Cipher text in C,How to repeat key characters

Question

• Explanation The ciphertext is generated from the plaintext by “adding” corresponding characters of the plaintext and the key together. If the plaintext is shorter than the key, only some of the key will be used. Similarly, if the plaintext is shorter than the key, the key will be used multiple times.

  For example, to encode the plaintext “HELLO” with the key “CAT”:

Plaintext: HELLO

Key: CATCA

Ciphertext: KFFOP

And to encode the plaintext “DOG” with the key “FIDO”:

Plaintext: DOG

Key: FID

Ciphertext: JXK

To add two letters together, use the following convention: A=1, B=2, …, Z=26. If the sum of two letters is greater than 26, subtract 26 from the sum. For example: A + E = 1 + 5 = 6 = F, and D + X = 4 + 24 = 28 = 2 = B.

• Now the problem with my code is that i am unable to repeat the key characters for further coding of plain text if key characters are less,how to repeat the key characters,so further coding can possible?

Help me guys.

Here is my code:

#include<stdio.h>
    #include<string.h>
    int main()
        {
            char str[100],k[50],str1[100];
            int i,n;
            gets(str);// Input plain text. 
            gets(str1);//Input key.
             for(i=0;str[i]!='\0';i++)
                    {
                         n=(str[i]-65+1)+(str1[i]-65+1);//Extracting the numerical position and adding them.
                         if(n>26) //if numerical value exceeds 26 then subtracting 26 from it and getting the numerical value.
                          {
                             n=n-26;
                          }
                         str[i]=n+64;//storing the ciphered character.
                    }

              for(i=0;str[i]!='\0';i++)//printing the ciphered characters.
              printf("%c",str[i]);
              return 0;

        }

Answer 1

While writing the loop, the variable used to index the key should be reset to 0 in order to repeat the key (if original text length is larger).

for(int i = 0, j = 0; input[i] != '\0'; ++i, ++j) {
    new_char = (input[i] - 64) + (key[j] - 64);
    new_char = adjust(new_char);
    cipher[i] = new_char + 64 ;
    if(j == (key_length - 2)) // if j is at the end, excluding null character, then make j = -1, which gets incremented to j = 0 in the next loop iteration
        j = -1;
}

Also use fgets for string input and dont use gets . Strings in C can be printed using the %s format specifier without writing an explicit loop to output the characters. For that make the last element of the cipher char array as the \\0 character.

Answer 2

You can use another loop variable an make the index of the key 0 every time it reaches its length. I have used variable j in this case. Try this code:

#include<stdio.h>
#include<string.h>
int main()
    {
        char str[100],k[50],str1[100];
        int i,n;
        gets(str);// Input plain text. 
        gets(str1);//Input key.
        int lenk=strlen(str1),j;   //calculate length of key
         for(i=0,j=0;str[i]!='\0';i++,j++)
    {
         if(j==lenk) j=j-lenk;      //make j=0
         n=(str[i]-65+1)+(str1[j]-65+1);    // add str1[j] instead
         if(n>26) 
          {
             n=n-26;
          }
         str[i]=n+64;//storing the ciphered character.

    }

          for(i=0;str[i]!='\0';i++)
          printf("%c",str[i]);
          return 0;

    }

NOTE THAT THIS WORKS ONLY FOR CAPITAL LETTER, YOU HAVE TO CHANGE YOUR CODE FOR SMALL LETTERS

Answer 3

You could also use modular arithmetic to repeat the characters.

for(i=0;str[i]='\0';i++)
{
    n = (str[i]-65+1 + (str1[i % lenk]-65 +1);    
    n = (n % 26) + 1;
    str[i] = n+64;
    //storing the ciphered character.

}

The expression i % 26 automatically rotates the value 0 through 25.

The same can be applied to rotate n from 1 to 26.