Convert Large Binary String to Decimal in Java

Question

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

Assignment, you can try out urself.. by the way, Java Class name is desired to be BinToDec and not binToDec — spiderman
– spiderman, Commented Jul 9, 2014 at 22:51

rgettman · Accepted Answer · 2014-07-09 22:54:55Z

3

There are 2 problems with your code.

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;
```
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);
```

answered Jul 9, 2014 at 22:54

rgettman

179k30 gold badges282 silver badges365 bronze badges

Sign up to request clarification or add additional context in comments.

4 Comments

Jashaszun Over a year ago

Your answer is better than mine, but one edit: it should be "there are more than 32 bits in your binary string".

rgettman Over a year ago

@Jashaszun In Java, ints are signed, so using that most significant 32nd bit will cause overflow (if it's set). In fact, using long just puts off the problem, as this will still have an overflow problem if there are more than 63 bits represented in the binary string.

MrSpyGuy Over a year ago

Thank you so much for answering and explaining! Got it to work perfectly!

Jashaszun Over a year ago

@rgettman Ah right, of course! I can't believe I didn't get that by myself.

Jashaszun · Accepted Answer · 2014-07-09 22:52:57Z

0

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

answered Jul 9, 2014 at 22:52

Jashaszun

9,2503 gold badges31 silver badges58 bronze badges

2 Comments

MrSpyGuy Over a year ago

what types can I use?

Jashaszun Over a year ago

Long, because it's 64-bit.

Collectives™ on Stack Overflow

Convert Large Binary String to Decimal in Java

2 Answers 2

4 Comments

2 Comments

Your Answer

Hot Network Questions

Collectives™ on Stack Overflow

2 Answers 2

4 Comments

2 Comments

Your Answer

Sign up or log in

Post as a guest

Related