What's wrong with the second method?

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The task is to write code to convert a decimal to a binary number and vice versa. I've kind of finally written the 'toBinary' method that works but the 'toDecimal' one keeps making an error. I still can't see what I'm doing wrong.

The public static toBinary(int) method must convert the integer received as an input parameter from the decimal numeral system to binary and return its string representation. And conversely, the public static toDecimal(String) method converts from the string representation of a binary number to a decimal number.

The methods only work with positive numbers and non-empty strings.
If the input parameter is less than or equal to 0, the toBinary(int) method returns an empty string.
If the input parameter is an empty string or null, then the toDecimal(String) method returns 0.

Your task is to implement these methods.

One algorithm for converting a decimal number to the string representation of a binary number is as follows:


while (the decimal number is not 0) {
the binary representation = remainder of the division of the decimal number by 2 + the binary representation
the decimal number = the decimal number / 2
}

One algorithm for converting the string representation of a binary number to a decimal number is as follows:


for (int i = 0; i < length of the binary representation; i++) {
the decimal number = the decimal number + digit in the binary representation * 2 to the power of i
}

Initially, the rightmost number is taken from the binary representation.
With each iteration of the loop, we take the next number closer to the beginning of the binary representation.

Hint: You can use the Math.pow(number, degree) method to raise a number to a power.
The main() method is not involved in testing.

Requirements:

The toBinary(int) method needs to be implemented as outlined in the task conditions.
The toDecimal(String) method needs to be implemented as outlined in the task conditions.
The Integer.toBinaryString(int) and Long.toBinaryString(int) methods cannot be used.
The Integer.parseInt(String, int) and Long.parseLong(String, int) methods cannot be used.
The Integer.toString(int, int) and Long.toString(long, int) methods cannot be used.
You cannot use a BigInteger object.
You cannot use a BigDecimal object.

package en.codegym.task.pro.task09.task0906; /* Binary converter */ public class Solution { public static void main(String[] args) { int decimalNumber = Integer.MAX_VALUE; System.out.println("Decimal number " + decimalNumber + " is equal to binary number " + toBinary(decimalNumber)); String binaryNumber = "1111111111111111111111111111111"; System.out.println("Binary number " + binaryNumber + " is equal to decimal number " + toDecimal(binaryNumber)); } public static String toBinary(int decimalNumber) { //write your code here String binaryNumber = ""; if (decimalNumber <= 0) { return binaryNumber; } while (decimalNumber != 0) { binaryNumber = String.valueOf(decimalNumber%2) + binaryNumber; decimalNumber = decimalNumber / 2; } return binaryNumber; } public static int toDecimal(String binaryNumber) { //write your code here if (binaryNumber == null || binaryNumber.isEmpty()) { return 0; } int decimalNumber = 0; for (int i = 0; i < binaryNumber.length(); i++) { decimalNumber = (int) (decimalNumber + Integer.valueOf(binaryNumber.charAt(i)) * Math.pow(2, i)); } return decimalNumber; } }

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Thomas

Level 41 , Bayreuth, Germany

12 April, 13:37

I see two problems. the first: char 0 == int 48, char 1 == int 49 (because that's ascii values and not numeral values, search ascii table in google and have a look at that) so you'd need to subtract 48 from the binary digit you currently work with The sedond problem... you start taking digits from the left, but that's (char 1) - 48 * Math.pow(2, binaryNumber.length()) and not Math.pow(2, 0) So you either start from the right or you change the Math.pow expression Smth. like that...

int currDigit = binaryNumber.charAt( binaryNumber.length() - i -1) - 48;
decimalNumber += (int) ( currDigit * Math.pow(2, i));

In a higher level exercise I have tried to explain that, too ;) Oh, and not only test with the given numbers, that overflow and you don't see the problem. Try small binary numbers like 11111111, 101, 111 or 100

Evgeniia Shabaeva

Level 11 , Budapest, Hungary

15 April, 11:24

Thank you very much. As for the second problem, I actually noticed I used the wrong order of taking the digits from the binary number and corrected it (but it still gave me wrong output with small binary numbers). Your advice to subtract 48 from that digit was really helpful. So, the code that finally worked in my case was:

public static int toDecimal(String binaryNumber) {
        if (binaryNumber == null || binaryNumber.isEmpty()) {
            return 0;
        }
        int decimalNum = 0;
        for (int i = 0; i < binaryNumber.length(); i++) {
            int tempIndex = binaryNumber.length() - 1 - i;
            decimalNum += (int) ((Integer.valueOf(binaryNumber.charAt(tempIndex)) - 48) * Math.pow(2, i));
        }
        return decimalNum;
    }

Thank you!