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Post by stiggity on Dec 3, 2021 6:13:08 GMT
I located some very valuable source, and have been able to make sense of it, except for this one instruction.
lda #"\z"
I'm using 64tass, and am not sure if this is a typo(i doubt it), or if someone could correct me. I'm not able to load both bytes, in one call. I can
lda #"\" or lda #"z"
Can someone explain this??
-Thank You.
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Post by bjonte on Dec 3, 2021 20:47:56 GMT
What assembler is supposed to be used to compile the source code?
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Post by stiggity on Dec 3, 2021 21:41:31 GMT
IM UNCERTAIN AS TO TE ORIGINAL
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Post by stiggity on Dec 3, 2021 21:42:05 GMT
I'm uncertain to the original compiler, but I'm using "64tass"
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Post by stiggity on Dec 3, 2021 21:46:12 GMT
Also, if im using 64tass, what exactly is this original source format doing?
;------------------------------- hex = 'b ;-------------------------------
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Post by stiggity on Dec 3, 2021 22:15:17 GMT
And!! LOL!! heres another original compiler syntax issue im trying to overcome..
err = -1 out = -2
my compiler won't work with negative numbers...
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Post by stiggity on Dec 3, 2021 23:50:54 GMT
figured everything out, except the lda #"\z"
any sort of aid, isa highly appreciated!!!
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Post by stiggity on Dec 4, 2021 4:28:21 GMT
if someone would be able to explain the conversions, I would appreciate it. I'm not sure im doing it right...
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Post by bjonte on Dec 4, 2021 4:57:52 GMT
In order to recognize the source format you need to post some more code. What does labels look like and assembler directives?
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Post by stiggity on Dec 4, 2021 16:18:25 GMT
Bjonte: The initialization begings with this..
hex = 'b bin = 'a
then continues with the following..
err = -1 timeout = -2 baddata = -3 cerr = -4 spedl = -5
my assembler cannot handle negative numbers, but I _think_ i should make it
err = 256-1 timeout = 256-2
and so on...
I appreciate the responses!
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