Ti asm in 28 days problem

[Xeda112358]

No problem XD Actually, in the next three days I have 4 exams and then I may or may not be without internet until mid January. :/ Good luck!

[jammasterz]

Hello everyone and merry Christmas!
I finally got some precious time for my asm study.

Im stuck on floating point numbers. Its in lesson 15, under "Constant division".
I pretty shure that an explanation in other words will help.

[Xeda112358]

I personally didn't bother with that section and I've never needed to use it. I always stick with fixed-point or integer math, anyways.

EDIT: Sorry, I didn't actually answer anything. I really couldn't follow it as it is.

[asdf]

I've never bothered with floating point numbers either as i've never really had a use for them. Actually, i just checked that section out and don't really understand it either. I'm not very good at math, though. I'll reread through the section above about fixed-point arithmetic and see if i can help out any. What in particular is it that you need help with?

[Xeda112358]

I usually only ever understand something after I have had to go through and figure it out myself :/ Unfortunately, I have found fixed-point math to be very useful and typically faster, so I haven't had much need to work with floating point.

[jammasterz]

Sorry i totally forgot about this topic . I think i figured it out few days ago, now im at the VAT.
There are few things that sound strange. Here is how i understood it:

OP registers are to hold fixed point numbers.
But you can also use them to make a variable by loading some tokens into it:

[Xeda112358]

[jammasterz]

Yeah sorry, i knew it was floating point, just a little mistake here. As allways all my questions answered, thank you!

I've been reading a little and now im at interrupts. Again there are some pretty strange things, that are hard to understand.

1.(Its actually Low-Level Key Input) If you want to get a keypress using OUT and IN, you have to first
1. Output group you want to read to port #1.
2. Get input(after 4 cycles i think).
But now, if i want to get any keypress like _GetKey does, how do i differ the keys if i dont know which group to send? For example if i press [Trace] it could be [Trace],[->], [1],[2],[3],[+] and .

2.Interrupts.

1.I register holds 8 upper bytes of the adress. Bit 7 and 6 seem to be random(really?), and the rest is allways %111111. Because those 2 bits are random you must ensure that whatever they are, the mashine jumps to the right ISR:

Code:


LD HL, $9DF0 ;adress of the ISR

LD (993F), HL
LD (997F), HL
LD (99BF), HL
LD (99FF), HL  ;because of the 2 random bits there are 4 possibilities where the mashine can jump.


2."Interrupts should execute as fast as possible. If an interrupt takes too long to complete it may very well be reinitiated and loop forever. To make sure it doesn't happen you must output 0 to port 3, to disable the hardware like: [ON] key interrupts, Timer and Linkport."- Why can this happen?

3. According to what i read you don't know when an interrupt is triggered. Thats why you must disable them to load the required data and then enable them again.

Code:


DI
LD HL, interrupt
LD ($993F), HL
LD ($997F), HL
LD ($99BF), HL
LD ($99FF), HL

LD A, $99
LD I, A ;loading the needed data with interrupts off

LD A, %00001101  ;enabling hardware before the interrupts are On.
IM 2
EI  ;As far as i know interrupts start triggering "randomly"(or with a certain interval) from this moment.


Why is he turning the hardware on if he said not to?

Wow thats a lot of questions . I hope it doesnt bother anyone. Thank you again for helping!

[Xeda112358]

Okay, here are some answers and some more info that I had to learn myself way back when...

For reading the keyport, the key groups are pretty simple to obtain. if you look at the binary, the first group is %11111101. the second group is %11111011, third is %11110111, et cetera. However, the group 7Fh does not exist (there are only 7 key groups. Keys also follow the same pattern, so if you want to make keytesting programs, you will get familiar with shifting, rotating, and cpl. CPL is a neat instruction that inverts the bits of A. So what are the inverted bits of the keygroups? 2,4,8,16... This is a lot simpler to remember to pass an argument through Ans to a program than 254,252,248...

So anyways, if you wanted to make a program that tested a key group and spit out the result for TI-BASIC users:

[jammasterz]

[Xeda112358]

[asdf]

Alright, so i started writing this last night but didn't finish. It looks like your questions were already answered, but just in case:

[jammasterz]

[asdf]

Ok, so when you are using direct input, it is a lot faster than using any TI-BASIC commands/_getkey/_getcsc. Usually, you will have a loop which will test for your key input. Say we want to use 2nd to confirm, Alpha to cancel, and the arrow keys to move. First check if 2nd or Alpha has been pressed.

[Xeda112358]

To give you an idea of just how fast direct input is, using the code above (which can be optimised >.>) it will take less than 200 cycles to run. What does that mean? In the slowest mode (6MHz), that code will loop 30 000 (thirty thousand) times a second. So pretty much, it is highly unlikely that the user will be able to press and release a key that fast. The code I gave to test which key is being pressed will take at most 746 cycles (that would be the [del] key, by the way) and that means it executes at over 8000 times per second. The OS uses timers and a counter as a key debouncer (I was actually the person to document the location and specifics of the debouncer on WikiTI).

[jammasterz]

Okay, so i finally found some time to read everything you said, but i still find it difficult with the out in instructions.

In the tutorial im reading the author said:
-You send the group you want to read from to port 1.
-Wait for 8 cycles(maybe less, but he said 2 NOPs)
-Read the keycode from the port 1.

I've done some tests and i found out that the delay is not needed.

[Xeda112358]

It depends on the situation for the delays and also the calculator. I have found programs to "glitch" and then I found out that it was because I had not enough delay. For example, you might notice that Enter registers as a Down keypress or vice-versa without a delay.

As for the code from asdf, I think it was a mistake. It should be DFh.

As for in and out, I had trouble with it for a long time. I learned it in hexadecimal and it was a lot easier. D301 would write to the port and DB01 would read from the port. Eventually I found a way to remember the syntax:

out will OUTput to the port. since ld a,b stores b to a, out stores a register to a port. So, out (),a.
in will get INput from a port. Since we are storing from a port, the port argument is last.

[jammasterz]

So i've been playing aroun a little with nterrupts and i wrote this code. Of course it doesnt work, and i cant find the error.
I could only see that the calc stops on HALT which means interrupts stop triggering after running my code.

[asdf]

Yeah, that was a mistake on my part, sorry. I'll look through the code later, but generally you set up a 256-byte table, i'm not sure if what they say in 28 Days about just the two upper bits being random is true or not. Also, the value in HL that you are loading doesn't really point anywhere (it points to your string). Try loading $9872 into hl first.

Again, i'll look through it more thoroughly later, if ThunderBolt or someone else doesn't beat me to it

[jammasterz]

I corrected it, but the problem persists.

EDIT: Besides that i jut noticed a problem with message- the adress has to be changed. BTW, i do not want to switch back to mode 1.
Even whn the program ends the calc has to display hey when i press enter.