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# Java Program to Add Two Binary Numbers

But before starting the program, let's first remind, how the addition of two binary number performs:

```1 + 0 = 1
0 + 1 = 1
0 + 0 = 0
1 + 1 = 10 (0 and carry 1)
1 + 1 + 1 = 10 + 1 = 11 (1 and carry 1)```

For example

```  1 1 1 0 1
+ 1 1 1 1 1
-----------
1 1 1 1 0 0```

## Binary Number Addition in Java using Integer

The question is, write a Java program to add two binary numbers. Both binary numbers must be entered by user at run-time. The answer to this question is given below.

The complete version of the binary number addition program in Java, is created at last of this article. But before creating the complete version of the program that performs addition on two binary numbers entered by user. Let's first create a very basic program, that provides easy-to-understand code. This program is created using int (integer) data type.

```import java.util.Scanner;

public class CodesCracker
{
public static void main(String[] args)
{
int binaryOne, binaryTwo, remOne, remTwo, i=0, carry=0;
int[] res = new int[10];
Scanner scan = new Scanner(System.in);

System.out.print("Enter the First Binary Number: ");
binaryOne = scan.nextInt();
System.out.print("Enter the Second Binary Number: ");
binaryTwo = scan.nextInt();

while(binaryOne!=0)
{
remOne = binaryOne%10;
remTwo = binaryTwo%10;
if(remOne==0 && remTwo==0)
{
if(carry==0)
{
res[i] = (res[i]*10) + 0;
}
else
{
res[i] = (res[i]*10) + 1;
carry = 0;
}
}
else if(remOne==0 && remTwo==1)
{
if(carry==0)
{
res[i] = (res[i]*10) + 1;
}
else
{
res[i] = (res[i]*10) + 0;
carry = 1;
}
}
else if(remOne==1 && remTwo==0)
{
if(carry==0)
{
res[i] = (res[i]*10) + 1;
}
else
{
res[i] = (res[i]*10) + 0;
carry = 1;
}
}
else if(remOne==1 && remTwo==1)
{
if(carry==0)
{
res[i] = (res[i]*10) + 0;
carry = 1;
}
else
{
res[i] = (res[i]*10) + 1;
carry = 1;
}
}
else
{
System.out.println("\nInvalid Input!");
return;
}
binaryOne = binaryOne/10;
binaryTwo = binaryTwo/10;
i++;
}
if(carry==1)
res[i] = (res[i]*10) + 1;
System.out.print("\nResult = ");
while(i>=0)
{
System.out.print(res[i]);
i--;
}
}
}```

Here is its sample run with user input 11101 as first and 11111 as second binary number:

Above program has limitations such as what if user enters two binary numbers in which one binary number has greater or lesser number of digits than other, or what if user enters a binary number whose size is bigger than int. Therefore let's modify the above program and re-create the same program using long data type. Also this program is basically the short version of previous.

```import java.util.Scanner;

public class CodesCracker
{
public static void main(String[] args)
{
long binaryOne, binaryTwo;
int remOne, remTwo, sumDigit, i=0, carry=0;
int[] sum = new int[10];
Scanner scan = new Scanner(System.in);

System.out.print("Enter the First Binary Number: ");
binaryOne = scan.nextLong();
System.out.print("Enter the Second Binary Number: ");
binaryTwo = scan.nextLong();

while(binaryOne!=0 || binaryTwo!=0)
{
remOne = (int)(binaryOne%10);
remTwo = (int)(binaryTwo%10);
sumDigit = remOne + remTwo + carry;
sum[i] = (int)(sumDigit%2);
carry = (int)(sumDigit/2);
binaryOne = binaryOne/10;
binaryTwo = binaryTwo/10;
i++;
}
if(carry==1)
sum[i] = carry;
System.out.print("\nResult = ");
while(i>=0)
{
System.out.print(sum[i]);
i--;
}
}
}```

Here is its sample run with user input 1010 as first and 111111 as second binary number:

Let's again shorten the above program. Here is the short version of the above program. The i++ is the post increment of i, that means the current value of i gets used, and then incremented its value by 1:

```import java.util.Scanner;

public class CodesCracker
{
public static void main(String[] args)
{
long binaryOne, binaryTwo;
int remOne, remTwo, sumDigit, i=0, carry=0;
int[] sum = new int[10];
Scanner scan = new Scanner(System.in);

System.out.print("Enter Two Binary Numbers: ");
binaryOne = scan.nextLong();
binaryTwo = scan.nextLong();

while(binaryOne!=0 || binaryTwo!=0)
{
sum[i++] = (int)((binaryOne%10 + binaryTwo%10 + carry)%2);
carry = (int)((binaryOne%10 + binaryTwo%10 + carry)/2);
binaryOne /= 10;
binaryTwo /= 10;
}
if(carry==1)
sum[i] = carry;
System.out.print("\nResult = ");
while(i>=0)
System.out.print(sum[i--]);
}
}```

Still the program is not perfect, because I've used array whose limit is 10. I know you can increase the limit by 100 or more. But what about the data type long, whose limit is 9,223,372,036,854,775,807.

Therefore, I must recommend you to go with string type to add any two binary numbers, without depending on its size or the number of digits, given binary numbers have.

## Binary Number Addition in Java using String

Before creating the complete version, let's first create a simple and basic version using string. This program assumes that the input of two binary numbers from user at run-time is of same number of digits:

```import java.util.Scanner;

public class CodesCracker
{
public static void main(String[] args)
{
String binaryOne, binaryTwo, resRev="", resOrig="";
char charOne, charTwo, carry='0';
int i;

Scanner scan = new Scanner(System.in);

System.out.print("Enter Two Binary Numbers: ");
binaryOne = scan.next();
binaryTwo = scan.next();

for(i=(binaryOne.length()-1); i>=0; i--)
{
charOne = binaryOne.charAt(i);
charTwo = binaryTwo.charAt(i);
if(charOne=='0' && charTwo=='0')
{
if(carry=='0')
resRev = resRev + "0";
else
{
resRev = resRev + "1";
carry = '0';
}
}
else if(charOne=='0' && charTwo=='1')
{
if(carry=='0')
resRev = resRev + "1";
else
{
resRev = resRev + "0";
carry = '1';
}
}
else if(charOne=='1' && charTwo=='0')
{
if(carry=='0')
resRev = resRev + "1";
else
{
resRev = resRev + "0";
carry = '1';
}
}
else if(charOne=='1' && charTwo=='1')
{
if(carry=='0')
{
resRev = resRev + "0";
carry = '1';
}
else
{
resRev = resRev + "1";
carry = '1';
}
}
else
{
System.out.println("\nInvalid Input!");
return;
}
}
if(carry=='1')
resRev = resRev + "1";
for(i=(resRev.length()-1); i>=0; i--)
resOrig = resOrig + resRev.charAt(i);
System.out.println("\nResult = " + resOrig);
}
}```

Here is its sample run with user input 11101 and 11111 as two binary numbers:

## Binary Number Addition in Java - Complete Version

Now let's create the complete version of the binary number addition program in Java. I'm saying this program as the complete version, because this program handles all types of errors, also handles binary numbers having any number of digits.

```import java.util.Scanner;

public class CodesCracker
{
public static void main(String[] args)
{
String binaryOne, binaryTwo, resRev="", resOrig="";
char charOne, charTwo, carry='0';
int binaryOneLen, binaryTwoLen, i, j, len;

Scanner scan = new Scanner(System.in);

System.out.print("Enter the First Binary Number: ");
binaryOne = scan.next();
System.out.print("Enter the Second Binary Number: ");
binaryTwo = scan.next();

binaryOneLen = binaryOne.length();
binaryTwoLen = binaryTwo.length();

if(binaryOneLen > binaryTwoLen)
{
for(i=(binaryOneLen-1), j=(binaryTwoLen-1); j>=0; i--, j--)
{
charOne = binaryOne.charAt(i);
charTwo = binaryTwo.charAt(j);
if(charOne=='0' && charTwo=='0')
{
if(carry=='0')
resRev = resRev + "0";
else
{
resRev = resRev + "1";
carry = '0';
}
}
else if(charOne=='0' && charTwo=='1')
{
if(carry=='0')
resRev = resRev + "1";
else
{
resRev = resRev + "0";
carry = '1';
}
}
else if(charOne=='1' && charTwo=='0')
{
if(carry=='0')
resRev = resRev + "1";
else
{
resRev = resRev + "0";
carry = '1';
}
}
else if(charOne=='1' && charTwo=='1')
{
if(carry=='0')
{
resRev = resRev + "0";
carry = '1';
}
else
{
resRev = resRev + "1";
carry = '1';
}
}
else
{
System.out.println("\nInvalid Input!");
return;
}
}
for(i=i; i>=0; i--)
{
charOne = binaryOne.charAt(i);
if(carry=='1')
{
if(charOne=='1')
{
resRev = resRev + "0";
carry = '1';
}
else if(charOne=='0')
{
resRev = resRev + "1";
carry = '0';
}
else
{
System.out.println("\nInvalid Input!");
return;
}
}
else
{
resRev = resRev + charOne;
}
}
}
else if(binaryOneLen < binaryTwoLen)
{
for(i=(binaryOneLen-1), j=(binaryTwoLen-1); i>=0; i--, j--)
{
charOne = binaryOne.charAt(i);
charTwo = binaryTwo.charAt(j);
if(charOne=='0' && charTwo=='0')
{
if(carry=='0')
resRev = resRev + "0";
else
{
resRev = resRev + "1";
carry = '0';
}
}
else if(charOne=='0' && charTwo=='1')
{
if(carry=='0')
resRev = resRev + "1";
else
{
resRev = resRev + "0";
carry = '1';
}
}
else if(charOne=='1' && charTwo=='0')
{
if(carry=='0')
resRev = resRev + "1";
else
{
resRev = resRev + "0";
carry = '1';
}
}
else if(charOne=='1' && charTwo=='1')
{
if(carry=='0')
{
resRev = resRev + "0";
carry = '1';
}
else
{
resRev = resRev + "1";
carry = '1';
}
}
else
{
System.out.println("\nInvalid Input!");
return;
}
}
for(i=j; i>=0; i--)
{
charTwo = binaryTwo.charAt(i);
if(carry=='1')
{
if(charTwo=='1')
{
resRev = resRev + "0";
carry = '1';
}
else if(charTwo=='0')
{
resRev = resRev + "1";
carry = '0';
}
else
{
System.out.println("\nInvalid Input!");
return;
}
}
else
{
resRev = resRev + charTwo;
}
}
}
else
{
for(i=(binaryOneLen-1); i>=0; i--)
{
charOne = binaryOne.charAt(i);
charTwo = binaryTwo.charAt(i);
if(charOne=='0' && charTwo=='0')
{
if(carry=='0')
resRev = resRev + "0";
else
resRev = resRev + "1";
}
else if(charOne=='0' && charTwo=='1')
{
if(carry=='0')
resRev = resRev + "1";
else
{
resRev = resRev + "0";
carry = '1';
}
}
else if(charOne=='1' && charTwo=='0')
{
if(carry=='0')
resRev = resRev + "1";
else
{
resRev = resRev + "0";
carry = '1';
}
}
else if(charOne=='1' && charTwo=='1')
{
if(carry=='0')
{
resRev = resRev + "0";
carry = '1';
}
else
{
resRev = resRev + "1";
carry = '1';
}
}
else
{
System.out.println("\nInvalid Input!");
return;
}
}
}
if(carry=='1')
resRev = resRev + "1";
len = resRev.length();
for(i=len-1; i>=0; i--)
resOrig = resOrig + resRev.charAt(i);
System.out.println("\nResult = " + resOrig);
}
}```

Here is its sample run with user input 1010110101011 as first and 11110001101111101 as second number:

I know the program is little longer, but I recommend you to take a deep concentrate on the program, to understand the logic used in the program. There are many other ways too, to create the same program. Since the article is getting longer than usual, therefore I'm letting you to do modify the program and create another way with yourself.

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