HP-25C Eumulator

[Chris Chung]

(10-23-2014 12:35 PM)Harald Wrote:  The only thing I found so far is that the register contents is lost when the calculator is turned off.
I wouldn't think it is a real problem, but it differs from the original.

It may matter to a purist. This is not a bug and it's really a compromise in this project.

For the 25, the continuous memory as I understand it is 16 register units, with the 1st 9 units being registers and the last 7 being program steps, each register unit requires 7 bytes (or 14 half-bytes to represent the bcd, original nonpareil code uses 14 bytes). That would make the whole continuous memory be 112 byte long.

In this project the memory retention is done by writing them to internal flash and reading it upon next power-on. The msp430g2533 mcu we have 4 banks of 64 byte flash blocks for the firmware and I can only use 3 of them (the other one is for factory clock calibration, etc).

So if I am running just the HP-25 ROM, I can put all the CM into 2 blocks (112 bytes), and if I want both HP-25 and HP-33 CM retention, I can only save part of it. Which is program steps only (49 bytes).

We are also using a 3rd block to save the default start-up settings + greeting messages. Also the flash blocks, being flash is not like RAM, and operations (read / write) must be done to the whole block. I.e. we can't share / split as we don't have enough buffer RAM to start w/.

Since I don't have the original hardware, I was guessing that the register contents are just a by-product of the CM and will be zeroised during power-on, thus to save flash space, I chose not to save them.

How do you / others think of this? Should I turn back and have all (registers and program steps) saved? This would mean the the HP-33 / HP-25 shares the same CM. I.e. they will step into each other. Please let me know your thoughts.

Also question for the HP-33 expert. I don't understand well enough the HP-33 (I don't even have the manual). From the description it is similar to the 25, w/ 8 registers, and 49 (7x7) steps. But from nonpareil source code it's RAM size is 32 (vs 16 on the HP-25). My implementation is to allocate 32 RAM units for it but when it comes to CM, I am only saving the 49 program steps. What I don't understand for the 33 is what does the last 16 unit of RAM are used for.

It would be nice if someone w/ HP-33C experience can advise, in particular the importance / impact on the CM retention.