HP-41 MCode Question

[Sylvain Cote]

Some basic questions for the HP-41 MCode gurus here.

After a pause of 30 years, I am starting to relearn HP-41 MCode again.

About the interaction between register C and RAM register 0 of a selected RAM chip ...

Available instructions

Code:

DEC   BIN         OCT   HEX  HP          J/DA        ZenROM      NEWT        Description
----  ----------  ----  ---  ----------  ----------  ----------  ----------  -----------------------------------------
                                                                             Copy RAM register 0 (T) to C
0040  0000101000  0050  028  REGN=C 0    WRIT 0(T)   REG=C 0/T   REGN=C      Copy C to RAM register 0 (T)
----  ----------  ----  ---  ----------  ----------  ----------  ----------  -----------------------------------------
0752  1011110000  1360  2F0  DATA=C      WRITE DATA  WDATA       DATA=C      Copy C to selected register
0056  0000111000  0070  038  C=DATA      READ DATA   RDATA       C=DATA      Copy selected register to C
----  ----------  ----  ---  ----------  ----------  ----------  ----------  -----------------------------------------

Note: interestingly opcode 0x038 take the same place in the Class 0 Subclass E group (0000-1110-00) as the would have been C=REG 0/T instruction.

Extract from MCODE for Beginners from Ken Emery page 53

Code:

Sometimes we don't know exactly where in a RAM chip we will be, and we can't have the RAMSLCT pointer being moved on us. 
How do we read or write to the selected RAM register without moving the RAMSLCT pointer? We use the READ DATA and WRITE DATA instructions. 
These instructions read and write data between the C register and RAM without modifying the RAMSLCT pointer.

The READ DATA instruction is sometimes listed as READ 0 by some disassemblers. THIS IS INCORRECT! There is no such thing as a READ 0 instruction. 
This was a mistake made by some of the early pioneers in the MCODE field, working without factory documentation that appeared later.

Live trace (extract) from Genesis-41 doing a RDN function who's implementation is actually doing 3 roll up !!!

Code:

,29DA 1 14ED  04E  N C=0       ALL     ; C[13:0] = 00000000000000, CARRY = 0
,29DC 1 14EE  270  p RAMSLCT           ; RAM reg = 000
,29DE 1 14EF  038  8 C=REG     0(T)    ; RAM read (C=08000000000000) from reg = 000 (subclass E)
,29E0 1 14F0  0AE    A<>C      ALL     ; A[13:0] = 08000000000000, C[13:0] = 000000000F0000, CARRY = 0
,29E2 1 14F1  078  x C=REG     1(Z)    ; RAM read (C=07000000000000) from reg = 001 (subclass E)
,29E4 1 14F2  028  ( REG=C     0(T)    ; RAM write (C=07000000000000) to reg 000 (subclass A)
,29E6 1 14F3  0B8    C=REG     2(Y)    ; RAM read (C=06000000000000) from reg = 002 (subclass E)
,29E8 1 14F4  068  h REG=C     1(Z)    ; RAM write (C=06000000000000) to reg 001 (subclass A)
,29EA 1 14F5  0F8    C=REG     3(X)    ; RAM read (C=05000000000000) from reg = 003 (subclass E)
,29EC 1 14F6  0A8    REG=C     2(Y)    ; RAM write (C=05000000000000) to reg 002 (subclass A)
,29EE 1 14F7  0AE    A<>C      ALL     ; A[13:0] = 05000000000000, C[13:0] = 08000000000000, CARRY = 0
,29F0 1 14F8  0E8    REG=C     3(X)    ; RAM write (C=08000000000000) to reg 003 (subclass A)

From the above trace, we clearly see at line 29DE, dissambled as C=REG 0(T) instead of RDATA, loading C register with register 0 of the selected RAM chip 0.

Question 1: could you explain why RDATA is not the equivalent of the inexistant C=REG 0(T)

Question 2: what are the differences between WDATA and REG=C 0/T

Thanks

Sylvain