Re: A89: hello world


[Prev][Next][Index][Thread]

Re: A89: hello world




In a message dated 98-11-23 20:34:48 EST, you write:

<< Woohoo some real code!!!
 okie... let the pro explain...
 
 > Hi, can someone please help me understand this supposedly simple code?
 > Let me start by stating what I know or what I think I know:
 > ***any help would be greatly appreciated***
 > -thanks-
 > Steven
 >
 > line 1: This does a jump to the utility library in order to clear
 > the screen
 > line 2: This pushes 2 onto the stack
 > line 3: 2 was pushed onto the stack because of the ti rom call
 > FontSetSys:the
 > 2 sets the font to normal size (?).  Is a7 used on all rom calls?
 > line 4: Because you pushed 2 onto the stack you must also pop the
 > 2 off the
 > stack: Why does this have to be done?  and why can't you just
 > add.w or add.b?
 
 The stack is 32-bit. Thus, you need to move the stack value back 2
 bytes. The stack value is 32-bit, thus use addq.l or add.l
 
 > line 5: This pushes 4 onto the stack
 > line 6: This loads the address of string to the a0 register: what is the
 > "(pc)" for?
 
 It's the way to show the "absolute" address of the string, so the rom
 call can find directly locate the string.
 
 > line 7: This pushes the address of the string to the stack: what
 > happened to
 > the 4 that was pushed to the stack?
 
 The four is still there...but now it's four bytes below... need me to
 explain stacks?
 
 okie... a little rundown on stacks...
 a7 is just a 32-bit number. there's nothing really important about the
 number. It could
 be $4440 or it could be $3300. However, the value (a7) is most important.
 Say, for example,
 a7 = $4440, (a7) is the value at $4440. So, when you do -(a7), A7 equals to
 either
  $443E or $443C.
 
 Okie... so how does this mix in? Well, say...
   at line 5,   a7 is equal to $4440.
    after line 5, a7 will equal to $443E, and ($443E)= $4, the color
   at line 7,   a7 is equal to $443E.
    after line 7, a7 will equal to $443A, and ($443A)= address of the string.
   at line 8,   a7 is equal to $443A,
    after line 8, a7 is equal to $4438, and ($4438) = 0, the horizontal
 position
   at line 9,   a7 is equal to $4438
    after line 9, a7 is equal to $4436, and ($4436) = 0, the vertical
 position
   okie, when calling that rom call, a7 is still equal to $4436.
   got that? this is what it looks in RAM
 
 | $4436 | $4438 | $443A | $443C | $443E | $4440 |
    0000    0000   string address   0004
   vert.   horiz.  string address  color
 
 > line 8: This pushes 0 onto the stack: what happened to the 4 and
 > the address
 > of the string that was previously pushed onto the stack?
 > line 9: This pushes 0 onto the stack again: why is it pushed again?
 > line 10: This does the rom call DrawStrXY which displays the
 > string: maybe i
 
 
 
 > think i know what is going on: the first thing that you need to do to call
 > this rom call is to push the address onto the stack then you must
 > push the 2
 > coordinates onto the stack:0,0 and the rom call takes care of the
 > rest?  But I
 > still don't know where the 4 comes in place.
 > line 11: It says restore the stack pointer...but where does the
 > 10 come from:
 > i only count 4 then 0 then 0 again=4...i don't know how that
 > works.  Why do
 > you even have to "restore" the stack...i remember with the z80
 > you didn't HAVE
 > to pop a register back after pushing it?
 
 Okie, so what is a7 after step 10? well, it is $4436, but before we did
 all this crap, it was $4440. So you subtract... $4440 - $4436 is equal to
 $0A or #10.
 
 > line 12: this does a jump to the utility library which basically
 > does a loop
 > waiting for a key to be pressed.
 > line 13: returns to the shell when a key is pressed...What other return
 > statements are there?
 
 that's it... pretty simple stuff... reply if u don't understand or if other
 ppl don't understand.  I like it when we're discussing code, not concepts...
 
 >
 > _main:
 >     ; Execution will start here, at the _main label
 >   1  jsr      util::clr_scr          ; clear the screen
 >   2 move.w   #2,-(a7)               ; use font #2 (large font)
 >   3  jsr      tios::FontSetSys       ; set the font
 >   4  add.l    #2,a7                  ; clean up the stack
 >   5  move.w   #4,-(a7)               ; move the first param to the stack
 >   6  lea      string(pc),a0          ; move adress of string to the stack
 >   7  move.l   a0,-(a7)
 >   8  move.w   #0,-(a7)               ; push position
 >   9  move.w   #0,-(a7)               ; push position
 >  10 jsr      tios::DrawStrXY        ; call the DrawStrXY ROM function
 >  11 add.l    #10,a7                 ; restore the stack pointer
 >  12 jsr      util::idle_loop        ; library routine which waits
 > until a key
 >                                     ; is pressed.
 >  13 rts
 >
 > string:
 >     dc.b "Hello, World!",0
 > _comment:
 >     dc.b "Hello World example",0
 >
 >     end
 > >>


You said:
""""""
" Okie... so how does this mix in? Well, say...
   at line 5,   a7 is equal to $4440.
    after line 5, a7 will equal to $443E, and ($443E)= $4, the color
   at line 7,   a7 is equal to $443E.
    after line 7, a7 will equal to $443A, and ($443A)= address of the string.
   at line 8,   a7 is equal to $443A,
    after line 8, a7 is equal to $4438, and ($4438) = 0, the horizontal
 position
   at line 9,   a7 is equal to $4438
    after line 9, a7 is equal to $4436, and ($4436) = 0, the vertical
 position
   okie, when calling that rom call, a7 is still equal to $4436.
   got that? this is what it looks in RAM
 
 | $4436 | $4438 | $443A | $443C | $443E | $4440 |
    0000    0000   string address   0004
   vert.   horiz.  string address  color"
"""""
So why would the stack be $443E after line 5...don't you just add 4 and
wouldn't that be $4444...and also how does the stack change if you are only
adding "0" to the stack through 7-9...


and can you explain what that table obove means...in english please...?


Thanks for replying...i'm slowly coming along...