计算长整数中的置位位数

本文介绍了计算长整数中的置位位数的处理方法，对大家解决问题具有一定的参考价值，需要的朋友们下面随着小编来一起学习吧！ 问题描述

此问题已在此处得到了回答.

This question has been answered here.

我的查询是，按照方法1起作用，但是它的变化，即方法2无效，而是给出了预期输出值的两倍.我不知道为什么.

My query is, following approach-1 works, however the variation of it, that is approach-2 does not, rather it gives double the value of expected output. I can not find out why.

方法1

public class Solution { public int numSetBits(long a) { int count = 0; long temp = 0; for(int i = 0 ; i < 64 ; i++) { // 64-bit for long data-type temp = 1; temp = temp << i; temp = a & temp; if((temp > 0)) count++; } return count; } }

方法2

public class Solution { public int numSetBits(long a) { int count=0; for(int i=0; i<64; i++) { if((a & (1 << i))>0) count++; } return count; } }

推荐答案

第二种方法失败，因为1 << i的结果是int，而不是long.因此，位掩码回绕了，因此a的低32位被扫描了两次，而a的高32位被忽略了.

The second approach fails because the result of 1 << i is an int, not a long. So the bit mask wraps around, and so the lower 32 bits of a get scanned twice, while the higher 32 bits of a are left uncounted.

因此，当i达到值32时，(1 << i)不会是2 32 ，而是2 0 (即1)，与当i为0时.类似地，当i为33时，(1 << i)不会是2 33 ，而是2 1 ，...等等.

So, when i reaches the value 32, (1 << i) will not be 232, but 20 (i.e. 1), which is the same as when i was 0. Similarly when i is 33, (1 << i) will not be 233, but 21, ...etc.

通过将常数1设为long来纠正此问题:

Correct this by making the constant 1 a long:

(1L << i)