在Java中将大二进制字符串转换为十进制

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

Answer 1

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); 

Answer 2

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