HP-29C: 98 Program Steps, So, why not an even 100?

[Andres]

As I don't have an HP-29C on hand (or 19C), the following ideas are mainly from memory (and old memories, I should say), so there may be some errors or missing details.

IIRC, 16 data registers were non-volatile, and 14 non-volatile registers were assigned to program space. Since each register amounts to 7 bytes (all numbers in these calculators take 56 bits), there are 14 x 7 = 98 program steps. Please note that each and every step occupies one byte.

Now, if the non-volatile registers are 32 at the hardware level, there are 16 registers for data, 14 for program steps, 1 for the X register (which was non-volatile on the 19C/29C) and possibly the remaining non-volatile register was used for keeping the calculator state (display mode, angular mode, position of the program counter, etc.).

It appears there were 16 volatile registers at the hardware level. There were 14 volatile data registers (indirectly addressed). The stack registers were internal to the ACT chip (CPU), but at least one external register was needed for LAST X. So at least 15 non-volatile registers were needed. Subroutine return addresses must be kept somewhere, so perhaps the last remaining register may be used for them.

To look for detailed information about these issues, I suggest looking at the invaluable work of Eric Smith including Nonpareil and related documents.