Java 将十六进制转换为带符号的 8 位代码

Question

我需要将表示为字符串的十六进制数字转换为带符号的 8 位字符串。

例如:给定以下代码片段:

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

输出应该是: “80 75 -32 -25”

所以我非常想用 Java 实现这个网站的一部分。 https://www.mathsisfun.com/binary-decimal-hexadecimal-converter.html

更新:解决方案需要适用于 JRE6，没有其他 Jars。

Answer 1

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);
    }

}

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);
    }

}

输出为:80 75 -32 -25