I don't understan why the toBinary function will not pass?

Question about the task

Binary to hexadecimal converter

Java Syntax, Level 11, Lesson 1

Resolved

When I run the verification, I keep failing the 2nd test as shown

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) { //write your code here //declare variables int leftIndex = 0, rightIndex = 4; String hexString = "", emptyStr = "", hexItem = ""; String binPart; String[] binaryArr = {"00000", "10001", "20010", "30011", "40100", "50101", "60110", "70111", "81000", "91001", "A1010", "B1011", "C1100", "D1101", "E1110", "F1111"}; boolean chklen; boolean isFound = false; int remainder = 0; if (binaryNumber == null || binaryNumber.isEmpty()) { return emptyStr; } else { //check if the binaryNumber is long enough (a multiple of 4) if (chklen = binaryNumber.length() % 4 == 0) { //loop through binaryNumber for (int i = 0; i < binaryNumber.length()/4; i++) { //get the next four digits of binaryNumber binPart = binaryNumber.substring(leftIndex, rightIndex); leftIndex += 4; rightIndex += 4; isFound = false; //find a match for the binary number in the binary array for (String binEle : binaryArr) { if (binEle.endsWith(binPart)) { //get the corresponding hex digit from the string element of the binary array hexItem = binEle.substring(0, 1); isFound = true; break; } } //if I loop through all the binary digits in the array and don't find a match, return an empty string if (!isFound) { return emptyStr; } //construct hexNumber String hexString += hexItem; } //if the binaryNumber is not long enough, add digits then perform operations } else { remainder = binaryNumber.length() % 4; switch (remainder) { case 1: { binaryNumber = "000" + binaryNumber; isFound = false; for (int i = 0; i < binaryNumber.length()/4; i++) { //get the next four digits of binaryNumber binPart = binaryNumber.substring(leftIndex, rightIndex); leftIndex += 4; rightIndex += 4; //find a match for the binary number in the binary array for (String binEle : binaryArr) { if (binEle.endsWith(binPart)) { //get the corresponding hex digit from the string element of the binary array hexItem = binEle.substring(0, 1); isFound = true; break; } } //if I loop through all the binary digits in the array and don't find a match, return an empty string if (!isFound) { return emptyStr; } //construct hexNumber String hexString += hexItem; } } case 2: { binaryNumber = "00" + binaryNumber; isFound = false; for (int i = 0; i < binaryNumber.length()/4; i++) { //get the next four digits of binaryNumber binPart = binaryNumber.substring(leftIndex, rightIndex); leftIndex += 4; rightIndex += 4; //find a match for the binary number in the binary array for (String binEle : binaryArr) { if (binEle.endsWith(binPart)) { //get the corresponding hex digit from the string element of the binary array hexItem = binEle.substring(0, 1); isFound = true; break; } } //if I loop through all the binary digits in the array and don't find a match, return an empty string if (!isFound) { return emptyStr; } //construct hexNumber String hexString += hexItem; } } default: { binaryNumber = "0" + binaryNumber; isFound = false; for (int i = 0; i < binaryNumber.length()/4; i++) { //get the next four digits of binaryNumber binPart = binaryNumber.substring(leftIndex, rightIndex); leftIndex += 4; rightIndex += 4; //find a match for the binary number in the binary array for (String binEle : binaryArr) { if (binEle.endsWith(binPart)) { //get the corresponding hex digit from the string element of the binary array hexItem = binEle.substring(0, 1); isFound = true; break; } } //if I loop through all the binary digits in the array and don't find a match, return an empty string if (!isFound) { return emptyStr; } //construct hexNumber String hexString += hexItem; } } } } return hexString; } } public static String toBinary(String hexNumber) { //write your code here //declare variables String emptyStr = ""; String binaryString = ""; String binaryItem = ""; String[] binaryRepr = {"00000", "10001", "20010", "30011", "40100", "50101", "60110", "70111", "81000", "91001", "A1010", "B1011", "C1100", "D1101", "E1110", "F1111"}; char hexChar; boolean isFound = false; //check if null or empty string has been provided, and return empty string if (hexNumber == null || hexNumber.isEmpty()) { return emptyStr; } else { //return the binary string representation //loop through the hex string representation for (int i = 0; i < hexNumber.length(); i++) { hexChar = hexNumber.charAt(i); //get the binary string from the location given by hexChar //compare hexChar to the beginning of each element in binaryRepr for (String binEle : binaryRepr) { if (binEle.startsWith((hexChar + "").toUpperCase())) { binaryItem = binEle.substring(1, 5); isFound = true; break; } } //if nothing is found, return empty string, else //concatenate binary String to binaryItem if (isFound == false) { return emptyStr; } else { binaryString += binaryItem; } } return binaryString; } } }

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Guadalupe Gagnon

Level 37 , Tampa, United States

10 March 2023, 15:06

I have found an error: "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." I tried this method with the String: "f2z" and it output: 111100100010

Thomas Yates

Level 16 , USA

10 March 2023, 17:24

Thank you, I was able to use your suggestion and also correct some other errors to pass. Once I made the correction you suggested, and moved the isFound = false inside the for loop and also I noticed that my strings contained uppercase A, B, C, D, E, and F.