Binary addition in java - eclipse

Question

I wrote a program for a binary addition in java. But the result is sometimes not right.
For example if i add 1110+111. The result should be 10101.
But my program throws out 10001.
Maybe one of you find the mistake.
import java.util.Scanner;
public class BinaryAdder {
public static String add(String binary1, String binary2) {
int a = binary1.length()-1;
int b = binary2.length()-1;
int sum = 0;
int carry = 0;
StringBuffer sb = new StringBuffer();
while (a >= 0 || b >= 0) {
int help1 = 0;
int help2 = 0;
if( a >=0){
help1 = binary1.charAt(a) == '0' ? 0 : 1;
a--;
} if( b >=0){
help2 = binary2.charAt(b) == '0' ? 0 : 1;
b--;
}
sum = help1 +help2 +carry;
if(sum >=2){
sb.append("0");
carry = 1;
} else {
sb.append(String.valueOf(sum));
carry = 0;
}
}
if(carry == 1){
sb.append("1");
}
sb.reverse();
String s = sb.toString();
s = s.replaceFirst("^0*", "");
return s;
}
public static void main(String[] args) {
Scanner scan = new Scanner(System.in);
System.out.print("First: ");
String input1 = scan.next("(0|1)*");
System.out.print("Second: ");
String input2 = scan.next("(0|1)*");
scan.close();
System.out.println("Result: " + add(input1, input2));
}
}

Answer 1

this function is much simpler :
public static String binaryAdd(String binary1,String binary2){
return Long.toBinaryString(Long.parseLong(binary1,2)+Long.parseLong(binary2,2));
}
you can change Long.parseLong into Integer.parseInt if you don't expect very large numbers, you can also replace parse(Long/Int) with parseUnsigned(Long/Int) since you don't expect your strings to have a minus sign do you ?

Answer 2

You are not considering the case when
help1 + help2 = 3
So your method String add(String binary1, String binary2) should be like this:
public static String add(String binary1, String binary2) {
int a = binary1.length()-1;
int b = binary2.length()-1;
int sum = 0;
int carry = 0;
StringBuffer sb = new StringBuffer();
while (a >= 0 || b >= 0) {
int help1 = 0;
int help2 = 0;
if( a >=0){
help1 = binary1.charAt(a) == '0' ? 0 : 1;
a--;
} if( b >=0){
help2 = binary2.charAt(b) == '0' ? 0 : 1;
b--;
}
sum = help1 +help2 +carry;
if (sum == 3){
sb.append("1");
carry = 1;
}
else if(sum ==2){
sb.append("0");
carry = 1;
} else {
sb.append(String.valueOf(sum));
carry = 0;
}
}
if(carry == 1){
sb.append("1");
}
sb.reverse();
String s = sb.toString();
s = s.replaceFirst("^0*", "");
return s;
}
I hope this could help you!

Answer 3

sum = help1 +help2 +carry;
if(sum >=2){
sb.append("0");
carry = 1;
} else {
sb.append(String.valueOf(sum));
carry = 0;
}
If sum is 2 then append "0" and carry = 1
What about when the sum is 3, append "1" and carry = 1
Will never be 4 or greater

Answer 4

Know I'm a bit late but I've just done a similar task so to anyone in my position, here's how I tackled it...
import java.util.Scanner;
public class Binary_Aids {
public static void main(String args[]) {
System.out.println("Enter the value you want to be converted");
Scanner inp = new Scanner(System.in);
int num = inp.nextInt();
String result = "";
while(num > 0) {
result = result + Math.floorMod(num, 2);
num = Math.round(num/2);
}
String flippedresult = "";
for(int i = 0; i < result.length(); i++) {
flippedresult = result.charAt(i) + flippedresult;
}
System.out.println(flippedresult);
}
}
This took an input and converted to binary. Once here, I used this program to add the numbers then convert back...
import java.util.Scanner;
public class Binary_Aids {
public static void main(String args[]) {
Scanner inp = new Scanner(System.in);
String decimalToBinaryString = new String();
System.out.println("First decimal number to be added");
int num1 = inp.nextInt();
String binary1 = decimalToBinaryString(num1);
System.out.println("Input decimal number 2");
int num2 = inp.nextInt();
String binary2 = decimalToBinaryString(num2);
int patternlength = Math.max[binary1.length[], binary2.length[]];
while(binary1.length() < patternlength) {
binary1 = "0" + binary2;
}
System.out.println(binary1);
System.out.println(binary2);
int carry = 0;
int frequency_of_one;
String result = "";
for(int i = patternlength -i; i >= 0; i--) {
frequency_of_one = carry;
if(binary1.charAt(i) == '1') {
frequency_of_one++;
}
if(binary2.charAt(i) == '1') {
frequency_of_one++;
}
switch(frequency_of_one) {
case 0 ;
carry = 0;
result = "1" + result;
break;
case 1 ;
carry = 0;
result = "1" + result;
break;
case 2;
carry = 1;
result = "0" + result;
breake;
case 3;
carry = 1;
result = "1" + result;
breake;
}
}
if(carry == 1) {
result = "1" + result;
}
System.out.println(result);
}
public static String decimalToBinaryString(int decimal1) {
String result = "";
while(decimal1 > 0) {
result = result + Math.floorMod(decimal1, 2);
decimal = Math.round(decimal1/2);
}
String flipresult = "";
for(int i = 0; i < result.length[]; i++) {
flipresult = result.charAt(i) + flippedresult;
}
return flippedresult;
}
}