如果我必须从直接bytebuffer读/写,最好首先读/写入线程本地字节数组,然后使用字节数组完全更新(用于写入)直接bytebuffer?
解决方法
Is get/put from a non-direct bytebuffer faster than get/put from direct bytebuffer ?
如果将堆缓冲区与不使用本机字节顺序的直接缓冲区进行比较(大多数系统是小端序列,而直接ByteBuffer的默认值是大端),性能非常相似.
如果使用本机有序字节缓冲区,则对于多字节值,性能可能会更好.对于字节,无论你做什么都没有什么区别.
在HotSpot / OpenJDK中,ByteBuffer使用Unsafe类,许多本机方法被视为intrinsics.这是依赖于JVM的,AFAIK是Android VM将其视为最新版本的内在函数.
如果您转储生成的程序集,可以看到Unsafe中的内在函数被转换为一个机器代码指令.即它们没有JNI呼叫的开销.
事实上,如果您进行微调,您可能会发现ByteBuffer getXxxx或setXxxx的大部分时间都花在边界检查中,而不是实际的内存访问.因此,当我必须达到最高性能时,我仍然直接使用Unsafe(注意:Oracle不鼓励这样做)
If I have to read / write from direct bytebuffer,is it better to first read /write in to a thread local byte array and then update ( for writes ) the direct bytebuffer fully with the byte array ?
我会讨厌看到比那更好. ;)听起来很复杂
通常最简单的解决方案更好,更快.
您可以使用此代码自行测试.
public static void main(String... args) { ByteBuffer bb1 = ByteBuffer.allocateDirect(256 * 1024).order(ByteOrder.nativeOrder()); ByteBuffer bb2 = ByteBuffer.allocateDirect(256 * 1024).order(ByteOrder.nativeOrder()); for (int i = 0; i < 10; i++) runTest(bb1,bb2); } private static void runTest(ByteBuffer bb1,ByteBuffer bb2) { bb1.clear(); bb2.clear(); long start = System.nanoTime(); int count = 0; while (bb2.remaining() > 0) bb2.putInt(bb1.getInt()); long time = System.nanoTime() - start; int operations = bb1.capacity() / 4 * 2; System.out.printf("Each putInt/getInt took an average of %.1f ns%n",(double) time / operations); }
版画
Each putInt/getInt took an average of 83.9 ns Each putInt/getInt took an average of 1.4 ns Each putInt/getInt took an average of 34.7 ns Each putInt/getInt took an average of 1.3 ns Each putInt/getInt took an average of 1.2 ns Each putInt/getInt took an average of 1.3 ns Each putInt/getInt took an average of 1.2 ns Each putInt/getInt took an average of 1.2 ns Each putInt/getInt took an average of 1.2 ns Each putInt/getInt took an average of 1.2 ns
我确定JNI通话需要的时间超过1.2 ns.
为了证明它不是“JNI”的呼叫,而是围绕它引起延迟的问题.您可以直接使用Unsafe编写相同的循环.
public static void main(String... args) { ByteBuffer bb1 = ByteBuffer.allocateDirect(256 * 1024).order(ByteOrder.nativeOrder()); ByteBuffer bb2 = ByteBuffer.allocateDirect(256 * 1024).order(ByteOrder.nativeOrder()); for (int i = 0; i < 10; i++) runTest(bb1,ByteBuffer bb2) { Unsafe unsafe = getTheUnsafe(); long start = System.nanoTime(); long addr1 = ((DirectBuffer) bb1).address(); long addr2 = ((DirectBuffer) bb2).address(); for (int i = 0,len = Math.min(bb1.capacity(),bb2.capacity()); i < len; i += 4) unsafe.putInt(addr1 + i,unsafe.getInt(addr2 + i)); long time = System.nanoTime() - start; int operations = bb1.capacity() / 4 * 2; System.out.printf("Each putInt/getInt took an average of %.1f ns%n",(double) time / operations); } public static Unsafe getTheUnsafe() { try { Field theUnsafe = Unsafe.class.getDeclaredField("theUnsafe"); theUnsafe.setAccessible(true); return (Unsafe) theUnsafe.get(null); } catch (Exception e) { throw new AssertionError(e); } }
版画
Each putInt/getInt took an average of 40.4 ns Each putInt/getInt took an average of 44.4 ns Each putInt/getInt took an average of 0.4 ns Each putInt/getInt took an average of 0.3 ns Each putInt/getInt took an average of 0.3 ns Each putInt/getInt took an average of 0.3 ns Each putInt/getInt took an average of 0.3 ns Each putInt/getInt took an average of 0.3 ns Each putInt/getInt took an average of 0.3 ns Each putInt/getInt took an average of 0.3 ns
因此,您可以看到本地电话比您期望的JNI呼叫快得多.这种延迟的主要原因可能是L2缓存速度.