6.1 built-in type obtained by the minimum and maximum bit operation
1 // Get the maximum int type
2 // get_int_max_1 () may be an integer constant overflow error
. 3 int get_int_max_1 () {
. 4 return ( 1 << 31 is ) - 1 ;
. 5 }
. 6 int get_int_max_2 () {
. 7 return ~ ( . 1 << 31 is );
. 8 }
. 9 // unsigned int is thirty-two number, a removing right sign bit
10 int get_int_max_3 () {
. 11 return ((unsigned int ) - . 1 ) >> . 1 ;
12 }
13 // Get the minimum value of type int
14 int get_int_min_1 () {
15 return . 1 << 31 is ;
16 }
6.2. The left one and a right of action
1 // thinking this way: 11110 1111 left one, the equivalent of every +1 in the original basis, which means that each bit doubled, converted to decimal time that is two decimal 1111 results obtained by adding a further
2 // shift left one corresponds to multiplying 2
. 3 int multiply_by_two ( int NUM) {
. 4 return NUM << . 1 ;
. 5 }
. 6 // right corresponds to a divided 2
. 7 int divide_by_two ( int NUM) {
. 8 return NUM >> . 1 ;
. 9 }
6.3. How is multiplied by 2 ^ n or divided by 2 ^ n
. 1 // m is multiplied by 2 ^ n
2 int multiply_by_two_power ( int m, int n) {
. 3 return m << n;
. 4 }
. 5 // divided by the m-th power of 2
. 6 int divide_by_two_power ( int m, int n ) {
. 7 return m << n-;
. 8 }
6.4 determining whether a number is a power of 2
1 // Analyzing a number is not a power of 2,
2 BOOL isFactorialofTwo ( int n-) {
. 3 return n-> 0 (n-& (n--? 1 )) == 0 : to false ;
. 4 }
. 5 / * If this is a power of 2 , n must be 100 ... n-1 is 1111 ....
6 so do the calculation result is 0 * /
6.5. From the original code to complement you should realize that, Bitwise +1 been the opposite of the original code!
1 // find a number of inverse number
2 int opposite_number ( int n-) {
. 3 return ~ + n- 1 ;
. 4 }
6.6.统计一个二进制数中0和1的个数 1 //统计一个二进制数中1和0的个数,n&(~n+1)可以得到最后一位为1的二进制数，n|=(n+1)可以使最后一位0置为1
1 int sum_0(int n) {
2 int count = 0;
3 while (n+1) //全部为1高位去除，结果为0
4 {
5 n |= (n + 1);
6 count++;
7 }
8 return count++;
9 }
6.7.判断一个数的奇偶性
1 //判断一个数的奇偶性
2 //高位去除，得到二进制的最后一位，0为偶数，1为奇
3 bool is_oddnumber(int n) {
4 return (n & 1) == 1;
5 }
6.8.不使用第三个数交换两个数
1 /交换两个数不使用第三个数，把一般运算符的方法也用上了
2 //一个数异或本身等于0，一个数异或0得到本身
3 void swap(int &a,int &b) {
4 a = a ^ b;
5 b = a ^ b;
6 a = a ^ b;
7 }
8 void swap_1(int &a, int &b) {
9 a = a - b;
10 b = a + b;
11 a = b - a;
12 }
6.9.不使用+，-，*，/完成整数相加
1 int Add(int num1, int num2)
2 {
3 int sum, carry;
4 do {
5 //将两个数异或，得到不进位的结果
6 sum = num1 ^ num2;
7 //将两个位运算，得到进位的结果
8 carry = (num1 & num2) << 1;
9 //很显然，我们只要把进位的结果+上不进位的结果就可以得到我们想要的，
10 //但是，我们不能使用+运算符，我们知道，进位的结果是1不断的左移，当进位的结果和不进位的结果不断异或，最左边的1
11 //超过了不进位的结果最左边的1，那么最后一次与运算显然结果为0，于是，最后一次异或运算就是我们想要的相加的结果
12 num1 = sum;
13 num2 = carry;
14 } while (num2 != 0); //将结果相加的过程就重复上述过程，直到进位为0
15
16 return sum;
17 }