Is my logic wrong?

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I haven't worked a lot with conversions, maybe I have some flaws in my logic. Here is my logic: 1. First I create a method that verifies if the input ins correct. It tries to parse the number to base 10 using Long.parseLong(value,radix). If the input is not correct it will throw an exception. 2.If the convertNumberToOtherNumberSystem method, I convert the number.getDigit() to a base 10 long using Long.parseLong(number.getDigit(), number.getNumberSystem().getNumberSystemIntValue()) 3. I create a string representation of the required base number from the decimal number that I created at step 2 using the method Long.ToString(decimalNumber, required base) 4. I return the new Nubmer.

Implement the convertNumberToOtherNumberSystem method, which should convert number.getDigit() from one number system (numberSystem) to another (expectedNumberSystem).
Throw a NumberFormatException if the passed number is invalid, e.g. the number "120" passed with the binary number system.
Check that number.getDigit() is not negative.
The main method is not tested.

Requirements:

There must be a convertNumberToOtherNumberSystem(Number, NumberSystem) method that returns a Number.
A NumberFormatException should be thrown if the number passed is invalid for the specified number system.
You need to correctly implement the convertNumberToOtherNumberSystem method: convert the number to the specified number system.
The NumberSystemType enum must support the NumberSystem interface.
The NumberSystemType enum must have 11 values representing number system bases. Here are the bases: 2, 3, 4, 5, 6, 7, 8, 9, 10, 12 and 16.

package com.codegym.task.task30.task3001; /* Numeral system converter */ public class Solution { public static void main(String[] args) { Number number = new Number(NumberSystemType._10, "6"); Number result = convertNumberToOtherNumberSystem(number, NumberSystemType._2); System.out.println(result); // 110 expected number = new Number(NumberSystemType._16, "6df"); result = convertNumberToOtherNumberSystem(number, NumberSystemType._8); System.out.println(result); // 3337 expected number = new Number(NumberSystemType._16, "abcdefabcdef"); result = convertNumberToOtherNumberSystem(number, NumberSystemType._16); System.out.println(result); // abcdefabcdef expected number = new Number(NumberSystemType._12, "7ab"); result = convertNumberToOtherNumberSystem(number, NumberSystemType._10); System.out.println(result); // 1139 expected } //I used this logic. First I convert the number given to a base 10 number, then I convert the number from base 10 to the required base. public static Number convertNumberToOtherNumberSystem(Number number, NumberSystem expectedNumberSystem) { //write your code here checkIfNumberFormatIsCorrect(number); long decimalNumber=Long.parseLong(number.getDigit(),number.getNumberSystem().getNumberSystemIntValue()); String convertedNumber = Long.toString(decimalNumber, expectedNumberSystem.getNumberSystemIntValue()); return new Number(expectedNumberSystem, convertedNumber); } //This methode checks if the number format is ok. If it is not, it throws and exception public static void checkIfNumberFormatIsCorrect (Number number){ switch (number.getNumberSystem().getNumberSystemIntValue()) { case 16:try{ long number1=Long.parseLong(number.getDigit(),16); } catch (NumberFormatException e){ throw new NumberFormatException("Not a good number format"); } break; case 12: try{ long number2=Long.parseLong(number.getDigit(),12); } catch (NumberFormatException e){ throw new NumberFormatException("Not a good number format"); } break; case 10: try{ long number3=Long.parseLong(number.getDigit(),10); } catch (NumberFormatException e){ throw new NumberFormatException("Not a good number format"); } break; case 9: try{ long number4=Long.parseLong(number.getDigit(),9); } catch (NumberFormatException e){ throw new NumberFormatException("Not a good number format"); } break; case 8: try{ long number5=Long.parseLong(number.getDigit(),8); } catch (NumberFormatException e){ throw new NumberFormatException("Not a good number format"); } break; case 7: try{ long number6=Long.parseLong(number.getDigit(),7); } catch (NumberFormatException e){ throw new NumberFormatException("Not a good number format"); } break; case 6: try{ long number7=Long.parseLong(number.getDigit(),6); } catch (NumberFormatException e){ throw new NumberFormatException("Not a good number format"); } break; case 5: try{ long number8=Long.parseLong(number.getDigit(),5); } catch (NumberFormatException e){ throw new NumberFormatException("Not a good number format"); } break; case 4: try{ long number1=Long.parseLong(number.getDigit(),4); } catch (NumberFormatException e){ throw new NumberFormatException("Not a good number format"); } break; case 3: try{ long number1=Long.parseLong(number.getDigit(),3); } catch (NumberFormatException e){ throw new NumberFormatException("Not a good number format"); } break; case 2: try{ long number1=Long.parseLong(number.getDigit(),2); } catch (NumberFormatException e){ throw new NumberFormatException("Not a good number format"); } break; } } }

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catalin1989

Level 37 , Romania

29 March, 11:38

Thanx guys. With your help, I got a much more elegant solution than CodeGym Solution. For validation I used Guadalupe's idea, and for calculating the number I used Thomas's idea. The calculation method looks like this. if(!checkIfNumberFormatIsCorrect(number,number.getNumberSystem().getNumberSystemIntValue())){ throw new NumberFormatException("Invalid number format!"); } BigInteger base10NUmber=new BigInteger(number.getDigit(),number.getNumberSystem().getNumberSystemIntValue()); String result= base10NUmber.toString(expectedNumberSystem.getNumberSystemIntValue()); return new Number(expectedNumberSystem,result); Much more simpler that CodeGym solution.

Guadalupe Gagnon

Level 37 , Tampa, United States

26 March, 14:49

useful

This is just a tip: Typically the try-catch structure is not supposed to be used as logic to the computer programs you write. In this code, Solution class lines 39-130, the logic relies on the possibility of a ParseException being thrown to indicate that an invalid number exists in the String. Here is a little bit of code (with a main method to show some test) I created using a simple regex to check whether a String contains anything other than the valid digits in the radix specified:

public class Solution {
    public static void main(String[] args) {
        String digits = "0123456789abcdef";
        checkString("f479d", 8);
        checkString("f479d", 16);
        checkString("f479dz", 8);
        checkString("f479dz", 16);
        checkString("479", 5);
        checkString("479", 9);
        checkString(null, 9);
    }

    static final String digits = "0123456789abcdef";
    private static void checkString(String s, int radix) {
        System.out.print("Checking string number '" + s + "' against the radix of "+radix+" digits: ");
        System.out.println(digits.substring(0,radix));
        /* regex: start with .* to signify any character any amount of times
            use the "digits" inside square brackets with ^ to signify not any of these
            for example [^123] would be anything not 1, 2, or 3
            end with .* for any character any amount of times
            so if anything is in the String that is not what is specified in the
            brackets then the matches() would result in true
        */
        if(s == null || s.toLowerCase().matches(".*[^"+digits.substring(0,radix)+"].*"))
            System.out.print(false);
        else System.out.print(true);
        System.out.println();
    }
}

You can modify this and use it if you want to replace lines 39 - 130 in the Solution class.

catalin1989

Level 37 , Romania

28 March, 17:15

Thanx. I admit that I really don't like to work with patterns. But I guess this is a more elegant solution. Like the idea of your validator.

Thomas

Level 41 , Bayreuth, Germany

26 March, 11:54

useful

not really... I'd try to replace Long with BigInteger. CG probably tests using very huge numbers. You also don't need switch case as you get the base from the enum. It's the name() -> substring(1) -> to int Or you may want to try to really implement the converter then one approach is as follows: That step 1 is not to hard to implement. Eg. base 2 has valid chars 0, 1 base 4: 0, 1, 2, 3 base 12: 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, a, b if you think of all valid chars for all conversion up to base 16 as a string or array like 0123456789abcdef then you can get the valid chars if you just invoke substring on that above string for base two substring(0, 2), for base 8 substring(0, 8) etc. When that's done you write to converters (if you want things to stay easy), one that converts the input to base 10 and then a converter from base 10 to the base you need. Use BigInteger to be able to hold huge numbers eg for converting to base 10 I used that formula num += pow(base, exp) * decVal of charAt(i) eg for hex: b2 16^1 * 11 (look up the position of b in the above all valid chars string to get the value) + 16^0 * 2 = 176 + 2 = 178 to convert from dec to base x modulus the base, add that to your result string and then / base eg 178 to hex = 178 % 16 = 2 -> lookup index 2 in the string -> 2 intermediate result 2 178 / 16 = 11 11 % 16 = 11 -> lookup index 11 in the string -> b final result b2

catalin1989

Level 37 , Romania

28 March, 17:23

You were right with the BigInteger.