GosuDa - 1 year ago 73

Java Question

EDIT: Since there are claims that this is a duplicate question, I will state my stance on why it is not.

- The other question had posed a problem regarding the conversion of a string input to an integer (the answer deemed correct). This is irrelevant here, as the problem in my case is one of logic.
- The solutions on the other question cannot correct the problem this question presents (finding the syntactical "charAt" error).
- Frankly, the context of the other question is beyond the scope of my knowledge in Java.

I have successfully created a program that converts binary numbers to decimal numbers.

`import java.util.Scanner;`

public class ProblemThree // to convert binary to decimal

{

public static void main (String[]args)

{

Scanner scan = new Scanner (System.in);

System.out.print("Enter binary");

int binary = scan.nextInt();

int answer = 0;

int length = String.valueOf(binary).length();

int[] number = new int[length];

int[] powerTwo = new int[length];

// To list every digit separately //

for (int n = 0; n <= length - 1; n++)

{

number[n] = (int) (binary / Math.pow(10, n)) % 10;

}

// To set the values in powerTwo to 2 //

for (int n = 0; n <= length - 1; n++)

{

powerTwo[n] = 2;

}

// To update the values in powerTwo //

for (int n = 0; n <= length - 1; n++)

{

if (n == 0)

powerTwo[0] = 1;

else if (n == 1)

powerTwo[1] = 2;

else

powerTwo[n] *= powerTwo[n-1];

}

// To add if current value is 1 //

for (int n = 0; n <= length - 1; n++)

{

if (number[n] == 1)

answer += powerTwo[n];

}

System.out.println(powerTwo[0]); // for testing

System.out.println(number[0]); // for testing

System.out.println("The decimal version of " + binary + " is " + answer + ".");

}

}

Seeing how some of the steps were redundant, I tried to simplify the code.

`import java.util.Scanner;`

public class ProblemThreeTestFunction

{

public static void main (String[]args)

{

Scanner scan = new Scanner (System.in);

System.out.print("Enter binary");

int binary = scan.nextInt();

int answer = 0;

int length = String.valueOf(binary).length();

String number = String.valueOf(binary);

for (int n = 0; n <= length - 1; n++)

{

if (number.charAt(n) == 1)

answer += (int)Math.pow(2, n);

}

System.out.println("The decimal version of " + binary + " is " + answer + ".");

}

}

I actually have no idea why the simplified code logic does not work. It uses the same concept as my original.

Thanks in advanced!

Recommended for you: Get network issues from **WhatsUp Gold**. **Not end users.**

Answer Source

You need to test against the character `'1'`

:

```
if (number.charAt(n) == '1')
```

not the number 1 that you are currently doing:

```
if (number.charAt(n) == 1) // WRONG!
```

Also, since `charAt`

counts from the left, and binary expansion counts from the right, you need to reverse the sense. Change

```
Math.pow(2, n)
```

to

```
Math.pow(2, length - 1 - n)
```

Recommended from our users: **Dynamic Network Monitoring from WhatsUp Gold from IPSwitch**. ** Free Download**