将二进制数转换为十进制数的Java程序。 输入是一串零和一串

Question

I have to create a java program that converts binary to decimal using the following steps. Being new at this I did something, but I don't know what I did wrong or how to continue.

public class BinaryToDecimal {
public static void main(String args[]){
    long sum = 0;
        int result;
        String s = "1001010101011010111001011101010101010101";
        for(int i = s.length()-1; i <= 0; i--){
            result = (int)Math.pow(2, i);
            if(s.charAt(i) == '1')
                sum=sum + result;
        }
        System.out.println(sum);
    }
    }

Use a loop to read (charAt()) each digit (0/1 char) in the input string, scanning from right to left;

Use the loop to build the required powers of 2;

Use a conditional statement to deal with 0 and 1 separately;

Debug using simple input, eg 1, 10, 101, and print intermediate values in the loop.

Use your program to find the decimal value of the following binary number:

1001010101011010111001011101010101010101

Answer 1

Do this only if your decimal value is at most 2147483647 or the maximum value an int can be in Java. If you don't know, just check the length of your string. If it's less than or equal to 32 ie 4 bytes, then you can use parseInt.:

int decimalValue = Integer.parseInt(s, 2);

Refer HERE for more info on the Integer.parseInt();

But if it's more, you can use your code. I modified your loop which is where your problem was:

 String s = "1001010101011010111001011101010101010101";
 long result = 0;
 for(int i = 0; i < s.length(); i++){
    result = (long) (result + (s.charAt(i)-'0' )* Math.pow(2, s.length()-i-1));
  }
    System.out.println(result);

Answer 2

The loop runs from i = s.length()-1 until i <= 0. This should be i>=0.
The next problem is "int result". It works fine with result as a long ;) (Reason: You calculate a 40-bit value at the MostSignificantBit, but Integers only use 32-bit)
Also: You start at the rightmost Bit with i=s.length()-1. But the power that you calculate for it is 2^(s.length()-1) though it should be 2^0=1.
The solution is: result = (long)Math.pow(2, s.length()-1-i)

Edit: I really like the solution of user2316981 because of its clear structure (without Math.pow, should be faster by using shift instead). And loops from 0 to MSB as I do with Double&Add algorithm. Can't comment on it yet, but thanks for the reminder ;)

Answer 3

The first thing I notice is that your binary number has more than 32 bits. This cannot be represented in the space of an int, and will result in overflow.

As a simpler answer, I ran the following and got the correct value at the end, it just uses simple bit shifts.

For each index in the string, if the character is 1, it sets the corresponding bit in the result.

public class BinaryToDecimal {
    public static void main(String[] args) {
        long sum;
        String bin = "1001010101011010111001011101010101010101";
        sum = 0;
        for (int i = 0; i < bin.length(); i++) {
            char a = bin.charAt(i);
            if (a == '1') {
               sum |= 0x01;
            }
            sum <<= 1;
        }
        sum >>= 1;
        System.out.println(sum);
    }
 }

Answer 4

import java.util.*;
import java.lang.Math;

class deci {

    int convert(int n) {
        int tem=1,power=0;
        int decimal=0;
        for (int j=0;j<n;j++) {
            if(n==0) {
                break;
            } else {
                while(n>0) {
                    tem=n%10;
                    decimal+=(tem*(Math.pow(2,power)));
                    n=n/10;
                    power++;
                }
            }    
        } 
        return decimal;
    }

    public static void main(String args[]) {
        System.out.print("enter the binary no");
        Scanner sc=new Scanner(System.in);
        int n=sc.nextInt();
        deci dc=new deci();
        int i=dc.convert(n);
        System.out.print(i);
    }

}