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Question about the task

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I'm stuck with the first and second requirement task....for binarynumber input ,I tried to use null as input....the result is null pointer exception....how to solve this task...some advice please

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 =null;//"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 String a=""; int x; int c=0 ; char[] ch= {'0','1','2','3','4','5','6','7','8','9','a','b','c','d','e','f'}; int b = binaryNumber.length()%4 ; if(binaryNumber.isEmpty()) {return a ;} else for(int i=0;i<binaryNumber.length();i++) { x=(int)Math.pow(2,binaryNumber.length()-1-i) ; if(binaryNumber.charAt(i)=='1' ) {c=c+x;} } while(c>0) { a=ch[c%16]+a; c=c/16; } return a; } public static String toBinary(String hexNumber) { //write your code here String a="" ; int b=0; String ch= "0123456789abcdef" ; if(hexNumber==null) {return a;} else for(int i=0;i<hexNumber.length();i++) { b=b+ch.indexOf(hexNumber.charAt(i)) *(int) Math.pow(16,hexNumber.length()-1-i) ; } while(b>0) { a=b%2+ a; b=b/2; } return a; } }

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Ajay

Level 2 , Madurai , India

20 February 2023, 18:03

The first line if (s == null) checks if the String "s" is null. If it is, it prints "0000" to the console and moves to the next iteration using the continue statement. If "s" is not null, the code checks whether the length of the string is less than 4 or not divisible by 4 using the condition s.length() < 4 || s.length() % 4 != 0. If the condition is true, the code enters a while loop and keeps adding "0" to the beginning of the String using the concat() method until the length of the string is a multiple of 4. Once the length of the string is a multiple of 4, the while loop ends, and the code continues with the next statement.

Mayank Chandratre

Level 20 , India

20 February 2023, 02:20

I think the requirement that you are missing is - If the input binary string have length less than four or its not a multiple of four then you have to add enough number of zeroes in front of it .... You may use code similar to this

if (s == null) {
        System.out.println("0000");
        continue;
      } else if (s.length() < 4 || s.length() % 4 != 0) {
        while (s.length() < 4 || s.length() % 4 != 0) {
          s = "0".concat(s);
        }
      }