Thread: ISBN Code Checking...!

You should run your program through a memory detector such as Valgrind and input several different ISBNs, because this loop:

Code:

for(i = 0; i < th2; i++)
            sum = sum + isbn[i]*(i+1);
        for(i = (th2 + 1); i <th1; i++)
            sum = sum + isbn[i]*(i+1);
        for(i = (th1+1); i <10; i++)
            sum = sum + isbn[i]*(i+1);
        sum = sum + 10*(th1+1);

never resets "i" and can easily go out of bounds of your array. Also, there's a chance that some of the loops won't run, especially the last one as going over 10 seems to be easy.


Also, there's a few areas that could be changed, but aren't as important:

Code:

 for(i = 0; i < 10; i++)
    {
          if(isbn[i] == '?')
        break;
    }
    return i;

could be simplified to:

Code:

for (i = 0; i < 10 && isbn[i] != '?'; i++)
    ;
return i;

and the same could be applied in a variety of places throughout your program.

Since I can't understand why the position of the X or ? in the ISBN values, makes a difference, the whole if(th1 > th2) block of code, is a mystery to me.

Here's some pseudo code that I'd suggest studying:

Code:

define X 10

This is all in main():
   integers isbn[10]={2,4,6,3,7,8,1,X,9,'?'}; //2463781X95
   integers i, valid=0, sum=0, mysteryNum=0, trial;

   for i=0;i<10;i++
      if(isbn[i]==10)
         print "%c ",'X';
      else if isbn[i]=='?'
         print"%c ",'?';
      else
         print "%d ",isbn[i];
   

   if isbn[0]==X 
      valid=0;
      print "\n Invalid Code: The ISBN first value, can't be an X.\n";
   

   this is the else to the if directly above, and it is the heart of the   
   program:

      for i=0,sum=0;i<10;i++   //multiply each digit by it's index +1, but
         if isbn[i] != '?'         //exclude the mystery number, if present
            sum += isbn[i] * (i+1);
         
         else 
            mysteryNum=i+1; //we have a mystery number at i, but it could be at index zero, (which we don't want, so +1
         
      
      print "\nOriginal ISBN sum: %d   Original sum mod 11: %d\n\n",sum, sum%11;

      if sum % 11==0  //we have a valid ISBN number
         if mysteryNum //with a mystery number
            mysteryNum=0; //so the mystery number if zero, gives us a valid ISBN number

         valid=1;  //and the ISBN number is valid
             
      else we need to test the mystery number  // ISBN number is invalid - test the mystery array[mysteryNum-1], for values. 
         for i=1;i<11;i++ { //i=0 has already been tested, and failed
            trial = sum + (i*mysteryNum);
            
            /* for debug only (I made an error down below, and this is
            a good way to show how you can find and fix errors)
            if i==2 
               printf("trial: %d  sum: %d  mysteryNum: %d\n",trial,sum, mysteryNum); getchar();
             
            */
            if trial % 11==0 
               valid=1;
               isbn[mysteryNum-1]=i;
               sum = trial;
               break;
            
         end of for
      end of else 
      if sum 
         print "\nFinal ISBN sum: %d   Final sum mod 11: %d\n\n",sum, sum % 11
         if valid
            print "The ISBN code is valid\n"
         else
            print "The ISBN code is invalid\n"
      
         for i=0;i<10;i++ 
            if isbn[i]==10
               isbn[i]='X';
            if isbn[i]=='?' OR isbn[i]=='X'
               print "%c ",isbn[i];
            else
               print "%d ",isbn[i];
         end of for
      end of if
   end of else
   
   printf("\n");
   return 0;
end of program

Originally Posted by Adak

Since I can't understand why the position of the X or ? in the ISBN values, makes a difference, the whole if(th1 > th2) block of code, is a mystery to me.

Here's some pseudo code that I'd suggest studying:

Code:

define X 10

This is all in main():
   integers isbn[10]={2,4,6,3,7,8,1,X,9,'?'}; //2463781X95
   integers i, valid=0, sum=0, mysteryNum=0, trial;

   for i=0;i<10;i++
      if(isbn[i]==10)
         print "%c ",'X';
      else if isbn[i]=='?'
         print"%c ",'?';
      else
         print "%d ",isbn[i];
   

   if isbn[0]==X 
      valid=0;
      print "\n Invalid Code: The ISBN first value, can't be an X.\n";
   

   this is the else to the if directly above, and it is the heart of the   
   program:

      for i=0,sum=0;i<10;i++   //multiply each digit by it's index +1, but
         if isbn[i] != '?'         //exclude the mystery number, if present
            sum += isbn[i] * (i+1);
         
         else 
            mysteryNum=i+1; //we have a mystery number at i, but it could be at index zero, (which we don't want, so +1
         
      
      print "\nOriginal ISBN sum: %d   Original sum mod 11: %d\n\n",sum, sum%11;

      if sum % 11==0  //we have a valid ISBN number
         if mysteryNum //with a mystery number
            mysteryNum=0; //so the mystery number if zero, gives us a valid ISBN number

         valid=1;  //and the ISBN number is valid
             
      else we need to test the mystery number  // ISBN number is invalid - test the mystery array[mysteryNum-1], for values. 
         for i=1;i<11;i++ { //i=0 has already been tested, and failed
            trial = sum + (i*mysteryNum);
            
            /* for debug only (I made an error down below, and this is
            a good way to show how you can find and fix errors)
            if i==2 
               printf("trial: %d  sum: %d  mysteryNum: %d\n",trial,sum, mysteryNum); getchar();
             
            */
            if trial % 11==0 
               valid=1;
               isbn[mysteryNum-1]=i;
               sum = trial;
               break;
            
         end of for
      end of else 
      if sum 
         print "\nFinal ISBN sum: %d   Final sum mod 11: %d\n\n",sum, sum % 11
         if valid
            print "The ISBN code is valid\n"
         else
            print "The ISBN code is invalid\n"
      
         for i=0;i<10;i++ 
            if isbn[i]==10
               isbn[i]='X';
            if isbn[i]=='?' OR isbn[i]=='X'
               print "%c ",isbn[i];
            else
               print "%d ",isbn[i];
         end of for
      end of if
   end of else
   
   printf("\n");
   return 0;
end of program

I think the position of X and ? do matters... the X could be anywhere except at the front... and also ? could be any where even at the front... So while Multiplying we have to skip the ? and also have to remember it's position so that the MystNumb we have found can be multiplied with the position and added to the sum of rest 9 numbers. and the new sum will be divisible by 11.

so what about this ex. I copied it from a books ISBN Code....
8121?2605X

now the number that I have hided from you is 9. So in place of ? it would be 9. We need to find the numb via C.

And very much thank you for your response.

In the pseudo code I posted, the position of X and ? are independent of each other.

It handles the 8121?2605X, just fine, finding the 9.

Although it's "pseudo code", it's as much code as pseudo code, probably.

And yes, it's all in C, when you add in the bits and pieces, it will compile!

You're welcome.

Originally Posted by Tipu Sultan

the X could be anywhere except at the front...

As Tim and I have already told you that's wrong. The X is only valid as the last digit (check digit).

From http://www.isbn.org/standards/home/i...ml/usmapp.htm:

...
3. Construction of an international standard book number
An international standard book number consists of ten digits[4] made up of the following parts
...
Footnote 4: The digits are the arabic numerals 0-9; in the case of the check digit only, an X can sometimes occur (see 2.4)

Bye, Andreas

Originally Posted by AndiPersti

As Tim and I have already told you that's wrong. The X is only valid as the last digit (check digit).

From http://www.isbn.org/standards/home/i...ml/usmapp.htm:


Bye, Andreas

THANKS A LOT... that reduces my work a lot more... now we just need to deal with 9 numbers and accordingly no need to deal with the position of the X... Thanks again....