Bit manipulation: keeping the common part at the left of the last different bit

Question

Consider two numbers written in binary (MSB at left):

X = x7 x6 x5 x4 x3 x2 x1 x0

and

Y = y7 y6 y5 y4 y3 y2 y1 y0

These numbers can have an arbitrary number of bits but both are of the same type. Now consider that x7 == y7, x6 == y6, x5 == y5, but x4 != y4.

How to compute:

Z = x7 x6 x5 0 0 0 0 0

or in other words, how to compute efficiently a number that keeps the common part at the left of the last different bit ?

template <typename T>
inline T f(const T x, const T y) 
{
    // Something here
}

For example, for:

x = 10100101
y = 10110010

it should return

z = 10100000

Note: it is for supercomputing purpose and this operation will be executed hundreds of billion times so scanning the bits one by one should be avoided...

Answer 1

My answer is based on @JerryCoffin's one.

int d = x ^ y;
d = d | (d >> 1);
d = d | (d >> 2);
d = d | (d >> 4);
d = d | (d >> 8);
d = d | (d >> 16);
int z = x & (~d);