I am trying to compile this code using Assembly Studion 8x, and I cannot find the file it creates.

compile this

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final code

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I used this code


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and it did this

What the hell are you trying to do? Whatever it is, you clearly have no idea what you're doing. Let's start with what exactly you're trying to do so I can explain what you're screwing up. Secondly, don't forget the .end / END at the end. Thirdly, and the reason I'm annoyed, stop triple-posting, it's rude and annoying.

sorry Kerm, we were just using this as a way to display code better. Second, Will_W was teaching me assembly, so anything I was having trouble with I would post it so he could see my source code.

That's fine, and that's a good way to do it. However, that's what the Edit button on each one of your posts is for. If it's less than 24 hours since you had the last post in the same topic, it's considered double-posting and poor netiquette. Now, what exactly are you trying to do with that 100-iteration piece of code?

While he was teaching me then he decided to test me to see what I knew. So he challenged me to create a programs that starts at 1 and counts to 100 and display it in the upper left hand corner. Here is what I have so far


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1) Because of the two-byte BB,6D token you need to start at $9d93, not $9d95
2) Why are you loading a into l after you inc l? Surely you should see why that's pointless
3) IIRC, DispHL trashed HL, so it might not be a bad idea to wrap a push hl/pop hl around it.
4) .DB is for a _B_yte, and .DW is for a _W_ord, or 16 bits or two bytes. Since hl is a 16-bit register, you can only load a two-byte value into it. However, even better would be xor a \ ld h,a \ ld l,a, which is three bytes and saves the two-byte .DW statement.
5) Indirection: What you actually meant was ld hl,(Variable), because you want the contents of the memory address Variable, not the actual numerical address. Indirection and stack manipulation are the two most vital z80 ASM concepts you need to master before you can start writing good, efficient code.


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did you modify that code any, besides the ".dw" and the pop and push parts?

Surely you can actually READ my post; not only did I explicitly talk about what I suggested before I made the changes, I posted code that you can use your perfectly good brain to compare with the code you posted above.

Sorry, I wasn't sure, when I ran the code it only showed 22-24. I saw your edited post and reread it a couple of times to make sure I understood it. Why do you xor a?

Good question. "xor a" is one byte, and it does the equivalent of "ld a,0", a two-byte instruction. It's a way of resetting the accumulator register with a smaller (though unfortunately no faster) instruction.

thanks, when I run your program exactly, it only shows numbers 1 2 and 3.

9D95 works whereas 9D93 doesn't. I have no idea why.

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It should be 9D93 for the sake of BB6D.

Exactly. We've established that executing the $BB,$6D bytes is harmless, but it's bad form to do so.