Bit Manipulation

• 12 : 1100 - 5 : 0101 = 0111
• inverse bits of 5 : 1010 and add 1 : 1011
• 1100+1011 = 0111 = 7
• & : AND
• | : OR
• ^ : XOR
• ~ : inverse
• >> : right shift | a>>1; divide by 2
• << : left shift | a<<1; multiple by 2
• bitwise OR is sum of 2 number
• a = 5 (101) and b = 2 (010)
• a|b = 111 which is 7 : if no carry is involved
• is carry is there, then a|b + a\&b

Divide and Multiply

int a = 6;
int ans1 = a << 1;
int ans2 = a >> 1;
cout << ans1 << " " << ans2 << endl;

Check if the kth bit is set or not

int num = 7; // 0111
int k = 3;

// bring ! in 0!11 to 011! and check for 1

int ans = num >> (k - 1);
if (ans & 1 == 1)
{
    cout << "yes" << endl;
}

Check If The Number is Odd or Even

// if last bit if 1, then odd, else even
int num = 5; // 101, 
if (num & 1 == 1)
    cout << "odd\n";
else
    cout << "even\n";

To swap two numbers

// it is what it is
int a = 5;
int b = 7;
a = a ^ b;
b = a ^ b;
a = a ^ b;
cout << b << " " << a << "\n";

Check if a number is a power of 2

// 8 = 1000
// 7 = 0111
// & = 0000
int x = 8;
if ((x & (x - 1)) == 0)
    cout << "Yes\n";
else
    cout << "No\n";

Count set bits

int num = 31;
int count = 0;
while (num > 0)
{
    if (num & 1 == 1) // like %2 for odd
    {
        ++count;
    }
    num = num>>1; // like divide by 2
}
cout << "Count : " << count << endl;

Find the odd occuring element

#include <stdio.h>

// Function to return the only odd
// occurring element
int findOdd(int arr[], int n)
{
    int res = 0, i;
    for (i = 0; i < n; i++)
        res ^= arr[i];
    return res;
}

int main(void)
{
    int arr[] = { 12, 12, 14, 90, 14, 14, 14 };
    int n = sizeof(arr) / sizeof(arr[0]);
    printf("The odd occurring element is %d ",
           findOdd(arr, n));
    return 0;
}

output :
The odd occurring element is 90