Another "It works in the IDE, but appears to fail on an unspecified parameter" issue

Question about the task

Under discussion

I'll admit this isn't the most elegant solution, but it works perfectly in IDE, including with binary numbers whose lengths are not multiples of 4. When I try to validate in CodeGym, though, I get a "String index 0 out of range" error in the toHex method, which shouldn't be possible.

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 { 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) { String hex = "0123456789abcdef"; while (binaryNumber.length() % 4 != 0) { binaryNumber = 0 + binaryNumber; } if (binaryNumber == "" || binaryNumber == null) { return ""; } int b = 0; b += (binaryNumber.charAt(0)-'0') * 8; b += (binaryNumber.charAt(1)-'0') * 4; b += (binaryNumber.charAt(2)-'0') * 2; b += (binaryNumber.charAt(3)-'0'); return hex.charAt(b) + toHex(binaryNumber.substring(4)); } public static String toBinary(String hexNumber) { String hex = "0123456789abcdef"; if (hexNumber == "" || hexNumber == null) { return ""; } String bin = ""; int n = hex.indexOf(Character.toString(hexNumber.charAt(0))); if (n>7) { bin += 1; n -= 8; } else { bin += 0; } if (n>3) { bin += 1; n -= 4; } else { bin += 0; } if (n>1) { bin += 1; n -= 2; } else { bin += 0; } bin += n; return bin + toBinary(hexNumber.substring(1)); } }

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Gellert Varga

Level 23 , Szekesfehervar, Hungary

26 October 2021, 12:29

I also got an StringIndexOutOfBoundsException. I interspersed Sys.o.pr.() all over your program to get some idea of what's going on. I think you should try your program as below, it might help you to find out what is happening and where it differs from what you wanted.

Gellert Varga

Level 23 , Szekesfehervar, Hungary

26 October 2021, 12:30

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));
    }

    public static String toHex(String binaryNumber) {
      String hex = "0123456789abcdef";
        while (binaryNumber.length() % 4 != 0) {
            binaryNumber = 0 + binaryNumber;
            System.out.println(binaryNumber);
        }
        if (binaryNumber == "" || binaryNumber == null) {
          return "";
        }
        int b = 0;
        System.out.println("line17, b= " + b);
        b += (binaryNumber.charAt(0)-'0') * 8;
        System.out.println("line19, b= " + b);
        b += (binaryNumber.charAt(1)-'0') * 4;
        System.out.println("line21, b= " + b);
        b += (binaryNumber.charAt(2)-'0') * 2;
        System.out.println("line23, b= " + b);
        b += (binaryNumber.charAt(3)-'0');
        System.out.println("line25, b= " + b);
        System.out.println("line26, binaryNumber= " + binaryNumber);
        return hex.charAt(b) + toHex(binaryNumber.substring(4));
    }
}

Gellert Varga

Level 23 , Szekesfehervar, Hungary

26 October 2021, 12:30

OUTPUT: line17, b= 0 line19, b= 8 line21, b= 8 line23, b= 8 line25, b= 9 line26, binaryNumber= 100111010000 line17, b= 0 line19, b= 8 line21, b= 12 line23, b= 12 line25, b= 13 line26, binaryNumber= 11010000 line17, b= 0 line19, b= 0 line21, b= 0 line23, b= 0 line25, b= 0 line26, binaryNumber= 0000 line17, b= 0 Exception in thread "main" java.lang.StringIndexOutOfBoundsException: String index out of range: 0 at java.base/java.lang.StringLatin1.charAt(StringLatin1.java:47) at java.base/java.lang.String.charAt(String.java:693) at Solution.toHex(Solution.java:18) at Solution.toHex(Solution.java:27) at Solution.toHex(Solution.java:27) at Solution.toHex(Solution.java:27) at Solution.main(Solution.java:4)

Thomas

Level 41 , Bayreuth, Germany

26 October 2021, 09:30

first fix

toBinary(new String(""));

(hint: == compares references) then

toHex(null);

maybe that's already enough to make it pass