RLCA/RRCA a range of values in memory

[puppydrum64]

I was playing around with RRD and RLD, and came up with something that I thought was quite clever. I haven't tested this on any Z80 hardware, this was all done by writing out what should happen on paper. (Do people still do that?)

EDIT: I have tested this on WinAPE and so far it seems to work.

EDIT 2: I can get it to work for rotating right, but not left...

Imagine memory address &1000 - &1003 contain the following, in order: &23, &45, &67, &89.

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LD HL,&1000
LD A,(HL)   		;A = &23
PUSH HL

Now, here's where the magic happens.

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	RRD		;A = &23, &1000 = &32 &45 &67 &89
	INC HL 			
	
	RRD		;A = &25, &1000 = &32 &34 &67 &89
	INC HL		
	
	RRD		;A = &27, &1000 = &32 &34 &56 &89
	INC HL
	
	RRD		;A = &29, &1000 = &32 &34 &56 &78
POP HL
PUSH HL
	RRD		;A = &22, &1000 = &93 &34 &56 &89
	INC HL
	
	RRD		;A = &24, &1000 = &93 &23 &56 &78
	INC HL
	
	RRD		;A = &26, &1000 = &93 &23 &45 &78
	INC HL
	
	RRD		;A = &28, &1000 = &93 &23 &45 &67
POP HL
RRD			;A = &23, &1000 = &89 &23 &45 &67

After all this is said and done, A equals what it did before any of the RRD commands took place, and the memory range has effectively been RRCA'd once. That is, the last byte is now the first, and the rest were shifted one byte over. We began with &23, &45, &67, &89, and ended with &89, &23, &45, &67.

This test worked for a 4 byte long range, but it seems like it can be generalized with the following (untested) code:

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RRCA_Range:
;input: 
;B = total size of memory range to rotate (measured in bytes)
;HL = address of zeroth element of the range.

LD A,(HL) ;ACCUMULATOR NEEDS TO EQUAL THE FIRST BYTE OF THE RANGE FOR THIS TO WORK
PUSH HL
PUSH BC
loop_RRCA_Range_firstloop:
	RRD
	INC HL
	DJNZ loop_RRCA_Range_firstloop
POP BC
POP HL
PUSH HL
PUSH BC
loop_RRCA_Range_secondloop:
	RRD
	INC HL
	DJNZ loop_RRCA_Range_secondloop
POP BC
POP HL
RRD
RET

And to properly call this routine you can use this macro:

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macro rrca_MemoryRange,addr,count
         LD HL,\addr
         LD b,count
         call RRCA_Range  ;the above subroutine
         endm

As I said before, I haven't fully tested this technique, but I believe this should work. If anyone wants to try it, feel free and please let me know if it works for you.