优化技巧：提前if判断帮助CPU分支预测

分支预测

在stackoverflow上有一个非常有名的问题：为什么处理有序数组要比非有序数组快？，可见分支预测对代码运行效率有非常大的影响。

现代CPU都支持分支预测(branch prediction)和指令流水线(instruction pipeline)，这两个结合可以极大提高CPU效率。对于像简单的if跳转，CPU是可以比较好地做分支预测的。但是对于switch跳转，CPU则没有太多的办法。switch本质上是据索引，从地址数组里取地址再跳转。

要提高代码执行效率，一个重要的原则就是尽量避免CPU把流水线清空，那么提高分支预测的成功率就非常重要。

那么对于代码里，如果某个switch分支概率很高，是否可以考虑代码层面帮CPU把判断提前，来提高代码执行效率呢？

Dubbo里ChannelEventRunnable的switch判断

在ChannelEventRunnable里有一个switch来判断channel state，然后做对应的逻辑：查看

一个channel建立起来之后，超过99.9%情况它的state都是ChannelState.RECEIVED，那么可以考虑把这个判断提前。

benchmark验证

下面通过jmh来验证下：

public class TestBenchMarks {
    public enum ChannelState {
        CONNECTED, DISCONNECTED, SENT, RECEIVED, CAUGHT
    }

    @State(Scope.Benchmark)
    public static class ExecutionPlan {
        @Param({ "1000000" })
        public int size;
        public ChannelState[] states = null;

        @Setup
        public void setUp() {
            ChannelState[] values = ChannelState.values();
            states = new ChannelState[size];
            Random random = new Random(new Date().getTime());
            for (int i = 0; i < size; i++) {
                int nextInt = random.nextInt(1000000);
                if (nextInt > 100) {
                    states[i] = ChannelState.RECEIVED;
                } else {
                    states[i] = values[nextInt % values.length];
                }
            }
        }
    }

    @Fork(value = 5)
    @Benchmark
    @BenchmarkMode(Mode.Throughput)
    public void benchSiwtch(ExecutionPlan plan, Blackhole bh) {
        int result = 0;
        for (int i = 0; i < plan.size; ++i) {
            switch (plan.states[i]) {
            case CONNECTED:
                result += ChannelState.CONNECTED.ordinal();
                break;
            case DISCONNECTED:
                result += ChannelState.DISCONNECTED.ordinal();
                break;
            case SENT:
                result += ChannelState.SENT.ordinal();
                break;
            case RECEIVED:
                result += ChannelState.RECEIVED.ordinal();
                break;
            case CAUGHT:
                result += ChannelState.CAUGHT.ordinal();
                break;
            }
        }
        bh.consume(result);
    }

    @Fork(value = 5)
    @Benchmark
    @BenchmarkMode(Mode.Throughput)
    public void benchIfAndSwitch(ExecutionPlan plan, Blackhole bh) {
        int result = 0;
        for (int i = 0; i < plan.size; ++i) {
            ChannelState state = plan.states[i];
            if (state == ChannelState.RECEIVED) {
                result += ChannelState.RECEIVED.ordinal();
            } else {
                switch (state) {
                case CONNECTED:
                    result += ChannelState.CONNECTED.ordinal();
                    break;
                case SENT:
                    result += ChannelState.SENT.ordinal();
                    break;
                case DISCONNECTED:
                    result += ChannelState.DISCONNECTED.ordinal();
                    break;
                case CAUGHT:
                    result += ChannelState.CAUGHT.ordinal();
                    break;
                }
            }
        }
        bh.consume(result);
    }
}

• benchSiwtch里是纯switch判断
• benchIfAndSwitch 里用一个if提前判断state是否ChannelState.RECEIVED

benchmark结果是：

Result "io.github.hengyunabc.jmh.TestBenchMarks.benchSiwtch":
  576.745 ±(99.9%) 6.806 ops/s [Average]
  (min, avg, max) = (490.348, 576.745, 618.360), stdev = 20.066
  CI (99.9%): [569.939, 583.550] 

# Run complete. Total time: 00:06:48

Benchmark                         (size)   Mode  Cnt     Score    Error  Units
TestBenchMarks.benchIfAndSwitch  1000000  thrpt  100  1535.867 ± 61.212  ops/s
TestBenchMarks.benchSiwtch       1000000  thrpt  100   576.745 ±  6.806  ops/s

可以看到提前if判断的确提高了代码效率，这种技巧可以放在性能要求严格的地方。
Benchmark代码：https://github.com/hengyunabc/jmh-demo

总结

• switch对于CPU来说难以做分支预测
• 某些switch条件如果概率比较高，可以考虑单独提前if判断，充分利用CPU的分支预测机制