java小数点数子进1_java – 如何计数1的数字将有二进制数？

而不是编写一个算法来做到这一点，它最好使用内置函数。

Integer.bitCount()

这使得这个特别高效的是JVM可以将其视为内在的。即在支持它的平台上用单个机器代码指令识别和替换整个事物。英特尔/ AMD

演示这种优化的有效性

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

perfTestIntrinsic();

perfTestACopy();

}

private static void perfTestIntrinsic() {

long start = System.nanoTime();

long countBits = 0;

for (int i = 0; i < Integer.MAX_VALUE; i++)

countBits += Integer.bitCount(i);

long time = System.nanoTime() - start;

System.out.printf("Intrinsic: Each bit count took %.1f ns, countBits=%d%n", (double) time / Integer.MAX_VALUE, countBits);

}

private static void perfTestACopy() {

long start2 = System.nanoTime();

long countBits2 = 0;

for (int i = 0; i < Integer.MAX_VALUE; i++)

countBits2 += myBitCount(i);

long time2 = System.nanoTime() - start2;

System.out.printf("Copy of same code: Each bit count took %.1f ns, countBits=%d%n", (double) time2 / Integer.MAX_VALUE, countBits2);

}

// Copied from Integer.bitCount()

public static int myBitCount(int i) {

// HD, Figure 5-2

i = i - ((i >>> 1) & 0x55555555);

i = (i & 0x33333333) + ((i >>> 2) & 0x33333333);

i = (i + (i >>> 4)) & 0x0f0f0f0f;

i = i + (i >>> 8);

i = i + (i >>> 16);

return i & 0x3f;

}

打印

Intrinsic: Each bit count took 0.4 ns, countBits=33285996513

Copy of same code: Each bit count took 2.4 ns, countBits=33285996513

使用固有版本和循环的每个位计数平均只需0.4纳秒。使用相同代码的副本需要6倍的时间(获得相同的结果)