这是一道面试题可以用以下的一些方案.第一种是很容易想到的采用循环的方式并且与1进行位与运算,具体代码如下. 1unsigned int GetBitNumOfOne_ByLoop1(unsigned int nValue) 2{ 3 const unsigned int nNumOfBitInByte = 8; 4 unsigned int nBitMask = 1; 5 unsigned int nBitNum = 0; 6 for(unsigned int i = 0 ; i < size
为何与0xff进行与运算 在剖析该问题前请看如下代码 public static String bytes2HexString(byte[] b) { String ret = ""; for (int i = 0; i < b.length; i++) { String hex = Integer.toHexString(b[ i ] & 0xFF); if (hex.length() == 1) { hex = '0' + hex; } ret += hex.toUp
Every non-negative integer N has a binary representation. For example, 5 can be represented as "101" in binary, 11 as "1011" in binary, and so on. Note that except for N = 0, there are no leading zeroes in any binary representation.