Question

c++

LUILIIR Exercise #3: The xOr Cipher Write the fol lowing functions: void integerToBinary(int, int [0): // Convers an integer number to binary and stores the binary // form in an array. I/ Converts a string to binary by converting the ASCII value of I/ each character into binary. l/ Calculates the xor of two binary numbers. void stringToßinary (string, int Il): void xOr(int, int O. int, int [I, int [l): The ASCII values of the characters 'A, 'B', C, Z are 65, 66, 67, , 90. The binary form of 66 is 1000010 stored in the array A as A[O] = 1, A[1 ] - 0 , A[ 2] 0, A[3] 0, A(4] 0, A[5 ] 1, A(6] 0. The binary form of each of 65, 66, 90 uses 7 bits. The binary form of THE is 101010010010001000101 stored in the array A as A[O] 0, A[4] 1, A[5] 0, A[6 ] =0, A[ 7] 1, A[8] = 0, . , A[ 19 ] 0, A[201 1. 1, A[ 1] 0, A[2] = 1 , A[3] 0 xor 0 0,0 xor 1 1,1 xor 0 1, 1 xor 1 Write a function main) that does the following: Reads a message in upper case letters without spaces to be encrypted. Reads the key as a text in upper case letters without spaces. Use the xor to produce the encrypted binary version of the message. Prints the encrypted binary version of the message. For example, if the message is DONOTCHEAT and the key is ISEEYOU, then The binary form of the message is 1000100100111110011101001111101010010000111001000100010110000011010100 The binary form of the key is 1001001101001110001011000101101100110011111010101 Because the key is shorter than the message then we repeat the key digits as needed. 1000100100111110011101001111101010010000111001000100010110000011010100

Answer 1

#include<stdio.h>
#include<iostream>
#include<string>
using namespace std;

void print_binary(int array[])
{
   int i;
   cout<<"\n\n";
   for(i=0;array[i]!=-1&&i<250;i++)
    {
cout<<array[i];
    }
cout<<"\n\n";
}
void rev_binary(int array[])
{
int count=-1,i,j;
for(;array[++count]!=-1;);
int array_temp[count];
for(i=0;array[i]!=-1&&i<250;i++)
{
array_temp[i]=array[i];
}
j=count-1;
for(i=0;i<count;i++)
{
array[i]=array_temp[j];
j--;
}
  
}
void IntegerToBinary(int num,int array[])
{
int r,d,i,count=-1;
d=num;
for(;count<250;)
{
r=d%2;
d=d/2;
  
count++;
array[count]=r;
if(d==1||d==0)
{
count++;
array[count]=d;
break;
}
}
array[count+1]=-1;
rev_binary(array);
}

void StringToBinary(string str,int array[])
{
   int i,j,count=0,temp;
   int array_temp[250];
  
   for(i=0;i<str.length();i++)
   {
       j=0;
       IntegerToBinary(str[i],array_temp);
       for(temp=count;array_temp[j]!=-1&&temp<250;temp++)
       {
           array[temp]=array_temp[j];
           j++;
           count++;
       }
   }
   array[count]=-1;
}

void XOr(int n1, int binary1[],int n2,int binary2[],int result[])
{
   int diff,i,temp;
   if(n1>n2)
   {
       temp=n2;
       diff=n1-n2;
       for(i=0;i<diff&&temp<250;i++)
       {
           binary2[temp]=binary2[i];
           temp++;
       }
       binary2[temp]=-1;
       n2=n1;
   }
   else if(n1<n2)
   {
       temp=n1;
       diff=n2-n1;
       for(i=0;i<diff&&temp<250;i++)
       {
           binary1[temp]=binary1[i];
           temp++;
       }
       binary1[temp]=-1;
       n1=n2;
   }
  
   for(i=0;i<n1;i++)
   {
       if(binary1[i]==binary2[i])
       {
           result[i]=0;
       }
       else
       {
           result[i]=1;
       }
   }
   result[i]=-1;
}
int GetLengthBinary(int array[])
{
   int i=-1;
   for(;array[++i]!=-1&&i<250;);
   return i;
}

int main()
{
   int array[250],i;

    IntegerToBinary(10,array);
    print_binary(array);

    StringToBinary("aaaaa",array);
    print_binary(array);
//    cout<<GetLengthBinary(array);

    int binary1[250],binary2[250],result[250],n1=-1,n2=-1;
    string st1,st2;
   
    cout<<"\nEnter String:";cin>>st1;
    cout<<"\nEnter Key:";cin>>st2;
   
    StringToBinary(st1,binary1);
    StringToBinary(st2,binary2);

   print_binary(binary1);
   print_binary(binary2);
  
   XOr(GetLengthBinary(binary1),binary1,GetLengthBinary(binary2),binary2,result);
  
   print_binary(result);
      
}