Assembly 32 bit left shifter

Question

I am having some trouble getting this procedure called DoLeftShift to work properly. The Shifter procedure is working, I know because I have test values and they all work, so the problem is in DoLeftShift.

For the shifter disabled, I am getting all 0's or all 1's, where it should be the same binary that was put in originally. Enable Shift and Shift Instruction also are not working.

Edit: Code removed because it was part of a homework assignment.

Don't vandalize your questions by removing the code. It removes the context for the answer. (If you have some reason for needing to remove some/all of the code, at least say so in the edit message or people are just going to roll it back.) — Peter Cordes, Oct 08 '18 at 20:44

Sep Roland · Accepted Answer · 2017-10-23T13:37:08.573

2

mov ecx, $parm2         ;ecx = enable/disable bit
call Shifter            ;call Shifter procedure

When you call the Shifter procedure, the enable/disable is set just fine (coming from $parm2).

But when you repeatedly call the Shifter procedure from within the loopTop loop, the ECX register no longer holds the correct info!

The Shifter procedure never preserves any registers and so the CL register is returned in a toggled state (xor cl, 1 ;cl = NOT cl).

Solution:

Reload the ECX from $parm2 each time before calling Shifter
Reverse the toggle of CL preferrably at the end of the Shifter procedure.
Preserve any registers that do not return a value.

edited Oct 23 '17 at 13:37

answered Oct 23 '17 at 13:19

Sep Roland

33,889
7
43
76

1

Third (and best IMHO) option: push bl and cl onto the stack at the start of the shifter proc and pop them just before returning. – JeremyP Oct 23 '17 at 13:26
@Jeremy: that would be option 3. – Rudy Velthuis Oct 23 '17 at 14:29
@RudyVelthuis Option 3 was added to the answer after I posted my comment, perhaps *because* I posted my comment. – JeremyP Oct 23 '17 at 14:41
@JeremyP: very well possible. So you can delete the comment. – Rudy Velthuis Oct 23 '17 at 14:49
Thanks so much, that was it. I wasn't pushing and popping ebx and ecx in the Shifter procedure before. – Justin Oct 23 '17 at 17:44

Assembly 32 bit left shifter

1 Answers1

Solution: