JVM大端判断

JVM采用大端方式存多字节的数据，判断方法如下：

     public static void bytesToInt() throws IOException {
         /**
          * 将字节数组（byte[]）转为整形(int)
          */
         byte[] byteAr = new byte[] { 0x78, 0x56, 0x34, 0x12 };
         ByteArrayInputStream bais = new ByteArrayInputStream(byteAr);
         DataInputStream dis = new DataInputStream(bais);
         System.out.println(Integer.toHexString(dis.readInt()));// output: 78563412，说明是大端
     }

     public static void intToBytes() throws IOException {
         /**
          * 将整形(int)转为字节数组（byte[]）
          */
         int a = 0x12345678;
         ByteArrayOutputStream baos = new ByteArrayOutputStream();
         DataOutputStream dos = new DataOutputStream(baos);
         dos.writeInt(a);
         byte[] b = baos.toByteArray();
         for (int i = 0; i < 4; i++) {
             System.out.print(Integer.toHexString(b[i]));
         }
         System.out.println();// Output: 12345678，说明是大端
     }

采用大小模式对数据进行存放的主要区别在于在存放的字节顺序，大端方式将高位存放在低地址，小端方式将高位存放在高地址。采用大端方式数据存放与阅读顺序一致，符合人类的正常思维，而采用小端方式进行数据存放利于计算机处理。到目前为止，采用大端或者小端进行数据存放，其孰优孰劣也没有定论。

参考资料：

Java's Virtual Machine's Endianness-*

The Java Virtual Machine Specification, Java SE 7 Edition, Chapter 4: The class File Format

A class file consists of a stream of 8-bit bytes. All 16-bit, 32-bit, and 64-bit quantities are constructed by reading in two, four, and eight consecutive 8-bit bytes, respectively. Multibyte data items are always stored in big-endian order, where the high bytes come first. In the Java SE platform, this format is supported by interfaces java.io.DataInput and java.io.DataOutput and classes such as java.io.DataInputStream and java.io.DataOutputStream.