Java:将二进制转换为十六进制时出现 NumberFormatException

Question

我有从 BIN 转换为 HEX 的作业，我编写了以下算法代码:

import java.util.Scanner;

public class BinaryToHex {

    public static void main(String[] args) {
        Scanner input = new Scanner(System.in);
        System.out.print("Binary number: ");
        String b = input.next();
        int bin = Integer.parseInt(b);
        int arrlength = b.length();

        while (arrlength%4  !=  0){
            arrlength++;
        }

        int[] arrbin =  new int [arrlength];
        int digit = 0;
        String hex = "";
        String str;
        int conv;

        for (int i = arrlength-1; i>=0; i--){
            digit = bin%10;
            arrbin[i]=digit;
            bin = bin/10;
        }

        System.out.print("Hex value = ");

        for (int index = 0; index < arrlength; index=index+4){
            str = "" + arrbin[index] + "" + arrbin[index+1] + "" + arrbin[index+2] + "" + arrbin[index+3];
            switch(str){
            case "0000": str = "0"; break;
            case "0001": str = "1"; break;
            case "0010": str = "2"; break;
            case "0011": str = "3"; break;
            case "0100": str = "4"; break;
            case "0101": str = "5"; break;
            case "0110": str = "6"; break;
            case "0111": str = "7"; break;
            case "1000": str = "8"; break;
            case "1001": str = "9"; break;
            case "1010": str = "A"; break;
            case "1011": str = "B"; break;
            case "1100": str = "C"; break;
            case "1101": str = "D"; break;
            case "1110": str = "E"; break;
            case "1111": str = "F"; break;
            }
            System.out.print(str);
        }
    }

}

问题是，当我尝试转换更大的数字时，它会抛出:

Exception in thread "main" java.lang.NumberFormatException: For input string: "10101010101010"
    at java.lang.NumberFormatException.forInputString(Unknown Source)
    at java.lang.Integer.parseInt(Unknown Source)
    at java.lang.Integer.parseInt(Unknown Source)
    at BinaryToHex.main(BinaryToHex.java:9)

我知道问题与 int 类型有关，但我不知道如何解决这个问题。我尝试使用 long 类型 - 结果是一样的。

如果你们能帮我更正这段代码，以便处理更大的数字，我将不胜感激。

Answer 1

在 Java 中，int 是一个带符号的 32 位，所以它的范围是

-2,147,483,648 to 2,147,483,647

所以你的值 10101010101010 超出了这个范围。

尝试使用更大的东西，比如 Long

long bin = Long.valueOf("10101010101010");
System.out.println(bin);

参见 Primitive Data Types