我有这段代码做范围最小查询.当t = 100000时,i和j总是在每个输入行中改变,它在Java 8u60中的执行时间约为12秒.
for (int a0 = 0; a0 < t; a0++) {
String line = reader.readLine();
String[] ls = line.split(" ");
int i = Integer.parseInt(ls[0]);
int j = Integer.parseInt(ls[1]);
int min = width[i];
for (int k = i + 1; k <= j; k++) {
if (min > width[k]) {
min = width[k];
}
}
writer.write(min + "");
writer.newLine();
}
Run Code Online (Sandbox Code Playgroud)
当我提取新方法以找到最小值时,执行时间快4倍(约2.5秒).
for (int a0 = 0; a0 < t; a0++) {
String line = reader.readLine();
String[] ls = line.split(" ");
int i = Integer.parseInt(ls[0]);
int j = Integer.parseInt(ls[1]);
int min = getMin(i, j);
writer.write(min + "");
writer.newLine();
}
private int getMin(int i, int j) {
int min = width[i];
for (int k = i + 1; k <= j; k++) {
if (min > width[k]) {
min = width[k];
}
}
return min;
}
Run Code Online (Sandbox Code Playgroud)
我一直认为方法调用很慢.但这个例子显示了相反的情况.Java 6也证明了这一点,但两种情况下的执行时间都要慢得多(17秒和10秒).有人可以对此提供一些见解吗?
apa*_*gin 12
TL; DR JIT编译器在第二种情况下有更多机会优化内部循环,因为堆栈内替换发生在不同的点.
我已经设法用简化的测试用例重现问题.
不涉及I/O或字符串操作,只有两个具有数组访问权限的嵌套循环.
public class NestedLoop {
private static final int ARRAY_SIZE = 5000;
private static final int ITERATIONS = 1000000;
private int[] width = new java.util.Random(0).ints(ARRAY_SIZE).toArray();
public long inline() {
long sum = 0;
for (int i = 0; i < ITERATIONS; i++) {
int min = width[0];
for (int k = 1; k < ARRAY_SIZE; k++) {
if (min > width[k]) {
min = width[k];
}
}
sum += min;
}
return sum;
}
public long methodCall() {
long sum = 0;
for (int i = 0; i < ITERATIONS; i++) {
int min = getMin();
sum += min;
}
return sum;
}
private int getMin() {
int min = width[0];
for (int k = 1; k < ARRAY_SIZE; k++) {
if (min > width[k]) {
min = width[k];
}
}
return min;
}
public static void main(String[] args) {
long startTime = System.nanoTime();
long sum = new NestedLoop().inline(); // or .methodCall();
long endTime = System.nanoTime();
long ms = (endTime - startTime) / 1000000;
System.out.println("sum = " + sum + ", time = " + ms + " ms");
}
}
Run Code Online (Sandbox Code Playgroud)
inline变体确实比它慢3-4倍methodCall.
我已经使用以下JVM选项来确认两个基准测试都是在最高层编译的,并且在两种情况下都成功地发生了OSR(堆栈内替换).
-XX:-TieredCompilation
-XX:CompileOnly=NestedLoop
-XX:+UnlockDiagnosticVMOptions
-XX:+PrintCompilation
-XX:+TraceNMethodInstalls
Run Code Online (Sandbox Code Playgroud)
251 46 % NestedLoop::inline @ 21 (70 bytes)
Installing osr method (4) NestedLoop.inline()J @ 21
Run Code Online (Sandbox Code Playgroud)
271 46 NestedLoop::getMin (41 bytes)
Installing method (4) NestedLoop.getMin()I
274 47 % NestedLoop::getMin @ 9 (41 bytes)
Installing osr method (4) NestedLoop.getMin()I @ 9
314 48 % NestedLoop::methodCall @ 4 (30 bytes)
Installing osr method (4) NestedLoop.methodCall()J @ 4
Run Code Online (Sandbox Code Playgroud)
这意味着JIT可以完成它的工作,但生成的代码必须是不同的.
我们来分析吧-XX:+PrintAssembly.
0x0000000002df4dd0: inc %ebp ; OopMap{r11=Derived_oop_rbx rbx=Oop off=114}
;*goto
; - NestedLoop::inline@53 (line 12)
0x0000000002df4dd2: test %eax,-0x1d64dd8(%rip) # 0x0000000001090000
;*iload
; - NestedLoop::inline@21 (line 12)
; {poll}
0x0000000002df4dd8: cmp $0x1388,%ebp
0x0000000002df4dde: jge 0x0000000002df4dfd ;*if_icmpge
; - NestedLoop::inline@26 (line 12)
0x0000000002df4de0: test %rbx,%rbx
0x0000000002df4de3: je 0x0000000002df4e4c
0x0000000002df4de5: mov (%r11),%r10d ;*getfield width
; - NestedLoop::inline@32 (line 13)
0x0000000002df4de8: mov 0xc(%r10),%r9d ; implicit exception
0x0000000002df4dec: cmp %r9d,%ebp
0x0000000002df4def: jae 0x0000000002df4e59
0x0000000002df4df1: mov 0x10(%r10,%rbp,4),%r8d ;*iaload
; - NestedLoop::inline@37 (line 13)
0x0000000002df4df6: cmp %r8d,%r13d
0x0000000002df4df9: jg 0x0000000002df4dc6 ;*if_icmple
; - NestedLoop::inline@38 (line 13)
0x0000000002df4dfb: jmp 0x0000000002df4dd0
Run Code Online (Sandbox Code Playgroud)
0x0000000002da2af0: add $0x8,%edx ;*iinc
; - NestedLoop::getMin@33 (line 36)
; - NestedLoop::methodCall@11 (line 27)
0x0000000002da2af3: cmp $0x1381,%edx
0x0000000002da2af9: jge 0x0000000002da2b70 ;*iload_1
; - NestedLoop::getMin@16 (line 37)
; - NestedLoop::methodCall@11 (line 27)
0x0000000002da2afb: mov 0x10(%r9,%rdx,4),%r11d ;*iaload
; - NestedLoop::getMin@22 (line 37)
; - NestedLoop::methodCall@11 (line 27)
0x0000000002da2b00: cmp %r11d,%ecx
0x0000000002da2b03: jg 0x0000000002da2b6b ;*iinc
; - NestedLoop::getMin@33 (line 36)
; - NestedLoop::methodCall@11 (line 27)
0x0000000002da2b05: mov 0x14(%r9,%rdx,4),%r11d ;*iaload
; - NestedLoop::getMin@22 (line 37)
; - NestedLoop::methodCall@11 (line 27)
0x0000000002da2b0a: cmp %r11d,%ecx
0x0000000002da2b0d: jg 0x0000000002da2b5c ;*iinc
; - NestedLoop::getMin@33 (line 36)
; - NestedLoop::methodCall@11 (line 27)
0x0000000002da2b0f: mov 0x18(%r9,%rdx,4),%r11d ;*iaload
; - NestedLoop::getMin@22 (line 37)
; - NestedLoop::methodCall@11 (line 27)
0x0000000002da2b14: cmp %r11d,%ecx
0x0000000002da2b17: jg 0x0000000002da2b4d ;*iinc
; - NestedLoop::getMin@33 (line 36)
; - NestedLoop::methodCall@11 (line 27)
0x0000000002da2b19: mov 0x1c(%r9,%rdx,4),%r11d ;*iaload
; - NestedLoop::getMin@22 (line 37)
; - NestedLoop::methodCall@11 (line 27)
0x0000000002da2b1e: cmp %r11d,%ecx
0x0000000002da2b21: jg 0x0000000002da2b66 ;*iinc
; - NestedLoop::getMin@33 (line 36)
; - NestedLoop::methodCall@11 (line 27)
0x0000000002da2b23: mov 0x20(%r9,%rdx,4),%r11d ;*iaload
; - NestedLoop::getMin@22 (line 37)
; - NestedLoop::methodCall@11 (line 27)
0x0000000002da2b28: cmp %r11d,%ecx
0x0000000002da2b2b: jg 0x0000000002da2b61 ;*iinc
; - NestedLoop::getMin@33 (line 36)
; - NestedLoop::methodCall@11 (line 27)
0x0000000002da2b2d: mov 0x24(%r9,%rdx,4),%r11d ;*iaload
; - NestedLoop::getMin@22 (line 37)
; - NestedLoop::methodCall@11 (line 27)
0x0000000002da2b32: cmp %r11d,%ecx
0x0000000002da2b35: jg 0x0000000002da2b52 ;*iinc
; - NestedLoop::getMin@33 (line 36)
; - NestedLoop::methodCall@11 (line 27)
0x0000000002da2b37: mov 0x28(%r9,%rdx,4),%r11d ;*iaload
; - NestedLoop::getMin@22 (line 37)
; - NestedLoop::methodCall@11 (line 27)
0x0000000002da2b3c: cmp %r11d,%ecx
0x0000000002da2b3f: jg 0x0000000002da2b57 ;*iinc
; - NestedLoop::getMin@33 (line 36)
; - NestedLoop::methodCall@11 (line 27)
0x0000000002da2b41: mov 0x2c(%r9,%rdx,4),%r11d ;*iaload
; - NestedLoop::getMin@22 (line 37)
; - NestedLoop::methodCall@11 (line 27)
0x0000000002da2b46: cmp %r11d,%ecx
0x0000000002da2b49: jg 0x0000000002da2ae6 ;*if_icmple
; - NestedLoop::getMin@23 (line 37)
; - NestedLoop::methodCall@11 (line 27)
0x0000000002da2b4b: jmp 0x0000000002da2af0
Run Code Online (Sandbox Code Playgroud)
编译后的代码完全不同; methodCall优化得更好.
width 字段缓存在寄存器中.相反,inline变体
width每次从内存加载数组;OSR编译的方法并不总是很好地优化,因为它们必须在转换点维护解释的堆栈帧的状态.这是同一问题的另一个例子.
堆栈替换通常发生在后向分支上(即在循环的底部).inline方法有两个嵌套循环,OSR在内循环内部发生,而methodCall只有一个外循环.外循环中的OSR转换更有利,因为JIT编译器可以更自由地优化内循环.这就是你的情况下究竟发生的事情.
| 归档时间: |
|
| 查看次数: |
1291 次 |
| 最近记录: |