为什么代码片段比代码片段B慢14倍？
@H_404_2@(在
Windows 7 64位上用jdk1.8.0_60进行测试)
代码段A：
import java.awt.geom.RoundRectangle2D;
public class Test {
private static final RoundRectangle2D.Double RECTANGLE = new RoundRectangle2D.Double(1,2,3,4,5,6);
public static void main(String[] args) {
int result = RECTANGLE.hashCode();
long start = System.nanoTime();
for (int i = 0; i < 100_000_000; i++) {
result += RECTANGLE.hashCode(); // <= Only change is on this line
}
System.out.println((System.nanoTime() - start) / 1_000_000);
System.out.println(result);
}
}
代码段B：
import java.awt.geom.RoundRectangle2D;
public class Test {
private static final RoundRectangle2D.Double RECTANGLE = new RoundRectangle2D.Double(1,6);
public static void main(String[] args) {
int result = RECTANGLE.hashCode();
long start = System.nanoTime();
for (int i = 0; i < 100_000_000; i++) {
result += new RoundRectangle2D.Double(1,6).hashCode();
}
System.out.println((System.nanoTime() - start) / 1_000_000);
System.out.println(result);
}
}
TL; DR：在循环中使用new关键字比访问静态final字段要快.
(注意：删除RECTANGLE上的最终关键字不会更改执行时间)
解决方法
在第一种情况(静态final)中,JVM需要从内存读取对象字段.
@H_404_2@在第二种情况下,已知值是常数.此外,由于对象不从循环中逸出,因此省略了分配.其字段被替换为局部变量.
package bench;
import org.openjdk.jmh.annotations.*;
import java.awt.geom.RoundRectangle2D;
@State(Scope.Benchmark)
public class StaticRect {
private static final RoundRectangle2D.Double RECTANGLE =
new RoundRectangle2D.Double(1,6);
@Benchmark
public long baseline() {
return 0;
}
@Benchmark
public long testNew() {
return new RoundRectangle2D.Double(1,6).hashCode();
}
@Benchmark
@Fork(jvmArgs = "-XX:-EliminateAllocations")
public long testNewNoEliminate() {
return new RoundRectangle2D.Double(1,6).hashCode();
}
@Benchmark
public int testStatic() {
return RECTANGLE.hashCode();
}
}
结果：
Benchmark Mode Cnt score Error Units StaticRect.baseline avgt 10 2,840 ± 0,048 ns/op StaticRect.testNew avgt 10 2,831 ± 0,011 ns/op StaticRect.testNewNoEliminate avgt 10 8,566 ± 0,036 ns/op StaticRect.testStatic avgt 10 12,689 ± 0,057 ns/op
testNew与返回一个常量一样快,因为对象分配被消除,并且hashCode在JIT编译期间是不变的.
当消除分配优化被禁用时,基准时间显着更高,但是hashCode的算术计算仍然是不变的.
在最后一个基准测试中,即使RECTANGLE被声明为final,它的字段可能在理论上被改变,所以JIT无法消除字段访问.
