How to copy the higher 16hits in the EAX to CX register?

You can't access the top 16-bits of an extended register directly.

So you must have a few lines of code.

push eax
shr eax, 16
mov cx, ax
pop eax

It's probably not the most efficient way of doing it, and it's certainly not the only way but it'll work.

Darwen.

True you could use rotate :

ror eax, 16
mov cx, ax
rol eax, 16

But I'm unsure if this is more or less efficient than using a push and a pop -

Darwen.

The "ror" solution seems to be fast, but is very slow with pentium pro, k6-2 and probably athlon, and P4.
Why?
Because it accesses a partial register after modifying the full register (it loses about 9 CPU cycles on pentium pro because of register renaming invalidation).
This code is better:

mov ecx,eax
shr ecx,16

Of course it destroy the high part of ecx, but it is probably better.
Moreover, the high part of ecx being zero, you can use ecx to do some operations (like additions, multiplications), that are faster than operations with cx, which need a prefix and can use a CPU cycle.