Java代码将字节转换为十六进制 [英] Java code To convert byte to Hexadecimal

问题描述

我有一个字节数组。
我希望将该数组的每个字节字符串转换为相应的十六进制值。

I have an array of bytes. I want each byte String of that array to be converted to its corresponding hexadecimal values.

Java中是否有任何函数将字节数组转换为十六进制数？

Is there any function in Java to convert a byte array to Hexadecimal ?

推荐答案

    byte[] bytes = {-1, 0, 1, 2, 3 };
    StringBuilder sb = new StringBuilder();
    for (byte b : bytes) {
        sb.append(String.format("%02X ", b));
    }
    System.out.println(sb.toString());
    // prints "FF 00 01 02 03 "

参见

• java.util.Formatter 语法
  
  
  
  
  • ％[flags] [width]转换
    
    
    
    
    • 标记'0' - 结果将为零填充
    
    
    • 宽度 2
    
    
    • 转换'X' - 结果格式为十六进制整数，大写
    
    

    See also

    • java.util.Formatter syntax
      • %[flags][width]conversion
        • Flag '0' - The result will be zero-padded
        • Width 2
        • Conversion 'X' - The result is formatted as a hexadecimal integer, uppercase

        查看问题的文本，这也可能是要求：

        Looking at the text of the question, it's also possible that this is what is requested:

            String[] arr = {"-1", "0", "10", "20" };
            for (int i = 0; i < arr.length; i++) {
                arr[i] = String.format("%02x", Byte.parseByte(arr[i]));
            }
            System.out.println(java.util.Arrays.toString(arr));
            // prints "[ff, 00, 0a, 14]"
        

        
        
        

        ---

        
        
        

        这里的几个答案使用 Integer.toHexString（int）的 ;这是可行的，但有一些警告。由于参数是 int ，因此对 byte 参数执行扩展的原始转换，该参数涉及符号扩展。 / p>
        
        

        ---

        Several answers here uses Integer.toHexString(int); this is doable, but with some caveats. Since the parameter is an int, a widening primitive conversion is performed to the byte argument, which involves sign extension.

            byte b = -1;
            System.out.println(Integer.toHexString(b));
            // prints "ffffffff"
        

        8位字节，用Java签名，符号扩展为32位 int 。要有效撤消此符号扩展，可以使用 0xFF 屏蔽字节。

        The 8-bit byte, which is signed in Java, is sign-extended to a 32-bit int. To effectively undo this sign extension, one can mask the byte with 0xFF.

            byte b = -1;
            System.out.println(Integer.toHexString(b & 0xFF));
            // prints "ff"
        

        使用 toHexString的另一个问题是它没有用零填充：

        Another issue with using toHexString is that it doesn't pad with zeroes:

            byte b = 10;
            System.out.println(Integer.toHexString(b & 0xFF));
            // prints "a"
        

        两个因素组合应该是 String.format 解决方案更优先。

        Both factors combined should make the String.format solution more preferrable.

        
        
        • JLS 4.2.1积分类型和值
          
          
          
          
          • 对于字节，来自 -128 到 127 ，包括
          
          
          • JLS 4.2.1 Integral Types and Values
            • For byte, from -128 to 127, inclusive

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