如何将大二进制转换为十进制？

Question

It works when I enter binary like 10, 1011, 1101. But it always prints "Not binary" when I enter 10011010010, whereas it should be 1234. How to modify it?

import java.util.Scanner;
public class Binary {

    public static int toDecimal(String b) {     
        int decimal = Integer.parseInt(b,2);
        return decimal;
    }


    public static boolean isBinary(String b) {
        int inputNum = Integer.parseInt(b);

        while(inputNum != 0){
            if(inputNum % 10 > 1){
                return false;
            }
            inputNum = inputNum / 10;
        }
        return true;
    }

    public static void main(String[] args) {

        System.out.print("Enter binary: ");
        Scanner in = new Scanner(System.in);
        String binaryNum = in.next();

        try{
            int intNum = Integer.parseInt(binaryNum);

            boolean isBinary = isBinary(binaryNum);
            if(isBinary){
                int outputDecimal = toDecimal(binaryNum);
                System.out.println("\n"+ outputDecimal +" in decimal");
            }else{
                System.out.println("\n" + "Not binary!");
            }
        }catch(Exception e){
            System.out.println("\n" + "Not binary!");
        }
    }
}

Answer 1

Why so many conversions from string to int? Keep the input in string form for checking if the input is binary or not. Then finally convert the string to number type. If you think your input will not exceed the maximum int value, you can use parseInt() otherwise use either long or BigInteger .

With input as string, your isBinary() may look like this:

public static boolean isBinary(String b) {
    for(int i=0; i<b.length(); i++) {
        if(b.charAt(i) != '0' && b.charAt(i) != '1') {
            return false;
        }
    }
    return true;
}

Then change all your Integer.parseInt() calls to long.parseLong() , and change the associated variables accordingly, ie from int to long. That should be enough to get what you expect, for now. What I mean is for arbitrarily large inputs, you have to use BigInteger (see below).

---

If you choose to use java.math.BigInteger , you can use the constructor with the radix like so:

public static BigInteger toDecimal(String b) {
    return new BigInteger(b, 2);
}

Answer 2

The problem is that you are trying to parse a number that exceeds
Integer.MAX_VALUE which is 2^31 - 1 :

int intNum = Integer.parseInt(binaryNum);

Try using Long.parseLong instead. A long has 64 bits and has a range of -9,223,372,036,854,775,808 to 9,223,372,036,854,775,807 which should be enough to cover your various inputs (although obviously will fail for longer numbers).

Answer 3

This works although it would be more efficient to process more than 1 bit at a time. Result for 1000000000000000000000000000000000000000000000000000000000000000 is 9,223,372,036,854,775,808

import java.util.regex.Pattern;
import java.math.BigDecimal;

...

public static BigDecimal toDecimal(String b) {     
    BigDecimal bd1 = new BigDecimal(b.charAt(0)=='1'?1:0);
    BigDecimal two = new BigDecimal(2);
    for (int i = 1; i<b.length(); i++) {
        bd1 = bd1.multiply(two);
        bd1 = bd1.add(new BigDecimal(b.charAt(i)=='1'?1:0));
    }
    return bd1;
}

public static boolean isBinary(String b) {
    return Pattern.compile("[01]+").matcher(b).matches();
}    

public static void main(String[] args) {
    System.out.print("Enter binary: ");
    Scanner in = new Scanner(System.in);
    String binaryNum = in.next();

    try{
        boolean isBinary = isBinary(binaryNum);
        if(isBinary){
            BigDecimal outputDecimal = toDecimal(binaryNum);
            System.out.println("\n"+ outputDecimal +" in decimal");
        }else{
            System.out.println("\n" + "Not binary!");
        }
    }catch(Exception e){
        System.out.println("\n" + "Not binary!");
    }
}

More minimal version with use of BigInteger

public static void main(String[] args) {
    System.out.print("Enter binary: ");
    Scanner in = new Scanner(System.in);
    String binaryNum = in.next();

    try{
        if(Pattern.compile("[01]+").matcher(binaryNum).matches()){
            BigInteger outputDecimal = new BigInteger(binaryNum, 2);
            System.out.println("\n"+ outputDecimal +" in decimal");
        }else{
            System.out.println("\n" + "Not binary!");
        }
    }catch(Exception e){
        System.out.println("\n" + "Not binary!");
    }
}