The toHex(String) method is not validated, but it works in IDE,

Question about the task

Binary to hexadecimal converter

Java Syntax, Level 10, Lesson 2

Resolved

Please help, I don't know where the problem is :(.

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 */ import java.util.regex.Pattern; public class Solution { public static void main(String[] args) { String binaryNumber = "100111010000"; System.out.println("Binary number " + binaryNumber + " is equal to hexadecimal number " + toHex(binaryNumber)); String hexNumber = "1"; System.out.println("Hexadecimal number " + hexNumber + " is equal to binary number " + toBinary(hexNumber)); } public static String toHex(String binaryNumber) { String start = "0"; String finalString = ""; int decimalNumberOf = 0; if (binaryNumber == null || binaryNumber.length() == 0) { return ""; } char[] array = binaryNumber.toCharArray(); for (int i = 0; i < array.length; i++) { if (array[i] != '0' && array[i] != '1') { return ""; } } for (int i = 0; i < binaryNumber.length(); i++) { if (binaryNumber.length() % 4 == 0) { break; } else { binaryNumber = start + binaryNumber; } } while (binaryNumber.length() > 0) { String binaryfours = binaryNumber.substring(0, 4); for (int i = 0; i < binaryfours.length(); i++) { int characterAt = Character.getNumericValue(binaryfours.charAt(binaryfours.length() - 1 - i)); int power2 = ((int) Math.pow(2, i)); decimalNumberOf = (characterAt * power2) + decimalNumberOf; } String addDecimalNumberOf = (decimalNumberOf >= 0 && decimalNumberOf <= 9) ? "" + decimalNumberOf : (decimalNumberOf == 10) ? "a" : (decimalNumberOf == 11) ? "b" : (decimalNumberOf == 12) ? "c" : (decimalNumberOf == 13) ? "d" : (decimalNumberOf == 14) ? "e" : "f"; finalString += addDecimalNumberOf; binaryNumber = binaryNumber.replace(binaryfours, ""); decimalNumberOf = 0; } return finalString; } public static String toBinary(String hexNumber) { String partialBinary = ""; String binaryNumberOf = ""; if ((hexNumber == null || hexNumber.length() == 0)) { return ""; } for (int i = 0; i < hexNumber.length(); i++) { partialBinary = (hexNumber.charAt(i) == '0') ? "0000" : (hexNumber.charAt(i) == '1') ? "0001" : (hexNumber.charAt(i) == '2') ? "0010" : (hexNumber.charAt(i) == '3') ? "0011" : (hexNumber.charAt(i) == '4') ? "0100" : (hexNumber.charAt(i) == '5') ? "0101" : (hexNumber.charAt(i) == '6') ? "0110" : (hexNumber.charAt(i) == '7') ? "0111" : (hexNumber.charAt(i) == '8') ? "1000" : (hexNumber.charAt(i) == '9') ? "1001" : (hexNumber.charAt(i) == 'a') ? "1010" : (hexNumber.charAt(i) == 'b') ? "1011" : (hexNumber.charAt(i) == 'c') ? "1100" : (hexNumber.charAt(i) == 'd') ? "1101" : (hexNumber.charAt(i) == 'e') ? "1110" : (hexNumber.charAt(i) == 'f') ? "1111" : "exit"; if (partialBinary.equals("exit")) { return ""; } binaryNumberOf += partialBinary; binaryNumberOf = binaryNumberOf.replaceFirst("^0{0,3}", ""); } return binaryNumberOf; } }

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Thomas

Level 41 , Bayreuth, Germany

28 September 2022, 22:35

try more numbers like:

for (int i = 16; i < 255; i++) {
    System.out.format("%s in hex: %s validation: %s %s%n", Integer.toBinaryString(i),
            toHex(Integer.toBinaryString(i)),
            Integer.toHexString(i),
            Integer.toHexString(i).equals(toHex(Integer.toBinaryString(i))) ? "" : "--- error ---");
}

Aska #11011302

Level 29 , Ceske Budejovice, Czech Republic

29 September 2022, 13:57

Thank you Thomas! I see it now. I was trying to make a shortcut to avoid complicated calculations, in order to not to make a mistake :). It was a big mistake :)