4 array. Life will be good.

Here's a sample of how to read in binary mode.

C++ Syntax (Toggle Plain Text)
#include<iostream>
#include <fstream>
using namespace std;
 
int main( )
{
   fstream f;
   unsigned char buff[16];
   int i, j;
 
   f.open( "ciphertext.txt", ios::binary | ios::in );
   if( !f )
   {
      cout << "error opening file, quitting" << endl;
      return -1;
   }
 
 
   for( j = 0; j < 10; j++ ) //we just want to see a sample of the file
   {
      f.read( (char *)buff, 16 );
      for( i = 0; i < 16; i++ )
         cout << int ( buff[i] ) << " " ; //display ASCII values
      cout << endl;
   }
 
 
   return 0;
}
#include<iostream>
#include <fstream>
using namespace std;

int main( )
{
   fstream f;
   unsigned char buff[16];
   int i, j;

   f.open( "ciphertext.txt", ios::binary | ios::in );
   if( !f )
   {
      cout << "error opening file, quitting" << endl;
      return -1;
   }


   for( j = 0; j < 10; j++ ) //we just want to see a sample of the file
   {
      f.read( (char *)buff, 16 );
      for( i = 0; i < 16; i++ )
         cout << int ( buff[i] ) << " " ; //display ASCII values
      cout << endl;
   }


   return 0;
}

This will give you a dump of the ASCII values for the 160 bytes, and that should be enough for you to get past the at least the first occurrence EOF (decimal 26) character.

For reading the full file, control the loop as:

C++ Syntax (Toggle Plain Text)
   while( f.read( (char *)buff, 16 ) )
   while( f.read( (char *)buff, 16 ) )

For this to work right, you must ensure your encoding process creates a file that is a multiple of 16 bytes in size.

vmanes

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Posting Virtuoso

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Feb 25th, 2008

Re: EOF problem

Various subtleties, but 'EOF character' is not a good idiom to maintain.
http://faq.cprogramming.com/cgi-bin/...&id=1043284351

Dave Sinkula

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Feb 25th, 2008

Re: EOF problem

Click to Expand / Collapse Quote originally posted by Dave Sinkula ...

Various subtleties, but 'EOF character' is not a good idiom to maintain.
http://faq.cprogramming.com/cgi-bin/...&id=1043284351

Using EOF to find end of file is not what this thread is about. It's NOT about reading till EOF char is encountered, but how to get past the EOF character.

Contrary to your reference, there is an EOF character (check your neighborhood ASCII chart) and reading it in text mode will cause end of input. The OP has file of encrypted data which may happen to include the value that equates to the EOF character, and that's what's causing problems.

vmanes

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Feb 25th, 2008

Re: EOF problem

Click to Expand / Collapse Quote originally posted by vmanes ...

Using EOF to find end of file is not what this thread is about. It's NOT about reading till EOF char is encountered, but how to get past the EOF character.

Contrary to your reference, there is an EOF character (check your neighborhood ASCII chart) and reading it in text mode will cause end of input. The OP has file of encrypted data which may happen to include the value that equates to the EOF character, and that's what's causing problems.

File systems vary, there is absolutely no need for a file to contain an 'EOF character'. On a common system, a particular character in a file opened in text mode might be translated into a value that causes the eof signal to be raised.

This is not because an 'EOF character' exists, it is because a binary file is being examined in text mode. Blurring such subtleties by repeating 'EOF character' to me is a bit of a disservice.

If the file is opened in the correct mode for the input, then there is no need to watch out for things that aren't there.

Dave Sinkula

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Feb 25th, 2008

Re: EOF problem

Not disputing what you say, Dave.

It's a matter of what the OP's problem is here, and how I (and at least one other) have been trying to help him around it.

In short, he read in a plain text file, encrypts it, stores it back - a collection of bytes in the range 0-255.

He (attempted) to open the encrypted file, as a text file, and read it back for decryption. This fails when a byte valued 26 (decimal) is encountered. We're trying to get the point across that, at least for reading the ciphertext, he must do so in binary mode. The resulting plaintext can be written back in text mode, assuming correct decryption.

vmanes

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Feb 25th, 2008

Re: EOF problem

Click to Expand / Collapse Quote originally posted by FireNet ...

P.S Refer to my reply to your mail.

Click to Expand / Collapse Quote originally posted by vmanes ...

As I'd said privately....

No one should be answering private requests for help. That's because no one should be asking for private help. Help is what these forums are for. Keep all questions and answers in the forums, please.

Click to Expand / Collapse Quote originally posted by vmanes ...

Not disputing what you say, Dave.

It's a matter of what the OP's problem is here, and how I (and at least one other) have been trying to help him around it.

And Dave is trying to clear up misinformation being passed to the OP. Though not directly related to his problem, telling him there's an EOF character is teaching him 'bad grammar' which he will pass on to the next person. So don't argue with Dave. He's trying to help, too.

WaltP

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Feb 26th, 2008

Re: EOF problem

c++ Syntax (Toggle Plain Text)
	char d;
	ifstream myfile1 ("myfile.txt");
	ofstream myfile3  ("ciphertext.txt");
 
	myfile1.get(d);  //priming read
 
  while(!myfile1.eof())
{
	  for ( j=0; j<BC; j++)
	  for ( i=0; i < 4; i++)        
	  {
		  a[i][j] = d; // plaintext
		  //cout<<a[i][j]; // to chk how many plaintext is copied
		  myfile1.get(d);  //reads first element of next block
	 }
		  Encrypt(a, rk);
		  counter1();
 
	  if (myfile3.is_open())
	  {
		  for(j=0; j< BC; j ++)
			  for ( i=0; i<4; i++)
				  myfile3<<a[i][j];
	  }   
	  else        
		cout << "Unable to open file";
  }  
   	myfile3.close();
	myfile1.close();
 
	unsigned char buff[4][4];
	fstream myfile4;
	myfile4.open( "ciphertext.txt", ios::binary | ios::in );
	if( !myfile4 )
	{
      cout << "error opening file, quitting" << endl;
      return -1;
	}
	ofstream myfile5 ("Text.txt");	
 
	  while(myfile4.read( (char *)buff, 16 ))
	  {
                    for( i = 0; i < 4; i++ )
		  for( j = 0; j < 4; j++ )
                               buff[i][j];
 
	  for ( j=0; j<BC; j++)
	  for ( i=0; i < 4; i++)        
	  {
		  a[i][j] = buff[i][j];
 
	  }	
		  Decrypt(a, rk);
		  counter2();
 
	if (myfile5.is_open())
		{
		    for(j=0; j<BC; j ++)
			for ( i=0; i<4; i++)
			myfile5<<a[i][j];
		}   
		  else        
			  cout << "Unable to open file";
  }		 
  	myfile5.close();
	myfile4.close();
	char d;
	ifstream myfile1 ("myfile.txt");
	ofstream myfile3  ("ciphertext.txt");

	myfile1.get(d);  //priming read

  while(!myfile1.eof())
{
	  for ( j=0; j<BC; j++)
	  for ( i=0; i < 4; i++)        
	  {
		  a[i][j] = d; // plaintext
		  //cout<<a[i][j]; // to chk how many plaintext is copied
		  myfile1.get(d);  //reads first element of next block
	 }
		  Encrypt(a, rk);
		  counter1();
	
	  if (myfile3.is_open())
	  {
		  for(j=0; j< BC; j ++)
			  for ( i=0; i<4; i++)
				  myfile3<<a[i][j];
	  }   
	  else        
		cout << "Unable to open file";
  }  
   	myfile3.close();
	myfile1.close();
	  
	unsigned char buff[4][4];
	fstream myfile4;
	myfile4.open( "ciphertext.txt", ios::binary | ios::in );
	if( !myfile4 )
	{
      cout << "error opening file, quitting" << endl;
      return -1;
	}
	ofstream myfile5 ("Text.txt");	
 
	  while(myfile4.read( (char *)buff, 16 ))
	  {
                    for( i = 0; i < 4; i++ )
		  for( j = 0; j < 4; j++ )
                               buff[i][j];

	  for ( j=0; j<BC; j++)
	  for ( i=0; i < 4; i++)        
	  {
		  a[i][j] = buff[i][j];

	  }	
		  Decrypt(a, rk);
		  counter2();

	if (myfile5.is_open())
		{
		    for(j=0; j<BC; j ++)
			for ( i=0; i<4; i++)
			myfile5<<a[i][j];
		}   
		  else        
			  cout << "Unable to open file";
  }		 
  	myfile5.close();
	myfile4.close();

Hello i am getting an error message, file cannot be open, quitting...
The code that Vmanes sent in the prevoius post is working in the main function and retrieving in binary...
I have modified the code in the aes main program and its not opening the file cipheretxt.txt to read .... please help me out how to solve this poblem...

kartouss

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Feb 26th, 2008

Re: EOF problem

From the code given, I don't see any reason the file open action should be failing. Can you step through the program and after the point at which ciphertext.txt should have been written, verify that it in fact exists?

Once it does open, I think you'll find the decryption will not work right. In all your 4x4 array accesses, you've done them in column major fashion, that is, working down a column, then move to the next column. Reading directly into the the buff array will store in a row major fashion, the natural way that C/C++ work. Your code in lines 43-46 does nothing.

To store from the buff to array a, in the order in which you've previously worked, you need to copy rows to columns.

C++ Syntax (Toggle Plain Text)
for ( j=0; j<BC; j++)
     for ( i=0; i < 4; i++)
    {
          a[i][j] = buff[j][i];
     }
for ( j=0; j<BC; j++)
     for ( i=0; i < 4; i++)
    {
          a[i][j] = buff[j][i];
     }

vmanes

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