## Issue

Function insertBits

X and Y are two integer values.

The function insertBits must insert all the bits of Y after the last set bit in X.

Then the function must return an integer representing the revised value of X.

```
Input:
10 14
Output:
188
Explanation:
10 -> 1010
14 -> 1110
After inserting 1110 after the last set bit in 1010 the binary representation of X becomes 10111100.
The decimal equivalent of 10111100 is 188.
Input:
152 9
Output:
2504
Explanation:
152 -> 10011000
9 -> 1001
After inserting 1001 after the last set bit in 10011000, the binary representation of X becomes 100111001000.
The decimal equivalent of 100111001000 is 2504.
```

This is my code:

```
#include <stdio.h>
#include <stdlib.h>
long bin(long n)
{
long d,r,binary=0;
d=n;
int temp=1;
while(n!=0)
{
r=n%2;
n=n/2;
binary=binary+r*temp;
temp=temp*10;
}
return binary;
}
int insertBits(int X, int Y)
{
bin(X);
bin(Y);
///
}
int main()
{
int X, Y;
scanf("%d %d", &X, &Y);
printf("%d", insertBits(X, Y));
return 0;
}
```

I converted decimal into binary. But I don’t know how to insert the bits in X.

## Solution

Don’t use decimal arithmetic. Binary is simpler and clearer:

```
uint32_t bitshift(uint32_t x, uint32_t y)
{
// put y in the result
uint32_t result = y;
// shift by the trailing 0s in x
while (x && !(x & 1)) {
x /= 2;
result *= 2;
}
// find the shift for x, which is larger than result
uint32_t shift = 1;
while (shift && shift <= result) shift *= 2;
// put x in the right place
result |= x * shift;
return result;
}
int main()
{
printf("10 14 = %5u\n",bitshift(10,14));
printf("152 9 = %5u\n",bitshift(152,9));
// Test some corner cases
printf("152 0 = %5u\n",bitshift(152,0));
printf("0 152 = %5u\n",bitshift(0,152));
printf("4 4 = %5u\n",bitshift(4,4));
printf("1 0xffffffff = %5u\n",bitshift(1,0xffffffffU));
}
```

Output:

```
10 14 = 188
152 9 = 2504
152 0 = 19
0 152 = 152
4 4 = 48
1 0xffffffff = 4294967295
```

Answered By – stark

Answer Checked By – Marie Seifert (BugsFixing Admin)