Java將十六進制轉換爲帶符號的8位代碼

Question

我需要將十六進制數字表示爲字符串轉換爲帶符號的8位字符串。

例如：鑑於此代碼段：

String hexideciaml = new String("50 4b e0 e7"); 
String signed8Bit = convertHexToSigned8Bit(hexideciaml); 
System.out.print(signed8Bit); 

輸出應該是： 「80 75 -32 -25」

所以我非常希望實現的一部分這個網站在Java中。 https://www.mathsisfun.com/binary-decimal-hexadecimal-converter.html

更新：解決方案需要JRE6沒有其他Jars。

Answer 1

的Java 1.8（流）

import java.util.Arrays; 

public class HexToDec { 

    public static String convertHexToSigned8Bit(String hex) { 
     return Arrays 
       .stream(hex.split(" +")) 
       .map(s -> "" + (byte) Integer.parseInt(s, 16)) 
       .reduce((s, s2) -> s + " " + s2) 
       .get(); 
    } 


    public static void main(String[] args) { 
     String hexidecimal = "50 4b e0 e7"; 
     String signed8Bit = convertHexToSigned8Bit(hexidecimal); 
     System.out.print(signed8Bit); 
    } 

} 

的Java < 1.8

import java.util.Arrays; 

public class HexToDec { 

    public static String convertHexToSigned8Bit(String hex) { 
     String[] tokens = hex.split(" +"); 
     StringBuilder result = new StringBuilder(); 
     for (int i = 0; i < tokens.length - 1; i++) { //append all except last 
      result.append((byte) Integer.parseInt(tokens[i], 16)).append(" "); 
     } 
     if (tokens.length > 1) //if more than 1 item in array, add last one 
      result.append((byte) Integer.parseInt(tokens[tokens.length - 1], 16)); 
     return result.toString(); 
    } 


    public static void main(String[] args) { 
     String hexidecimal = "50 4b e0 e7"; 
     String signed8Bit = convertHexToSigned8Bit(hexidecimal); 
     System.out.print(signed8Bit); 
    } 

} 

輸出爲：80 75 -32 -25