﻿ Java exercises: Multiply two binary numbers - w3resource # Java Exercises: Multiply two binary numbers

## Java Basic: Exercise-18 with Solution

Write a Java program to multiply two binary numbers.

In digital electronics and mathematics, a binary number is a number expressed in the base-2 numeral system or binary numeral system. This system uses only two symbols: typically 1 (one) and 0 (zero).

Test Data:
Input first binary number: 110
Input second binary number: 101

Sample Solution:

Java Code:

``````import java.util.Scanner;
public class Exercise18 {
public static void main(String[] args)
{
long binary1, binary2, multiply = 0;
int digit, factor = 1;
Scanner in = new Scanner(System.in);
System.out.print("Input the first binary number: ");
binary1 = in.nextLong();
System.out.print("Input the second binary number: ");
binary2 = in.nextLong();
while (binary2 != 0)
{
digit = (int)(binary2 % 10);
if (digit == 1)
{
binary1 = binary1 * factor;
multiply = binaryproduct((int) binary1, (int) multiply);
}
else
{
binary1 = binary1 * factor;
}
binary2 = binary2 / 10;
factor = 10;
}
System.out.print("Product of two binary numbers: " + multiply+"\n");
}
static int binaryproduct(int binary1, int binary2)
{
int i = 0, remainder = 0;
int[] sum = new int;
int binary_prod_result = 0;

while (binary1 != 0 || binary2 != 0)
{
sum[i++] = (binary1 % 10 + binary2 % 10 + remainder) % 2;
remainder = (binary1 % 10 + binary2 % 10 + remainder) / 2;
binary1 = binary1 / 10;
binary2 = binary2 / 10;
}
if (remainder != 0)
{
sum[i++] = remainder;
}
--i;
while (i >= 0)
{
binary_prod_result = binary_prod_result * 10 + sum[i--];
}
return binary_prod_result;
}
}
```
```

Sample Output:

```Input the first binary number: 110
Input the second binary number: 101
Product of two binary numbers: 11110
```

Flowchart: Flowchart : binaryproduct() Java Code Editor:

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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