JavaNIOBuffer分配详解：allocate与allocateDirect的区别与应用

一、nio之Buffer

1 ByteBuffer.allocate分配的内存在JVM里面&＃xff0c;返回ByteBuffer的子类HeapByteBuffer&＃xff0c;初始值0&＃xff1b;

2 ByteBuffer.allocateDirect分配的内存在系统里面&＃xff0c;返回MappedByteBuffer的子类DirectByteBuffer&＃xff0c;初始值0&＃xff1b;

3 系统内存分配速度慢&＃xff0c;但使用效率更高&＃xff0c;并且不参与垃圾回收&＃xff1b;

4 buffer的四个属性&＃xff0c;0 <&＃61; mark <&＃61; position <&＃61; limit <&＃61; capacity&＃xff0c;position决定了read和write的起始位置&＃xff1b;

5 mark用于reset使用&＃xff0c;没有mark调用reset会抛异常&＃xff0c;初始状态无标记&＃xff0c;只有position小于mark时标记被丢弃&＃xff1b;

6 buffer.get()和buffer.put(byte)都是相对方法&＃xff0c;都会导致position&＃43;&＃43;&＃xff1b;

7 buffer.get(int)和buffer.put(int, byte)都是绝对方法&＃xff0c;不会导致position变化&＃xff1b;

8 clear只是将position设置为0&＃xff0c;limit设置为capacity&＃xff0c;并不清除数据&＃xff0c;做好put准备&＃xff1b;

9 flip将limit设置为position&＃xff0c;position设置为0&＃xff0c;做好get准备&＃xff1b;

10 rewind将position设置为0&＃xff0c;做好重新get准备&＃xff1b;

11 remaining表示position到limit之间的元素数量&＃xff1b;

12 compact将position到limit之间的元素移动到数组开始&＃xff1b;

13 多线程访问是不安全的&＃xff0c;需要使用同步手段进行控制&＃xff1b;

14 ByteOrder.nativeOrder()获取本机字节序列&＃xff1b;

15 视图和allocateDirect情况下hasArray返回false&＃xff1b;

16 array()方法返回底层数组&＃xff1b;

1 属性测试&＃xff1a;

ByteBuffer buffer &＃61; ByteBuffer.allocate(12);

System.out.println(buffer);

System.out.println(buffer.position(2).mark());

System.out.println(buffer.position(4));

System.out.println(buffer.reset());

System.out.println(buffer.position(0));

buffer.reset(); //抛出InvalidMarkException异常

输出&＃xff1a;

java.nio.HeapByteBuffer[pos&＃61;0 lim&＃61;12 cap&＃61;12]

java.nio.HeapByteBuffer[pos&＃61;2 lim&＃61;12 cap&＃61;12]

java.nio.HeapByteBuffer[pos&＃61;4 lim&＃61;12 cap&＃61;12]

java.nio.HeapByteBuffer[pos&＃61;2 lim&＃61;12 cap&＃61;12]

java.nio.HeapByteBuffer[pos&＃61;0 lim&＃61;12 cap&＃61;12]

2 数据操作&＃xff0c;默认大端模式&＃xff1a;

ByteBuffer buffer &＃61; ByteBuffer.allocate(12);

byte by &＃61; 1;

int in &＃61; 32;

buffer.put(by);

buffer.putInt(in);

System.out.println(buffer);

System.out.println(buffer.get(4));

输出&＃xff1a;

java.nio.HeapByteBuffer[pos&＃61;5 lim&＃61;12 cap&＃61;12]

32

3duplicate复制原buffer所有状态&＃xff0c;slice将position和limit之间的内容映射成新buffer&＃xff0c;两者都使用原来内存&＃xff1a;

1、duplicate复制mark值&＃xff0c;slice清除mark值&＃xff1b;

2、asReadOnlyBuffer与duplicate完成相同&＃xff0c;除了缓冲区只读以外&＃xff1b;

ByteBuffer buffer &＃61; ByteBuffer.allocate(12);

byte by &＃61; 1;

int in &＃61; 32;

buffer.put(by);

buffer.putInt(in);

System.out.println(buffer);

System.out.println(buffer.array());

System.out.println(buffer.duplicate());

System.out.println(buffer.duplicate().array());

buffer.position(4);

System.out.println(buffer.slice());

System.out.println(buffer.slice().array());

System.out.println(buffer.slice().arrayOffset());

输出&＃xff1a;

java.nio.HeapByteBuffer[pos&＃61;5 lim&＃61;12 cap&＃61;12]

[B&＃64;1c0ec97

java.nio.HeapByteBuffer[pos&＃61;5 lim&＃61;12 cap&＃61;12]

[B&＃64;1c0ec97

java.nio.HeapByteBuffer[pos&＃61;0 lim&＃61;8 cap&＃61;8]

[B&＃64;1c0ec97

4

4 数组转换到ByteBuffer

byte[] bytes &＃61; new byte[12];

ByteBuffer buf &＃61; ByteBuffer.wrap(bytes, 1, 1);

System.out.println(buf);

System.out.println(bytes);

System.out.println(buf.array());

System.out.println(buf.arrayOffset());

输出&＃xff1a;

java.nio.HeapByteBuffer[pos&＃61;1 lim&＃61;2 cap&＃61;12]

[B&＃64;ecb281

[B&＃64;ecb281

0

5 大小端

ByteBuffer buf &＃61; ByteBuffer.allocate(12);

System.out.println(buf.order());

System.out.println(buf.order(ByteOrder.LITTLE_ENDIAN).order());

输出&＃xff1a;

BIG_ENDIAN

LITTLE_ENDIAN

6 IntBuffer测试

ByteBuffer buffer &＃61; ByteBuffer.allocate(16);

buffer.position(4);

IntBuffer intBuf &＃61; buffer.asIntBuffer();

System.out.println(intBuf);

System.out.println(intBuf.order());

System.out.println(intBuf.hasArray());

intBuf &＃61; IntBuffer.allocate(4);

System.out.println(intBuf);

System.out.println(intBuf.order());

System.out.println(intBuf.hasArray());

输出&＃xff1a;

java.nio.ByteBufferAsIntBufferB[pos&＃61;0 lim&＃61;3 cap&＃61;3]

BIG_ENDIAN

false

java.nio.HeapIntBuffer[pos&＃61;0 lim&＃61;4 cap&＃61;4]

LITTLE_ENDIAN

true

7 compareTo对所有元素进行比较

ByteBuffer buf1 &＃61; ByteBuffer.allocate(4);

ByteBuffer buf2 &＃61; ByteBuffer.allocateDirect(4);

IntBuffer buf3 &＃61; IntBuffer.allocate(1);

System.out.println(buf1.compareTo(buf2));

System.out.println(buf1.asIntBuffer().compareTo(buf3));

输出&＃xff1a;

0

0