char x = 'A'; // 01000001
char y = 'L'; // 01001100
char z = x & y;// 01000000
3.Or operation
char x = 'A'; // 01000001
char y = 'L'; // 01001100
char z = x | y;// 01001101
4.XOR operation
兩個同時等於0, 兩個不同時等於1
b ^ 11111111 = ~b
b ^ b = 0
char x = 'A'; // 01000001
char y = 'L'; // 01001100
char z = x ^ y;// 00001101
5.Shift operation
char x = 'A'; // 01000001
char z = x >> 3;// 00001000 shift each bit 3 places to the right
char y = x << 3;// 00001000 shift each bit 3 places to the left
6.Get the Nth bit
用Nth為1的check bit做and運算, 再shift left
(b & 00010000) << 4
int get_nth_bit(int num, int n)
{
int check_bit = 1 << n;
int and_bit = num & check_bit;
return and_bit == check_bit;
}
7.Set the Nth bit to 1
int set_nth_bit_to_1(int num, int n)
{
int set_bit = 1 << n;
return num | set_bit;
}
8.Print bits an integer
Shift it left sizeof(int)*8 - 1 times
void print_bits(int num)
{
unsigned int check_bit = 1 << (sizeof(int) * 8 - 1);
while(check_bit != 0)
{
int result = num & check_bit;
if (result == check_bit)
{
printf("%d\n", 1);
}
else
{
printf("%d\n", 0);
}
check_bit = check_bit >> 1;
}
printf("\n");
}
9.Count the number of 1 bits
The simple way is very similar to the print bits example: O(N)
binary subtraction: n & (n - 1): O(1)
int count_ls_optimized(int num)
{
int count = 0;
while(num != 0)
{
num = num & (num - 1);
count++;
}
return count++;
}
10.Reverse the bits in an integer
int reverse_bits(int num)
{
int reverse_num = 0;
unsigned int count = sizeof(int)*8 -1;
while(num != 0)
{
int last_bit = num & 1;
reverse_num = reverse_num | last_bit;
reverse_num = reverse_num << 1;
num = num >> 1;
count--;
}
reverse_num = reverse_num << count;
return reverse_num;
}
