java - 在 Java 中从字节数组创建 48 位有符号值

Question

这个问题已经有答案了:

How to convert a byte array to its numeric value (Java)? (9 个回答)

已关闭 4 年前。

我试图从字节数组中提取 6 个字节并将它们转换为 48 位有符号整数值(即 Java long)。如何才能做到这一点？

编辑:例如，如果字节数组具有以下之一:

byte[] minInt48 = new byte[] { (byte)0x80, 0, 0, 0, 0, 0 };
byte[] maxInt48 = new byte[] { (byte)0x7F, (byte)0xFF, (byte)0xFF, (byte)0xFF, (byte)0xFF, (byte)0xFF };

如何将其解析为 Java 原语(即 Java long)以便保留符号值？

Answer 1

首先您需要了解符号扩展。您需要将每个字节视为无符号值。

      long v = 0;
      byte q = -2; // unsigned value of 254
      v = v + q;
      System.out.println(v); // prints -2 which is not what you want.
      v = 0;
      v = v + (q & 0xFF); // mask off sign extension from q.
      System.out.println(v); // prints 254 which is correct.

这是一种方法。

     long val = 0;
     byte[] bytes = { -1, 12, 99, -121, -3, 123
     };
     for (int i = 0; i < bytes.length; i++) {
        // shift the val left by 8 bits first.
        // then add b.  You need to mask it with 0xFF to
        // eliminate sign extension to a long which will
        // result in an incorrect conversion.
        val = (val << 8) | ((i == 0 ? bytes[i]
              : bytes[i] & 0xFF));
     }
     System.out.println(val);