CODEWITHSUNDEEP
javabinarydecimal
MrSpyGuy
MrSpyGuy

Reputation: 13

Convert Large Binary String to Decimal in Java

Need a bit of help here:

I was assigned a project to convert a Hex values in a file to Decimal using Java without using methods available in Java that can do this directly(we have to do it the long way).

I have converted my Hex values to Binary successfully but am having trouble converting the Binary strings to Decimal. The output is not correct; I only need help in this section so I just put the first binary value I need to convert to make things simple to understand and explain. Here is what I have so far

public class binToDec {

    public static void main(String[] args) {
        int decimal = 0;
        String binary = "101010111100110111101111101010111100";
        for (int pow = (binary.length()-1); pow > -1; pow--) {
            if (binary.charAt(pow)=='1'){
            decimal += (Math.pow(2, pow));
            }
        }
        System.out.print(decimal);
    }
}

run: 2147483647 //this is incorrect. it should be 46118402748

Thank you so much for your help

Upvotes: 0

Views: 3406

Answers (2)

rgettman
rgettman

Reputation: 178263

There are 2 problems with your code.

  1. Overflow is certainly occurring, because an int can hold only a value up to 2^31 - 1, and there are more than 31 bits in your binary string. Declare decimal to be a long.

    long decimal = 0;

  2. You are applying the exponent from the wrong end of your loop when adding to decimal. The 0 char is the most significant, but you are sending in an exponent of 0, as if it was the least significant. Try

    decimal += (Math.pow(2, (len - pow - 1)));

    (This assumes len is declared as int len = binary.length();.)

    Using Math.pow could be considered overkill. You can also try

    decimal += 1L << (len - pow - 1);

Upvotes: 3

Jashaszun
Jashaszun

Reputation: 9270

It won't work because int in Java is 32-bit, and your binary string there has 36 bits. This will result in overflow.

Upvotes: 0

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