Wht its fail

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public static int toDecimal(String binaryNumber) { //write your code here if(binaryNumber==null || binaryNumber.isEmpty()) { return 0; } int decNo=0; int temp=Integer.valueOf(binaryNumber); for(int i=0;i<binaryNumber.length();i++) { int n=temp%10; decNo=(int) (decNo+(n*(Math.pow(2, i)))); temp=temp/10; } return decNo; }

The public static toBinary(int) method must convert the integer received as an input parameter from the decimal numeral system to binary and return its string representation. And conversely, the public static toDecimal(String) method converts from the string representation of a binary number to a decimal number.

The methods only work with positive numbers and non-empty strings.
If the input parameter is less than or equal to 0, the toBinary(int) method returns an empty string.
If the input parameter is an empty string or null, then the toDecimal(String) method returns 0.

Your task is to implement these methods.

One algorithm for converting a decimal number to the string representation of a binary number is as follows:

while (the decimal number is not 0) {
the binary representation = remainder of the division of the decimal number by 2 + the binary representation
the decimal number = the decimal number / 2
}

One algorithm for converting the string representation of a binary number to a decimal number is as follows:

for (int i = 0; i < length of the binary representation; i++) {
the decimal number = the decimal number + digit in the binary representation * 2 to the power of i
}

Initially, the rightmost number is taken from the binary representation.
With each iteration of the loop, we take the next number closer to the beginning of the binary representation.

Hint: You can use the Math.pow(number, degree) method to raise a number to a power.
The main() method is not involved in testing.

Requirements:

The toBinary(int) method needs to be implemented as outlined in the task conditions.
The toDecimal(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.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.task0906; /* Binary converter */ public class Solution { public static void main(String[] args) { int decimalNumber = Integer.MAX_VALUE; System.out.println("Decimal number " + decimalNumber + " is equal to binary number " + toBinary(decimalNumber)); String binaryNumber = "110011"; System.out.println("Binary number " + binaryNumber + " is equal to decimal number " + toDecimal(binaryNumber)); } public static String toBinary(int decimalNumber) { //write your code here if(decimalNumber<=0) { return ""; } String BinaryNo=""; int i=0; int remender; while (decimalNumber!=0) { remender=decimalNumber%2; BinaryNo= Integer.toString(remender) +BinaryNo; decimalNumber=decimalNumber/2; i++; } return BinaryNo; } public static int toDecimal(String binaryNumber) { //write your code here if(binaryNumber==null || binaryNumber.isEmpty()) { return 0; } int decNo=0; int temp=Integer.valueOf(binaryNumber); for(int i=0;i<binaryNumber.length();i++) { int n=temp%10; decNo=(int) (decNo+(n*(Math.pow(2, i)))); temp=temp/10; } return decNo; } }

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Guadalupe Gagnon

Level 37 , Tampa, United States

1 September 2023, 15:50

I tried your toDecimal with the String "1111111111111111111111111111111". This resulted in a number format exception because 31 one's is WAY larger than the max an int can hold of 2147483647. Anything larger than that max will result in the same exception.