If I want to round an unsigned integer to the closest smaller or equal even integer, can I divide by 2 then multiply by 2?

Question

For example :

f(8)=8 f(9)=8 

Can I do x = x/2*2; ? Is there a risk that the compiler will optimize away such expression ?

Answer 1

The compiler is allowed to make any optimisiations it likes so long as it does not introduce any side effects into the program. In your case it can't cancel the '2's as then the expression will have a different value for odd numbers.

x / 2 * 2 is evaluated strictly as (x / 2) * 2, and x / 2 is performed in integer arithmetic if x is an integral type.

This, in fact, is an idiomatic rounding technique.

Answer 2

Since you specified the integers are unsigned, you can do it with a simple mask:

x & (~1u) 

Which will set the LSB to zero, thus producing the immediate even number that is no larger than x. That is, if x has a type that is no wider than an unsigned int.

You can of course force the 1 to be of the same type as a wider x, like this:

x & ~((x & 1u) | 1u) 

But at this point, you really ought to look at this approach as an exercise in obfuscation, and accept Bathsheba's answer.

---

I of course forgot about the standard library. If you include stdint.h (or cstdint, as you should in C++ code). You can let the implementation take care of the details:

uintmax_t const lsb = 1; x & ~lsb; 

or

x & ~UINTMAX_C(1)