所以我的代码目前看起来像这样
public boolean in(TransactionType... types)
{
if (types == null || types.length == 0)
return false;
for (int i = 0; i < types.length; ++i)
if (types[i] != null && types[i] == this)
return true;
return false;
}
我把它改成了
public boolean in(TransactionType... types)
{
if (types == null || types.length == 0)
return false;
for (int i = 0; i < types.length; ++i)
if (types[i] == this)
return true;
return false;
}
（TransactionType是一个包含大约30个值的枚举）
结果震惊了我。在我的所有测试中，第二个测试速度提高了一个数量级。我预计可能快2倍，但不是一个数量级。为什么不同？这是一个非常慢的nullcheck，或者是额外的数组访问会发生什么奇怪的事情？
我的基准代码如下所示
public class App
{
public enum TransactionType
{
A(1, "A", "A"),
B(3, "B", "B"),
C(5, "C", "C"),
D(6, "D", "D"),
E(7, "E", "E"),
F(8, "F", "F"),
G(9, "G", "G"),
H(10, "H", "H"),
I(11, "I", "I"),
J(12, "J", "J"),
K(13, "K", "K"),
L(14, "L", "L"),
M(15, "M", "M"),
N(16, "N", "N"),
O(17, "O", "O"),
P(18, "P", "P"),
Q(19, "Q", "Q"),
R(20, "R", "R"),
S(21, "S", "S"),
T(22, "T", "T"),
U(25, "U", "U"),
V(26, "V", "V"),
W(27, "W", "W"),
X(28, "X", "X"),
Y(29, "Y", "Y"),
Z(30, "Z", "Z"),
AA(31, "AA", "AA"),
AB(32, "AB", "AB"),
AC(33, "AC", "AC"),
AD(35, "AD", "AD"),
AE(36, "AE", "AE"),
AF(37, "AF", "AF"),
AG(38, "AG", "AG"),
AH(39, "AH", "AH"),
AI(40, "AI", "AI"),
AJ(41, "AJ", "AJ"),
AK(42, "AK", "AK"),
AL(43, "AL", "AL"),
AM(44, "AM", "AM"),
AN(45, "AN", "AN"),
AO(46, "AO", "AO"),
AP(47, "AP", "AP");
public final static TransactionType[] aArray =
{
O, Z, N, Y, AB
};
public final static TransactionType[] bArray =
{
J, P, AA, L, Q, M, K, AE, AK,
AF, AD, AG, AH
};
public final static TransactionType[] cArray =
{
S, U, V
};
public final static TransactionType[] dArray =
{
A, B, D, G, C, E,
T, R, I, F, H, AC,
AI, AJ, AL, AM, AN,
AO
};
private int id;
private String abbrev;
private String name;
private TransactionType(int id, String abbrev, String name)
{
this.id = id;
this.abbrev = abbrev;
this.name = name;
}
public boolean in(TransactionType... types)
{
if (types == null || types.length == 0)
return false;
for (int i = 0; i < types.length; ++i)
if (types[i] == this)
return true;
return false;
}
public boolean inOld(TransactionType... types)
{
if (types == null || types.length == 0)
return false;
for (int i = 0; i < types.length; ++i)
{
if (types[i] != null && types[i] == this)
return true;
}
return false;
}
}
public static void main(String[] args)
{
for (int i = 0; i < 10; ++i)
bench2();
for (int i = 0; i < 10; ++i)
bench1();
}
private static void bench1()
{
final TransactionType[] values = TransactionType.values();
long runs = 0;
long currTime = System.currentTimeMillis();
while (System.currentTimeMillis() - currTime < 1000)
{
for (TransactionType value : values)
{
value.inOld(TransactionType.dArray);
}
++runs;
}
System.out.println("old " + runs);
}
private static void bench2()
{
final TransactionType[] values = TransactionType.values();
long runs = 0;
long currTime = System.currentTimeMillis();
while (System.currentTimeMillis() - currTime < 1000)
{
for (TransactionType value : values)
{
value.in(TransactionType.dArray);
}
++runs;
}
System.out.println("new " + runs);
}
}
以下是基准运行的结果
new 20164901
new 20084651
new 45739657
new 45735251
new 45757756
new 45726575
new 45413016
new 45649661
new 45325360
new 45380665
old 2021652
old 2022286
old 2246888
old 2237484
old 2246172
old 2268073
old 2271554
old 2259544
old 2272642
old 2268579
这是使用Oracle JDK 1.7.0.67
答案 0 :(得分：3)
空检查没有完成任何事情，我也很惊讶它会产生这样的差异。但我相信你的评论基本上回答了你自己的问题。
@Cogman写道：
...迭代数组涉及非常少的分支并且是高度本地化的操作（意味着它可能会占用CPU缓存的很多优势）。在大多数现代CPU中，分支的类型也是高度可预测和优化的......
如果您编译您的类并使用javap打印出这两种方法的反汇编字节代码，您将看到：
public boolean in(App$TransactionType...);
Code:
0: aload_1
1: ifnull 9
4: aload_1
5: arraylength
6: ifne 11
9: iconst_0
10: ireturn
11: iconst_0
12: istore_2
13: iload_2
14: aload_1
15: arraylength
16: if_icmpge 34
19: aload_1
20: iload_2
21: aaload
22: aload_0
23: if_acmpne 28
26: iconst_1
27: ireturn
28: iinc 2, 1
31: goto 13
34: iconst_0
35: ireturn
还有：
public boolean inOld(App$TransactionType...);
Code:
0: aload_1
1: ifnull 9
4: aload_1
5: arraylength
6: ifne 11
9: iconst_0
10: ireturn
11: iconst_0
12: istore_2
13: iload_2
14: aload_1
15: arraylength
16: if_icmpge 40
19: aload_1
20: iload_2
21: aaload
22: ifnull 34
25: aload_1
26: iload_2
27: aaload
28: aload_0
29: if_acmpne 34
32: iconst_1
33: ireturn
34: iinc 2, 1
37: goto 13
40: iconst_0
41: ireturn
您的新方法删除了六个操作和一个潜在的分支站点。
之前的循环很紧，现在它非常紧张。
我原本以为Java会将这两种方法JIT同样基本相同。你的时间数字另有说明。
一些随机数字：
1.6.33 32b：646100 vs 727173
1.6.33 64b：1667665 vs 2668513
1.7.67 32b：661003 vs 716417
1.7.07 64b：1663926 vs 32493989
1.7.60 64b：1700574 vs 32368506
1.8.20 64b：1648382 vs 32222823
所有64位JVM都比32位版本更快地执行这两种实现。