链接到文档：http://download.oracle.com/javase/6/docs/api/java/lang/Long.html#numberOfTrailingZeros%28long%29
以下是Java实现源代码：
/**
* Returns the number of zero bits following the lowest-order ("rightmost")
* one-bit in the two's complement binary representation of the specified
* <tt>long</tt> value. Returns 64 if the specified value has no
* one-bits in its two's complement representation, in other words if it is
* equal to zero.
*
* @return the number of zero bits following the lowest-order ("rightmost")
* one-bit in the two's complement binary representation of the
* specified <tt>long</tt> value, or 64 if the value is equal
* to zero.
* @since 1.5
*/
public static int numberOfTrailingZeros(long i) {
// HD, Figure 5-14
int x, y;
if (i == 0) return 64;
int n = 63;
y = (int)i; if (y != 0) { n = n -32; x = y; } else x = (int)(i>>>32);
y = x <<16; if (y != 0) { n = n -16; x = y; }
y = x << 8; if (y != 0) { n = n - 8; x = y; }
y = x << 4; if (y != 0) { n = n - 4; x = y; }
y = x << 2; if (y != 0) { n = n - 2; x = y; }
return n - ((x << 1) >>> 31);
}
该算法将长整型分解为两个整型，并处理每个整型。我的问题是，为什么不使用y=x<这是我的版本：
public static int bit(long i)
{
if (i == 0) return 64;
long x = i;
long y;
int n = 63;
y = x << 32; if (y != 0) { n -= 32; x = y; }
y = x << 16; if (y != 0) { n -= 16; x = y; }
y = x << 8; if (y != 0) { n -= 8; x = y; }
y = x << 4; if (y != 0) { n -= 4; x = y; }
y = x << 2; if (y != 0) { n -= 2; x = y; }
return (int) (n - ((x << 1) >>> 63));
}
我测试了两种方法并取平均值实现时间：595，我的版本时间：593。也许最初的实现在32位系统上更快，因为我使用的是Windows764位。至少java应该在x64 sdk中使用类似我的版本。有什么想法吗？
最佳答案
几乎每个应用程序都可以忽略0.5%的性能差异。如果您在一个应用程序上工作，这个方法需要peek性能，您可以自己实现它。
关于java - Long.numberOfTrailingZeros()的Java实现，我们在Stack Overflow上找到一个类似的问题：https://stackoverflow.com/questions/6506356/