CodeGym/Help with Java Tasks/toBinary(String hexNumber) is Failing, but passing in the...

Level 17

United States of America

17.03.2022
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toBinary(String hexNumber) is Failing, but passing in the IDE

Question about the task

Binary to hexadecimal converter

Java Syntax, Level 9, Lesson 2

Under discussion

What is wrong with my toBinary() method? It looks right to me when I run the code.

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

The methods only work with non-empty strings.
If the input parameter is an empty string or null, then both methods return an empty string.
If the input parameter of the toHex(String) method contains any character other than 0 or 1, then the method returns an empty string.
If the input parameter of the toBinary(String) method contains any character other than digits from 0 to 9 or lowercase Latin letters from a to f, then the method returns an empty string.

Your task is to implement these methods.

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

We check the length of the string received in the input parameter. It must be a multiple of 4.
If this is not the case, then add the required number of 0s to the beginning of the string.
We take every four characters (bits) and check which hexadecimal character they correspond to.

For example:

the binary representation is "100111010000", where "1001" is "9", "1101" is "d", and "0000" - "0"
the hexadecimal representation is "9d0".

One algorithm for converting the string representation of a hexadecimal number to the string representationof a binary number is as follows:
We take each character and check which binary number (4 bits) it corresponds to.

For example:

the hexadecimal representation is "9d0", where "9" is "1001", "d" is "1101", and "0" is "0000"
the binary representation is "100111010000".

The main() method is not involved in testing.

Requirements:

The toHex(String) method needs to be implemented as outlined in the task conditions.
The toBinary(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.toHexString(int) and Long.toHexString(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.task0908; /* Binary to hexadecimal converter */ public class Solution { private static final String HEX = "0123456789abcdef"; public static void main(String[] args) { String binaryNumber = "100111010000"; System.out.println("Binary number " + binaryNumber + " is equal to hexadecimal number " + toHex(binaryNumber)); String hexNumber = "9d0"; System.out.println("Hexadecimal number " + hexNumber + " is equal to binary number " + toBinary(hexNumber)); } public static String toHex(String binaryNumber) { //write your code here if (binaryNumber == "" || binaryNumber == null) return ""; String testNumber = binaryNumber; while (testNumber.length() % 4 != 0) { testNumber = '0' + testNumber; } int binaryToDecimal = toDecimal(testNumber); String decimalToHex = toHex(binaryToDecimal); return decimalToHex; } public static String toBinary(String hexNumber) { //write your code here if (hexNumber == "" || hexNumber == null) return ""; for (int i = 0; i < hexNumber.length(); i++) { Character letter = hexNumber.charAt(i); if (HEX.indexOf(letter) == -1) return ""; } int decimalNumber = hexToDecimal(hexNumber); String binaryNumber = decimalToBinary(decimalNumber); return binaryNumber; } public static int toDecimal(String binaryNumber) { //write your code here int decimalNumber = 0; if (binaryNumber == "" || binaryNumber == null) return 0; for (int i = 0; i < binaryNumber.length(); i++) { char lastChar = binaryNumber.toCharArray()[binaryNumber.length()-1-i]; int lastDigit = Character.getNumericValue(lastChar); decimalNumber = decimalNumber + lastDigit * (int)Math.pow(2.0, i); } return decimalNumber; } public static String toHex(int decimalNumber) { //write your code here String hexNumber = ""; if (decimalNumber <= 0) return ""; while (decimalNumber != 0) { int idx = decimalNumber % 16; char digit = HEX.charAt(idx); hexNumber = digit + hexNumber; decimalNumber = decimalNumber / 16; } return hexNumber; } public static int hexToDecimal(String hexNumber) { //write your code here int decimalNumber = 0; if (hexNumber == "" || hexNumber == null) return 0; for (int i = 0; i < hexNumber.length(); i++) { char letter = hexNumber.charAt(i); int idx = HEX.indexOf(letter); decimalNumber = 16 * decimalNumber +idx; } return decimalNumber; } public static String decimalToBinary(int decimalNumber) { //write your code here String binaryNumber = ""; if (decimalNumber <= 0) return ""; while(decimalNumber != 0) { binaryNumber = decimalNumber % 2 + binaryNumber; decimalNumber /= 2; } return binaryNumber; } }

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Lisa L

Level 47 , Nuremberg, Germany

17 March 2022, 10:30

useful

that's what I guess... 0x0 to binary: expected 0000, result 0x1 to binary: expected 0001, result 1 and when you do a check for null, then null is alway the first you test for. This just doesn't implode here cause the String class internally converts null to the String "null". If that's not done then myString == "" would throw a NullPointerException (all other objects other that string would do). The next problem is the == with that you compare references only. Again with String class' interning feature that my work most times. But if it won't once you've a hard time finding the problem. So just do it that way..

if (hexNumber == null || hexNumber.equals(""))
    return "";

if CG does some strict testing this also could make the task fail.

Maven

Level 17 , United States of America, USA

17 March 2022, 17:20

My toHex() method fails if I change hexNumber == "" to hexNumber.equals(""). It only passes if I use the double equals

Maven

Level 17 , United States of America, USA

17 March 2022, 17:25

I added the code to add the extra 0s to the beginning of the binary

while (binaryNumber.length() % 4 != 0) {
    binaryNumber = '0' + binaryNumber;
}

Maven

Level 17 , United States of America, USA

17 March 2022, 17:25

Good point! Yes, my result was messed up for 0 and 1 hex. I needed to add the '0's to the beginning of the binary results. I fixed that, but it's still failing

Lisa L

Level 47 , Nuremberg, Germany

17 March 2022, 18:32

for strings you better may use

if (hexNumber == null || hexNumber.isEmtpy()) { ...
// or
if (hexNumber == null || hexNumber.length() == 0) { ...

-2

Maven

Level 17 , United States of America, USA

18 March 2022, 03:46

Both of them doesn't work. It gives a null pointer Exception error

if (hexNumber == null || hexNumber.isEmtpy()) { ...
// or
if (hexNumber == null || hexNumber.length == 0) { ...

Maven

Level 17 , United States of America, USA

18 March 2022, 03:47

What I have is working. I don't think that's causing it to fail

if (hexNumber == "" || hexNumber == null) return "";

Lisa L

Level 47 , Nuremberg, Germany

18 March 2022, 09:06

where should give this a NullPointerException? I even tried all three suggestions ant the three necessary places and they work perfectly. As said, the way you code works only with strings cause they internally create a string for null, a "null" string. If null is tested and true, then the check will return an empty string. No further problems with that. Maybe post current code Works perfectly...

public static String toBinary(String hexNumber) {
    if (hexNumber == null || hexNumber.isEmpty())
        return "";

    for (String c : hexNumber.split(""))
        if (!HEX.contains(c))
            return "";

    return decimalToBinary(hexToDecimal(hexNumber));
}

Lisa L

Level 47 , Nuremberg, Germany

23 March 2022, 14:01

OK, I just finished that task myself now and it is really mean. If you just glue together the methods from the previous ones you for sure will miss some requirements. If you go the route CG suggests you do not have any problems. In addition to the checks for a-f and 0-9 you need to a) take care that the zero appeares 0 -> 0000 (in the prev task 0 should return "") b) if a string like 0f gets passed you need to output -> 0000ffff (the trailing zero gets lost when converting to dec... so you need to come up with another idea). If you move the functionality of the both prev written methods hexToDec and DecToBin to the toBinary method, then it isn't to difficult anymore. The toHex method is easy, you just add the 0|1 check in the binaryToDecimal method and then call it without any further check like. The checking is done inside the previously written methods.

public static String toHex(String binaryNumber) {
    return decimalToHex(binaryToDecimal(binaryNumber));
}