AFAIK每个类都知道它扩展的所有类以及实现的接口.这些可以存储在提供O(1)查找时间的哈希集中.

当代码经常采用同一分支时,由于几乎可以消除成本,因为CPU可以在确定是否应该采用该分支之前在分支中执行代码,从而使成本几乎为零.

由于微基准测试是在4年前进行的,我希望最新的CPU和JVM会更快.

public static void main(String... args) {

Object[] doubles = new Object[100000];

Arrays.fill(doubles, 0.0);

doubles[100] = null;

doubles[1000] = null;

for (int i = 0; i < 6; i++) {

testSameClass(doubles);

testSuperClass(doubles);

testInterface(doubles);

}

}

private static int testSameClass(Object[] doubles) {

long start = System.nanoTime();

int count = 0;

for (Object d : doubles) {

if (d instanceof Double)

count++;

}

long time = System.nanoTime() - start;

System.out.printf("instanceof Double took an average of %.1f ns%n", 1.0 * time / doubles.length);

return count;

}

private static int testSuperClass(Object[] doubles) {

long start = System.nanoTime();

int count = 0;

for (Object d : doubles) {

if (d instanceof Number)

count++;

}

long time = System.nanoTime() - start;

System.out.printf("instanceof Number took an average of %.1f ns%n", 1.0 * time / doubles.length);

return count;

}

private static int testInterface(Object[] doubles) {

long start = System.nanoTime();

int count = 0;

for (Object d : doubles) {

if (d instanceof Serializable)

count++;

}

long time = System.nanoTime() - start;

System.out.printf("instanceof Serializable took an average of %.1f ns%n", 1.0 * time / doubles.length);

return count;

}

最终打印

instanceof Double took an average of 1.3 ns

instanceof Number took an average of 1.3 ns

instanceof Serializable took an average of 1.3 ns

如果我改变“双打”

for(int i=0;i

doubles[i] = "";

我懂了

instanceof Double took an average of 1.3 ns

instanceof Number took an average of 1.6 ns

instanceof Serializable took an average of 2.2 ns

注意：如果我改变

if (d instanceof Double)

至

if (d != null && d.getClass() == Double.class)

表现是一样的.