Given that you know that every other bit is 0 in your application, you can do it like this:
x = (x | (x >> 1)) & 0x33333333;
x = (x | (x >> 2)) & 0x0f0f0f0f;
x = (x | (x >> 4)) & 0x00ff00ff;
x = (x | (x >> 8)) & 0x0000ffff;
The first step looks like this:
0a0b0c0d0e0f0g0h0i0j0k0l0m0n0o0p x
| 00a0b0c0d0e0f0g0h0i0j0k0l0m0n0o0 x >> 1
--------------------------------
= 0aabbccddeeffgghhiijjkkllmmnnoop x | (x >> 1)
& 00110011001100110011001100110011 0x33333333
--------------------------------
= 00ab00cd00ef00gh00ij00kl00mn00op (x | (x >> 1)) & 0x33333333
Then the second step works with two bits at a time, and so on.
与恶龙缠斗过久,自身亦成为恶龙;凝视深渊过久,深渊将回以凝视…