Left shift using %cl

Question

Let's say I have a number in binary. 1010 which is 10 in decimal.

I understand that shifting left by 1 bit is essentially multiplying the number by 2.

Theres a line in a textbook that's got me confused.

salq %cl, %rdx

%rdx is a number and %salq is a left shift. What I'm confused about is the %cl.

I've read that CL is 8 bits, does that mean I'm multiplying by 2^8?

Govind Parmar · Answer 1 · 2018-10-10T20:24:38.897

3

The cl partial register (which is really the lowest 8 bits of the register rcx) contains the value by which rdx will be shifted left. It's eight bits long, but the amount shifted is whatever is actually in there:

movb $10, %cl
salq %cl, %rdx ; rdx is shifted 10 bits left.

edited Oct 10 '18 at 20:24

answered Oct 10 '18 at 19:30

Govind Parmar

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It's not a "pseudo"-register. It's 100% a real register, but yes it aliases the low byte of RCX. Maybe the term you're looking for is "partial register", as in [Why doesn't GCC use partial registers?](https://stackoverflow.com/q/41573502). – Peter Cordes Oct 10 '18 at 20:19
Updated my answer. – Govind Parmar Oct 10 '18 at 20:24

score 1 · Answer 2 · answered Oct 10 '18 at 19:30

CL is the shift count. In C, it's rdx <<= cl, or rdx = rdx << cl. http://felixcloutier.com/x86/SAL:SAR:SHL:SHR.html.

Or more precisely, rdx <<= (cl&63), because x86 masks shift counts.

You're multiplying RDX by 2^cl.

8 is the bit width of CL, meaning it can represent values from 0..255.

Left shift using %cl

2 Answers2