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A project log for Gigatron Forth

Attempting to implement Forth on the Gigatron

PeterPeter 11/23/2019 at 00:110 Comments

[This log relates to the state of the project as of  commit b58f04048e080c8397b921c5705d2b3de66b05e4]

I can't believe that I've been working on this for 24 days, with such a small amount to show for it. I guess they weren't joking when they said implementing Forth in the native instruction set would be hard! It's not that I haven't been putting any effort into it - I really have!

I've finally finished implementing the various NEXT routines, with pretty good tests. However I'm now convinced that my original design was just bad. All of the fail-fast tests, and jumping from one routine to another leads to dispatch having a very large cost - whilst also causing me to do a huge amount of thinking just to work out what the code should do. As someone who is not good at arithmetic, and gets easily confused by pointers, this has been proving to be a challenge.

https://docs.google.com/spreadsheets/d/1FN_2EKFwPx6gkxwS5qprr5ikTHDJjQV4124Q-Ay5Tx4/edit?usp=sharing

My original design goal was to maximise the number of instructions a native-forth word can be without breaking the time window. The image above is from a spreadsheet I have made to find out how much the overhead is going to be, and if the system will even work in terms of the timings (does the cost of dispatch dwarf the instructions a word can have). What happens if you enter a time-slice with different length words in the 'W' register (the X-axis in the chart). In this particular case I'm examining the '---D line 0' time-slice, which is the shortest at 114 cycles, and we're assuming that we're executing code from ROM, and the word ends with a call to NEXT. All of these can be changed in the spreadsheet.

As you can see, it works: If we run a word with 0 - 32 "real-work" instructions, we'll finish it, dispatch to the next word, and possibly start another word all within our window. If our word is between 33 and 48 instructions, we will load the W register with the address of the next word, but won't dispatch to it (the steps are various points it can fail along the path), If our word is 48 to 82 cycles long we will complete the work, and queue the dispatch, but won't attempt it. Greater than that, and we can't do the required book keeping, so we don't enter the word.

I definitely want to come back to this area of the code, and get that big red stripe narrower, but for now, it's going to be good enough.

Next DOCOL and EXIT. I'm thinking of writing some Python utilities to do the code-counting for me - doing it by hand is pretty tedious!

P.S. Spreadsheets are a trap. It's so easy to get started, but a nightmare to keep track once they get a little bit big.

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