﻿ Java exercises: Add two numbers without using any arithmetic operators - w3resource # Java Exercises: Add two numbers without using any arithmetic operators

## Java Basic: Exercise-111 with Solution

Write a Java program to add two numbers without using any arithmetic operators.

Given x = 10 and y = 12; result = 22

Sample Solution:

Java Code:

``````import java.util.Scanner;
public class Example111 {
public static void main(String[] arg)
{
int x, y ;
Scanner in = new Scanner(System.in);
System.out.print("Input first number: ");
x = in.nextInt();
System.out.print("Input second number: ");
y = in.nextInt();
while(y != 0){
int carry = x & y;
x = x ^ y;
y = carry << 1;
}
System.out.print("Sum: "+x);
System.out.print("\n");
}
}
```
```

Sample Output:

```Input first number: 10
Input second number: 12
Sum: 22
```

Pictorial Presentation: Flowchart: Java Code Editor:

What is the difficulty level of this exercise?

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## Java: Tips of the Day

Try and catch:

Java is excellent at catching errors, but it can only recover gracefully if you tell it what to do. The cascading hierarchy of attempting to perform an action in Java starts with try, falls back to catch, and ends with finally. Should the try clause fail, then catch is invoked, and in the end, there's always finally to perform some sensible action regardless of the results. Here's an example:

```try {
cmd = parser.parse(opt, args);

if(cmd.hasOption("help")) {
HelpFormatter helper = new HelpFormatter();
helper.printHelp("Hello ", opt);
System.exit(0);
}
else {
if(cmd.hasOption("shell") || cmd.hasOption("s")) {
String target = cmd.getOptionValue("tgt");
} // else
} // fi
} catch (ParseException err) {
System.out.println(err);
System.exit(1);
} //catch
finally {
new Hello().helloWorld(opt);
} //finally
} //try
```

It's a robust system that attempts to avoid irrecoverable errors or, at least, to provide you with the option to give useful feedback to the user. Use it often, and your users will thank you!

Ref: https://red.ht/3EZc9OC