在Java中读取长二进制数

Question

110011000111100111001100011110011100110001111001110011000111100111001100011110011100110001111001

Answer 1

That's not a valid way to check a binary number. You're converting to an int in base 10, then checking that each of the digits in base 10 is zero or one. The conversion itself will fail on long enough input strings, and if it doesn't the checking will fail as well.

You shouldn't be converting it all all, you should be checking the input string itself.

EDIT Actually you don't have to do anything. Integer.parseInt() will check it for you and throw NumberFormatException if the string isn't a number in the specified radix.

Answer 2

You are parsing your binary digit string as a decimal integer first. If it has more than 10 significant digits then its decimal interpretation is too big to fit in an int , so the decimal conversion fails.

When you are going to parse the digit string as a binary number, simply avoid first parsing it as a decimal one. For instance, most of what you posted could be reduced to this:

int inputNumber = Integer.parseInt(returnEnterNumber(), 
        binaryToDecimal.isSelected() ? 2 : 10);

Answer 3

Take a look at Integers MAX values:

public class MainClass {

public static void main(String[] arg) {
System.out.println(Integer.MAX_VALUE);   

System.out.println(Integer.MIN_VALUE);   

 }
}

the output will be:

2147483647
-2147483648

This means that if you have more than 10 digits, you have exceeded the max number for the Integer data type. Try Using BigInteger on your binary value or consider returning it as String

Answer 4

这是一行代码，可以满足您的需求

System.out.println(new BigInteger("10101010101010111101010101001101010101010101001010101010101001011101010101010101",2).toString());