How to convert a binary String to a decimal string in Java

Question

I was working on a homework and thought I had finished but the teacher told me that it wasn't what he was looking for so I need to know how I can convert a binary number that is stored as a String to a decimal string without using any built-in function outside of length(), charAt(), power function, and floor/ceiling in Java.

This is what I had on the beginning.

import java.util.Scanner;

public class inclass2Fall15Second {
    public static void convertBinaryToDecimalString() {
        Scanner myscnr = new Scanner(System.in);

        int decimal = 0;

        String binary;
        System.out.println("Please enter a binary number: ");
        binary = myscnr.nextLine();
        decimal = Integer.parseInt(binary, 2);
        System.out.println("The decimal number that corresponds to " + binary + " is " + decimal);
    }

    public static void main (String[] args) {
        convertBinaryToDecimalString();
    }
}

Show us what you have tried already. Without any code we can't help you. — Krzysztof Krasoń
– Krzysztof Krasoń, Commented Oct 16, 2015 at 5:34
What is the language you are coding in? Without this information, we'll be unable to help you. — David Harris
– David Harris, Commented Oct 16, 2015 at 5:40
I updated my post with the information. I apologize for not being specific. — B D M
– B D M, Commented Oct 16, 2015 at 6:17

Nayeem Zen · Accepted Answer · 2015-10-17 03:58:05Z

1

To convert a base 2 (binary) representation to base 10 (decimal), multiply the value of each bit with 2^(bit position) and sum the values.

e.g. 1011 -> (1 * 2^0) + (1 * 2^1) + (0 * 2^2) + (1 * 2^3) = 1 + 2 + 0 + 8 = 11

Since binary is read from right-to-left (i.e. LSB (least significant bit) is on the rightmost bit and MSB (most-significant-bit) is the leftmost bit), we traverse the string in reverse order.

To get the bit value, subtract '0' from the char. This will subtract the ascii value of the character with the ascii value of '0', giving you the integer value of the bit.

To calculate 2^(bit position), we can keep a count of the bit position, and increment the count on each iteration. Then we can just do 1 << count to obtain the value for 2 ^ (bit position). Alternatively, you could do Math.pow(2, count) as well, but the former is more efficient, since its just a shift-left instruction.

Here's the code that implements the above:

public static int convertBinStrToInt(String binStr) {
    int dec = 0, count = 0;
    for (int i = binStr.length()-1; i >=0; i--) {
        dec += (binStr.charAt(i) - '0') * (1 << count++);
    }

    return dec;
}

edited Oct 17, 2015 at 3:58

answered Oct 16, 2015 at 5:43

Nayeem Zen

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1 Comment

B D M Over a year ago

Thank you so much @zindigo. It makes sense now and I appreciate the detailed explanation, it helped me understand the code better.

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