Question

我在基准测试中注意到，Java 中的算术复合运算符总是优于常规赋值：

    d0 *= d0;           //faster
    //d0 = d0 * d0;     //slower

    d0 += d0;           //faster
    //d0 = d0 + d0;     //slower

有人可以对上述观察结果发表评论并解释原因。我假设字节码级别的一些差异是加速的原因吗？提前谢谢。

这是我的基准，更充分：

public long squaring() {
    long t0 = System.currentTimeMillis();
    double d0 = 0;

    for (int k = 0; k < 100_000_000; k++){

        //check bytecode for below to see why timing differs
        d0 *= d0;           //faster
        //d0 = d0 * d0;     //slower
    }

    long t1 = System.currentTimeMillis();
    long took = (t1 - t0);
    System.out.println("took: "+took + " ms");
    System.out.println("result: " +d0);

    return took;
}

@Test
    public void testSquaring() {
        int repetitions = 10;
        long sum = 0;
        for (int i = 0; i < repetitions; i++) {
            sum += cut.squaring();
            System.out.println("accumulated: "+ sum + "\n-------------------");
        }
        double avg = sum/repetitions;
        System.out.println("average: "+avg);

    }

以下是结果：

took: 244 ms
result: 0.0
accumulated: 244
-------------------
took: 302 ms
result: 0.0
accumulated: 546
-------------------
took: 0 ms
result: 0.0
accumulated: 546
-------------------
took: 0 ms
result: 0.0
accumulated: 546
-------------------
took: 0 ms
result: 0.0
accumulated: 546
-------------------
took: 0 ms
result: 0.0
accumulated: 546
-------------------
took: 0 ms
result: 0.0
accumulated: 546
-------------------
took: 0 ms
result: 0.0
accumulated: 546
-------------------
took: 0 ms
result: 0.0
accumulated: 546
-------------------
took: 0 ms
result: 0.0
accumulated: 546
-------------------
average: 54.0

Answer 1

尝试反编译这两种情况：

static void a(double d0) {
  d0 *= d0;
}

static void b(double d0) {
  d0 = d0 * d0;
}

反编译为：

  static void a(double);
    Code:
       0: dload_0
       1: dload_0
       2: dmul
       3: dstore_0
       4: return

  static void b(double);
    Code:
       0: dload_0
       1: dload_0
       2: dmul
       3: dstore_0
       4: return

即。它们在字节码级别上是相同的，因此不存在内在的性能差异。此代码之外的因素会影响运行时间。

Answer 2

a *= b;生成与a = a * b;完全相同的字节码。这意味着您的性能测试存在问题。

同样适用于+，-，/等。

Java中算术复合赋值运算符的性能

2 个答案: