我正在尝试使用Unsafe迭代内存而不是遍历byte []中的值。使用unsafe分配内存块。内存足以容纳65536个字节值。
我在尝试这个：
char aChar = some character
if ((byte) 0 == (unsafe.getByte(base_address + aChar) & mask)){
// do something
}
INSTEAD OF：
char aChar = some character
if ((byte) 0 == ( lookup[aChar] & mask )){
// do something
}
我以为不安全可以比使用常规数组访问更快地访问内存，并对每个索引执行索引检查...
只是一厢情愿地认为jvm会有一个特殊的操作（不安全），它会以某种方式使常规数组访问和迭代更快。在我看来，jvm在正常的byte []迭代中运行良好，并且可以使用普通的，纯粹的，纯粹的java代码快速完成它们。
@millimoose击中了众所周知的“钉在头上”
“不安全可能对很多事情有用，但这种微观优化程度不是其中之一。”毫米“
在非常严格的有限情况下使用不安全更快：
使用Unsafe（在我的测试中）速度较慢：
server jvm选项，也会变慢
在32位jvm（64位甚至更慢？）的代码中，不安全从9％或更多（1_GB数组和UnsafeLookup_8B（最快的一个）慢）
这有什么理由吗？**
当我运行下面发布的代码yellowB（检查1GB字节[]）时，正常情况仍然是最快的：
C:\Users\wilf>java -Xms1600m -Xprof -jar "S:\wilf\testing\dist\testing.jar"
initialize data...
initialize data done!
use normalLookup()...
Not found '0'
time : 1967737 us.
use unsafeLookup_1B()...
Not found '0'
time : 2923367 us.
use unsafeLookup_8B()...
Not found '0'
time : 2495663 us.
Flat profile of 26.35 secs (2018 total ticks): main
Interpreted + native Method
0.0% 1 + 0 test.StackOverflow.main
0.0% 1 + 0 Total interpreted
Compiled + native Method
67.8% 1369 + 0 test.StackOverflow.main
11.7% 236 + 0 test.StackOverflow.unsafeLookup_8B
11.2% 227 + 0 test.StackOverflow.unsafeLookup_1B
9.1% 184 + 0 test.StackOverflow.normalLookup
99.9% 2016 + 0 Total compiled
Stub + native Method
0.0% 0 + 1 sun.misc.Unsafe.getLong
0.0% 0 + 1 Total stub
Flat profile of 0.00 secs (1 total ticks): DestroyJavaVM
Thread-local ticks:
100.0% 1 Blocked (of total)
Global summary of 26.39 seconds:
100.0% 2023 Received ticks
C:\Users\wilf>java -version
java version "1.7.0_07"
Java(TM) SE Runtime Environment (build 1.7.0_07-b11)
Java HotSpot(TM) Client VM (build 23.3-b01, mixed mode, sharing)
在带有Windows 7_64,3位java的4核AMD64上运行测试：
initialize data...
initialize data done!
use normalLookup()...
Not found '0'
time : 1631142 us.
use unsafeLookup_1B()...
Not found '0'
time : 2365214 us.
use unsafeLookup_8B()...
Not found '0'
time : 1783320 us.
在带有Windows 7_64,64位java的4核AMD64上运行测试：
use normalLookup()...
Not found '0'
time : 655146 us.
use unsafeLookup_1B()...
Not found '0'
time : 904783 us.
use unsafeLookup_8B()...
Not found '0'
time : 764427 us.
Flat profile of 6.34 secs (13 total ticks): main
Interpreted + native Method
23.1% 3 + 0 java.io.PrintStream.println
23.1% 3 + 0 test.StackOverflow.unsafeLookup_8B
15.4% 2 + 0 test.StackOverflow.main
7.7% 1 + 0 java.io.DataInputStream.<init>
69.2% 9 + 0 Total interpreted
Compiled + native Method
7.7% 0 + 1 test.StackOverflow.unsafeLookup_1B
7.7% 0 + 1 test.StackOverflow.main
7.7% 0 + 1 test.StackOverflow.normalLookup
7.7% 0 + 1 test.StackOverflow.unsafeLookup_8B
30.8% 0 + 4 Total compiled
Flat profile of 0.00 secs (1 total ticks): DestroyJavaVM
Thread-local ticks:
100.0% 1 Blocked (of total)
Global summary of 6.35 seconds:
100.0% 14 Received ticks
42.9% 6 Compilation
答案 0 :(得分：4)
我认为您发布的两个函数基本相同，因为它们只读取1个字节，然后将其转换为int并进行进一步的比较。
每次读取4字节int或8字节更有效。我写了两个函数来做同样的事情：比较两个字节[]的内容，看看它们是否相同：
功能1：
public static boolean hadoopEquals(byte[] b1, byte[] b2)
{
if(b1 == b2)
{
return true;
}
if(b1.length != b2.length)
{
return false;
}
// Bring WritableComparator code local
for(int i = 0;i < b1.length; ++i)
{
int a = (b1[i] & 0xff);
int b = (b2[i] & 0xff);
if (a != b)
{
return false;
}
}
return true;
}
功能2：
public static boolean goodEquals(byte[] b1,byte[] b2)
{
if(b1 == b2)
{
return true;
}
if(b1.length != b2.length)
{
return false;
}
int baseOffset = UnSafe.arrayBaseOffset(byte[].class);
int numLongs = (int)Math.ceil(b1.length / 8.0);
for(int i = 0;i < numLongs; ++i)
{
long currentOffset = baseOffset + (i * 8);
long l1 = UnSafe.getLong(b1, currentOffset);
long l2 = UnSafe.getLong(b2, currentOffset);
if(0L != (l1 ^ l2))
{
return false;
}
}
return true;
}
我在笔记本电脑上运行了这两个功能（corei7 2630QM，8GB DDR3,64bit win 7,64bit Hotspot JVM），比较两个400MB字节[]，结果如下：
功能1：~670ms
功能2：~80ms
2更快。
所以我的建议是每次读取8字节并使用XOR运算符（^）：
long l1 = UnSafe.getLong(byteArray, offset); //8 byte
if(0L == l1 ^ 0xFF) //if the lowest byte == 0?
/* do something */
if(0L == l1 ^ 0xFF00) //if the 2nd lowest byte == 0?
/* do something */
/* go on... */
=============================================== =============================
嗨Wilf， 我使用你的代码制作一个测试类，如下所示，这个类比较了查找字节数组中第一个0的3个函数之间的速度：
package test;
import java.lang.reflect.Field;
import sun.misc.Unsafe;
/**
* Test the speed in looking up the 1st 0 in a byte array
* Set -Xms the same as -Xms to avoid Heap reallocation
*
* @author yellowb
*
*/
public class StackOverflow
{
public static Unsafe UnSafe;
public static Unsafe getUnsafe() throws SecurityException,
NoSuchFieldException, IllegalArgumentException,
IllegalAccessException
{
Field theUnsafe = Unsafe.class.getDeclaredField("theUnsafe");
theUnsafe.setAccessible(true);
Unsafe unsafe = (Unsafe) theUnsafe.get(null);
return unsafe;
}
/**
* use 'byte[index]' form to read 1 byte every time
* @param buf
*/
public static void normalLookup(byte[] buf)
{
for (int i = 0; i < buf.length; ++i)
{
if ((byte) 0 == buf[i])
{
System.out.println("The 1st '0' is at position : " + i);
return;
}
}
System.out.println("Not found '0'");
}
/**
* use Unsafe.getByte to read 1 byte every time directly from the memory
* @param buf
*/
public static void unsafeLookup_1B(byte[] buf)
{
int baseOffset = UnSafe.arrayBaseOffset(byte[].class);
for (int i = 0; i < buf.length; ++i)
{
byte b = UnSafe.getByte(buf, (long) (baseOffset + i));
if (0 == ((int) b & 0xFF))
{
System.out.println("The 1st '0' is at position : " + i);
return;
}
}
System.out.println("Not found '0'");
}
/**
* use Unsafe.getLong to read 8 byte every time directly from the memory
* @param buf
*/
public static void unsafeLookup_8B(byte[] buf)
{
int baseOffset = UnSafe.arrayBaseOffset(byte[].class);
//The first (numLongs * 8) bytes will be read by Unsafe.getLong in below loop
int numLongs = buf.length / 8;
long currentOffset = 0L;
for (int i = 0; i < numLongs; ++i)
{
currentOffset = baseOffset + (i * 8); //the step is 8 bytes
long l = UnSafe.getLong(buf, currentOffset);
//Compare each byte(in the 8-Byte long) to 0
//PS:x86 cpu is little-endian mode
if (0L == (l & 0xFF))
{
System.out.println("The 1st '0' is at position : " + (i * 8));
return;
}
if (0L == (l & 0xFF00L))
{
System.out.println("The 1st '0' is at position : " + (i * 8 + 1));
return;
}
if (0L == (l & 0xFF0000L))
{
System.out.println("The 1st '0' is at position : " + (i * 8 + 2));
return;
}
if (0L == (l & 0xFF000000L))
{
System.out.println("The 1st '0' is at position : " + (i * 8 + 3));
return;
}
if (0L == (l & 0xFF00000000L))
{
System.out.println("The 1st '0' is at position : " + (i * 8 + 4));
return;
}
if (0L == (l & 0xFF0000000000L))
{
System.out.println("The 1st '0' is at position : " + (i * 8 + 5));
return;
}
if (0L == (l & 0xFF000000000000L))
{
System.out.println("The 1st '0' is at position : " + (i * 8 + 6));
return;
}
if (0L == (l & 0xFF00000000000000L))
{
System.out.println("The 1st '0' is at position : " + (i * 8 + 7));
return;
}
}
//If some rest bytes exists
int rest = buf.length % 8;
if(0 != rest)
{
currentOffset = currentOffset + 8;
//Because the length of rest bytes < 8,we have to read them one by one
for(; currentOffset < (baseOffset + buf.length); ++currentOffset)
{
byte b = UnSafe.getByte(buf, (long)currentOffset);
if (0 == ((int) b & 0xFF))
{
System.out.println("The 1st '0' is at position : " + (currentOffset - baseOffset));
return;
}
}
}
System.out.println("Not found '0'");
}
public static void main(String[] args) throws SecurityException,
NoSuchFieldException, IllegalArgumentException,
IllegalAccessException
{
UnSafe = getUnsafe();
int len = 1024 * 1024 * 1024; //1G
long startTime = 0L;
long endTime = 0L;
System.out.println("initialize data...");
byte[] byteArray1 = new byte[len];
for (int i = 0; i < len; ++i)
{
byteArray1[i] = (byte) (i % 128 + 1); //No byte will equal to 0
}
//If you want to set one byte to 0,uncomment the below statement
// byteArray1[2500] = (byte)0;
System.out.println("initialize data done!");
System.out.println("use normalLookup()...");
startTime = System.nanoTime();
normalLookup(byteArray1);
endTime = System.nanoTime();
System.out.println("time : " + ((endTime - startTime) / 1000) + " us.");
System.out.println("use unsafeLookup_1B()...");
startTime = System.nanoTime();
unsafeLookup_1B(byteArray1);
endTime = System.nanoTime();
System.out.println("time : " + ((endTime - startTime) / 1000) + " us.");
System.out.println("use unsafeLookup_8B()...");
startTime = System.nanoTime();
unsafeLookup_8B(byteArray1);
endTime = System.nanoTime();
System.out.println("time : " + ((endTime - startTime) / 1000) + " us.");
}
}
输出是：
initialize data...
initialize data done!
use normalLookup()...
Not found '0'
time : 1271781 us.
use unsafeLookup_1B()...
Not found '0'
time : 716898 us.
use unsafeLookup_8B()...
Not found '0'
time : 591689 us.
结果表明，每次通过Unsafe.getByte（）读取1个字节比定期迭代byte []要快得多。读取8字节长是最快的。
答案 1 :(得分：1)
我认为Unsafe可以比使用常规数组访问更快地访问内存，并对每个索引执行索引检查...
范围检查可能不是因素的一个可能的原因是JIT编译器的优化器。由于数组的大小永远不会改变，优化器可能会“提升”所有范围检查并在循环开始时执行一次。
相比之下，JIT编译器可能无法优化（例如内联）Unsafe.getByte（）调用。或者getByte方法可能有读屏障......）
然而这是猜测。可以肯定的是让JVM为这两种情况转储出JIT编译的本机代码，并按指令进行比较。
答案 2 :(得分：0)
不安全的方法可能被标记为本机，但这并不意味着它们必然是JNI。几乎所有的Unsafe方法都是内在函数（参见这里的简短帖子：http://psy-lob-saw.blogspot.co.uk/2012/10/java-intrinsics-are-not-jni-calls.html），对于Sun JVM，它们将被转换为单个汇编指令（在许多情况下），对于其他JVM，它们可能会或可能不会很好在处理内在函数时，可以将它们转换为JNI调用或普通java调用。据我所知，JRockit倾向于采用JNI方式，Android JVM也是如此。