在IntelliJ IDEA里面Ctrl+Alt+M用来拆分方法。选中一段代码，敲下这个组合，非常简单。Eclipse也用类似的快捷键，使用 Alt+Shift+M。我讨厌长的方法，提起这个下面这个方法我就觉得太长了：
public void processOnEndOfDay(Contract c) {
if (DateUtils.addDays(c.getCreated(), 7).before(new Date())) {
priorityHandling(c, OUTDATED_FEE);
notifyOutdated(c);
log.info("Outdated: {}", c);
} else {
if (sendNotifications) {
notifyPending(c);
}
log.debug("Pending {}", c);
}
}首先，它有个条件判断可读性很差。先不管它怎么实现的，它做什么的才最关键。我们先把它拆分出来：
public void processOnEndOfDay(Contract c) {
if (isOutDate(c)) {
priorityHandling(c, OUTDATED_FEE);
notifyOutdated(c);
log.info("Outdated: {}", c);
} else {
if (sendNotifications) {
notifyPending(c);
}
log.debug("Pending {}", c);
}
}
private boolean isOutDate(Contract c) {
return DateUtils.addDays(c.getCreated(), 7).before(new Date());
}public void processOnEndOfDay(Contract c) {
if (c.isOutDate()) {
priorityHandling(c, OUTDATED_FEE);
notifyOutdated(c);
log.info("Outdated: {}", c);
} else {
if (sendNotifications) {
notifyPending(c);
}
log.debug("Pending {}", c);
}
}public void processOnEndOfDay(Contract c) {
if (c.isOutDate()) {
handleOutdated(c);
} else {
if (sendNotifications) {
notifyPending(c);
}
log.debug("Pending {}", c);
}
}
private void handleOutdated(Contract c) {
priorityHandling(c, OUTDATED_FEE);
notifyOutdated(c);
log.info("Outdated: {}", c);
}public void processOnEndOfDay(Contract c) {
if (c.isOutDate()) {
handleOutdated(c);
} else {
stillPending(c);
}
}
private void stillPending(Contract c) {
if (sendNotifications) {
notifyPending(c);
}
log.debug("Pending {}", c);
}
private void handleOutdated(Contract c) {
priorityHandling(c, OUTDATED_FEE);
notifyOutdated(c);
log.info("Outdated: {}", c);
}方法内联
如果JVM监测到一些小方法被频繁的执行，它会把方法的调用替换成方法体本身。比如说下面这个：
private int add4(int x1, int x2, int x3, int x4) {
return add2(x1, x2) + add2(x3, x4);
}
private int add2(int x1, int x2) {
return x1 + x2;
}private int add4(int x1, int x2, int x3, int x4) {
return x1 + x2 + x3 + x4;
}JVM更清楚运行的目标环境 ，CPU，内存，体系结构，它可以更积极的进行优化。 JVM可以发现你代码运行时的特征，比如，哪个方法被频繁的执行，哪个虚方法只有一个实现，等等。 旧编译器编译的.class在新版本的JVM上可以获取更快的运行速度。更新JVM和重新编译源代码，你肯定更倾向于后者。
我们对这些假设做下测试。我写了一个小程序，它有着分治原则的最糟实现的称号。add128方法需要128个参数并且调用了两次add64方法——前后两半各一次。add64也类似，不过它是调用了两次add32。你猜的没错，最后会由add2方法来结束这一切，它是干苦力活的。有些数字我给省略了，免得亮瞎了你的眼睛：
public class ConcreteAdder {
public int add128(int x1, int x2, int x3, int x4, ... more ..., int x127, int x128) {
return add64(x1, x2, x3, x4, ... more ..., x63, x64) +
add64(x65, x66, x67, x68, ... more ..., x127, x128);
}
private int add64(int x1, int x2, int x3, int x4, ... more ..., int x63, int x64) {
return add32(x1, x2, x3, x4, ... more ..., x31, x32) +
add32(x33, x34, x35, x36, ... more ..., x63, x64);
}
private int add32(int x1, int x2, int x3, int x4, ... more ..., int x31, int x32) {
return add16(x1, x2, x3, x4, ... more ..., x15, x16) +
add16(x17, x18, x19, x20, ... more ..., x31, x32);
}
private int add16(int x1, int x2, int x3, int x4, ... more ..., int x15, int x16) {
return add8(x1, x2, x3, x4, x5, x6, x7, x8) + add8(x9, x10, x11, x12, x13, x14, x15, x16);
}
private int add8(int x1, int x2, int x3, int x4, int x5, int x6, int x7, int x8) {
return add4(x1, x2, x3, x4) + add4(x5, x6, x7, x8);
}
private int add4(int x1, int x2, int x3, int x4) {
return add2(x1, x2) + add2(x3, x4);
}
private int add2(int x1, int x2) {
return x1 + x2;
}
} public class InlineAdder {
public int add128n(int x1, int x2, int x3, int x4, ... more ..., int x127, int x128) {
return x1 + x2 + x3 + x4 + ... more ... + x127 + x128;
}
}public abstract class Adder {
public abstract int add128(int x1, int x2, int x3, int x4, ... more ..., int x127, int x128);
public abstract int add64(int x1, int x2, int x3, int x4, ... more ..., int x63, int x64);
public abstract int add32(int x1, int x2, int x3, int x4, ... more ..., int x31, int x32);
public abstract int add16(int x1, int x2, int x3, int x4, ... more ..., int x15, int x16);
public abstract int add8(int x1, int x2, int x3, int x4, int x5, int x6, int x7, int x8);
public abstract int add4(int x1, int x2, int x3, int x4);
public abstract int add2(int x1, int x2);
} public class VirtualAdder extends Adder {
@Override
public int add128(int x1, int x2, int x3, int x4, ... more ..., int x128) {
return add64(x1, x2, x3, x4, ... more ..., x63, x64) +
add64(x65, x66, x67, x68, ... more ..., x127, x128);
}
@Override
public int add64(int x1, int x2, int x3, int x4, ... more ..., int x63, int x64) {
return add32(x1, x2, x3, x4, ... more ..., x31, x32) +
add32(x33, x34, x35, x36, ... more ..., x63, x64);
}
@Override
public int add32(int x1, int x2, int x3, int x4, ... more ..., int x32) {
return add16(x1, x2, x3, x4, ... more ..., x15, x16) +
add16(x17, x18, x19, x20, ... more ..., x31, x32);
}
@Override
public int add16(int x1, int x2, int x3, int x4, ... more ..., int x16) {
return add8(x1, x2, x3, x4, x5, x6, x7, x8) + add8(x9, x10, x11, x12, x13, x14, x15, x16);
}
@Override
public int add8(int x1, int x2, int x3, int x4, int x5, int x6, int x7, int x8) {
return add4(x1, x2, x3, x4) + add4(x5, x6, x7, x8);
}
@Override
public int add4(int x1, int x2, int x3, int x4) {
return add2(x1, x2) + add2(x3, x4);
}
@Override
public int add2(int x1, int x2) {
return x1 + x2;
}
}
具体的环境信息：
看起来慢的机器上JVM更倾向于进行方法内联。不仅是简单的私有方法调用的版本，虚方法的版本也一样。为什么会这样？因为JVM发现Adder只有一个子类，也就是说每个抽象方法都只有一个版本。如果你在运行时加载了另一个子类（或者更多），你会看到性能会直线下降，因为无能再进行内联了。先不管这个了，从测试中来看，
这些方法的调用并不是开销很低，是根本就没有开销！
方法调用（还有为了可读性而加的文档）只存在于你的源代码和编译后的字节码里，运行时它们完全被清除掉了（内联了）。
我对第二个结果也不太理解。看起来性能高的机器B运行单个方法调用的时候要快点，另两个就要慢些。也许它倾向于延迟进行内联？结果是有些不同，不过差距也不是那么的大。就像 优化栈跟踪信息生成 那样——如果你为了优化代码性能，手动进行内联，把方法越搞越庞大，越弄越复杂，那你就真的错了。
ps：64bit 机器之所以运行慢有可能是因为 JVM 内联的要求的方法长度较长。
文章原文来源于：
http://www.javacodegeeks.com/2013/02/how-aggressive-is-method-inlining-in-jvm.html
http://it.deepinmind.com/java/2014/03/01/JVM的方法内联.html
原文地址：http://blog.csdn.net/zq602316498/article/details/40266633
