Optimising bitwise operations in C

Question

I had a problem in hand as this : "Exercise 2-6. Write a function setbits(x,p,n,y) that returns x with the n bits that begin at position p set to the rightmost n bits of y, leaving the other bits unchanged."

I've written a function for this as below. This is working as expected.

int func_setx(int x,int p,int n,int y)
{
    int a_t= ~0 << (p+n);
    int b_t= ~a_t >> n;
    int x_t= x& (a_t | b_t);    // a temporary x which has the bits to be changed as 0 , rest of the bits unchanged.

    int mask= (y << p) & (~(~0 << (p+n)));     // a mask which has required bits from y in positions to be set , rest all bits 0.

    printf("\nCheckpoint : mask= %x  x_t= %x\n",mask,x_t);

    int result= mask|x_t;
    return result;
}

But I somehow feel the logic is long and can be optimized, but can't think of any other way yet. Can anyone suggest any optimization to this please?

Answer 1

To make an n bit mask:

mask_y = (1U << n) - 1;

To start it at bit p:

mask_x = mask_y << p;

Clear the appropriate bits in x:

x &= ~mask_x;

Extract the bits from y:

y &= mask_y;

Upshift them to position p:

y <<= p;

Put it all together:

result = x | y;

Or in a more compact form:

mask = (1U << n) - 1;
result = x & ~(mask << p);
result |= (y & mask) << p;