# Merge Binary Strings And Insert Bits In C

## 19 February 2022 - 1 answer

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.

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
``````