mirror of
https://github.com/dlidstrom/hp48page
synced 2024-09-29 05:37:01 +02:00
265 lines
7.2 KiB
HTML
265 lines
7.2 KiB
HTML
|
<HTML>
|
||
|
|
<HEAD>
|
||
|
|
<TITLE>HP48 Assembly-Programming</TITLE>
|
||
|
|
<BODY BGCOLOR="#FFFFFF">
|
||
|
|
</HEAD>
|
||
|
|
<BODY>
|
||
|
|
<H1> Primer</H1>
|
||
|
|
This is a short description of some of the registers used while programming in saturn assembly.<P>
|
||
|
|
|
||
|
|
<H2> Data pointers </H2>
|
||
|
|
D0: Instruction pointer. <BR>
|
||
|
|
D1: Stack pointer. <BR>
|
||
|
|
|
||
|
|
<H2> Working registers </H2>
|
||
|
|
|
||
|
|
A: Can be used freely. <BR>
|
||
|
|
B: Pointer to top of return stack. <BR>
|
||
|
|
C: Can be used freely. <BR>
|
||
|
|
D: Amount of free memory between return stack and stack. <BR>
|
||
|
|
|
||
|
|
<H2> Scratch registers </H2>
|
||
|
|
|
||
|
|
R0, R1, R2, R3, R4: Used to temporarily store data, addresses etc. <BR>
|
||
|
|
|
||
|
|
<H2> Others </H2>
|
||
|
|
P: Pointer used to point into C and for loops. I must add one thing about this register. Its value
|
||
|
|
determines where loading instructions start their loading. For example, if P=15 then loading A or C
|
||
|
|
will load that value into field S (see below for architecture). P must be 0 when exiting your program!<BR>
|
||
|
|
IN: Input, used in keyscanning. <BR>
|
||
|
|
OUT: Output, used to generate tones and in keyscanning. <BR>
|
||
|
|
ST: Status bits. The lower 12 can be freely used while the upper four are used by the operating
|
||
|
|
system. <BR>
|
||
|
|
Return stack: An 8-level stack used to temporarily hold addresses. Note that you must never use
|
||
|
|
more than 6 levels as the top 2 are used by the interrupt system. <BR>
|
||
|
|
|
||
|
|
<HR>
|
||
|
|
|
||
|
|
<H1> Example usage </H1>
|
||
|
|
<H2> Your first program </H2>
|
||
|
|
|
||
|
|
When you start programming in ml you need to make sure you restore the rpl pointers when the program finishes. You
|
||
|
|
can either save them at the beginning and restore them at the end. The most common entries to do this are:
|
||
|
|
<PRE>
|
||
|
|
=SAVPTR save the pointers
|
||
|
|
=GETPTRLOOP restore them and continue with rpl
|
||
|
|
</PRE>
|
||
|
|
You could also however, decide not to use B[A], D[A], D0, and D1 (or restore them
|
||
|
|
yourself). Then you would exit similar to this (we save D1 to show an example):
|
||
|
|
<PRE>
|
||
|
|
CD1EX swap D1 and C[A]
|
||
|
|
RSTK=C save D1 on returnstack
|
||
|
|
.
|
||
|
|
.
|
||
|
|
.
|
||
|
|
C=RSTK retrieve D1 from return stack
|
||
|
|
D1=C restore D1 from C[A]
|
||
|
|
* continue with rpl
|
||
|
|
D0=D0+ 5 point to next instruction
|
||
|
|
A=DAT0 A read it into A[A]
|
||
|
|
PC=(A) execute next instruction
|
||
|
|
</PRE>
|
||
|
|
You could also exit with a command of your choice:
|
||
|
|
<PRE>
|
||
|
|
LC(5) =UNCOERCE
|
||
|
|
A=C A
|
||
|
|
PC=(A)
|
||
|
|
</PRE>
|
||
|
|
Also remember to ensure P=0 and cpu is in HEXMODE.
|
||
|
|
|
||
|
|
<H2> The working registers </H2>
|
||
|
|
|
||
|
|
<TABLE BORDER CELLPADDING=4 WIDTH="537" bordercolor="#000000" >
|
||
|
|
<TR>
|
||
|
|
<TD COLSPAN="16">
|
||
|
|
<CENTER>The Arhitecture of A, B, C, and D
|
||
|
|
</CENTER>
|
||
|
|
</TD>
|
||
|
|
</TR>
|
||
|
|
<TR>
|
||
|
|
<TD WIDTH="6%">
|
||
|
|
<CENTER><FONT SIZE=+1>15</FONT></CENTER>
|
||
|
|
</TD>
|
||
|
|
<TD WIDTH="6%">
|
||
|
|
<CENTER><FONT SIZE=+1>14</FONT></CENTER>
|
||
|
|
</TD>
|
||
|
|
<TD WIDTH="6%">
|
||
|
|
<CENTER><FONT SIZE=+1>13</FONT></CENTER>
|
||
|
|
</TD>
|
||
|
|
<TD WIDTH="6%">
|
||
|
|
<CENTER><FONT SIZE=+1>12</FONT></CENTER>
|
||
|
|
</TD>
|
||
|
|
<TD WIDTH="6%">
|
||
|
|
<CENTER><FONT SIZE=+1>11</FONT></CENTER>
|
||
|
|
</TD>
|
||
|
|
<TD WIDTH="6%">
|
||
|
|
<CENTER><FONT SIZE=+1>10</FONT></CENTER>
|
||
|
|
</TD>
|
||
|
|
<TD WIDTH="6%">
|
||
|
|
<CENTER><FONT SIZE=+1>9</FONT></CENTER>
|
||
|
|
</TD>
|
||
|
|
<TD WIDTH="6%">
|
||
|
|
<CENTER><FONT SIZE=+1>8</FONT></CENTER>
|
||
|
|
</TD>
|
||
|
|
<TD WIDTH="6%">
|
||
|
|
<CENTER><FONT SIZE=+1>7</FONT></CENTER>
|
||
|
|
</TD>
|
||
|
|
<TD WIDTH="6%">
|
||
|
|
<CENTER><FONT SIZE=+1>6</FONT></CENTER>
|
||
|
|
</TD>
|
||
|
|
<TD WIDTH="6%">
|
||
|
|
<CENTER><FONT SIZE=+1>5</FONT></CENTER>
|
||
|
|
</TD>
|
||
|
|
<TD WIDTH="6%">
|
||
|
|
<CENTER><FONT SIZE=+1>4</FONT></CENTER>
|
||
|
|
</TD>
|
||
|
|
<TD WIDTH="6%">
|
||
|
|
<CENTER><FONT SIZE=+1>3</FONT></CENTER>
|
||
|
|
</TD>
|
||
|
|
<TD WIDTH="6%">
|
||
|
|
<CENTER><FONT SIZE=+1>2</FONT></CENTER>
|
||
|
|
</TD>
|
||
|
|
<TD WIDTH="6%">
|
||
|
|
<CENTER><FONT SIZE=+1>1</FONT></CENTER>
|
||
|
|
</TD>
|
||
|
|
<TD WIDTH="6%">
|
||
|
|
<CENTER><FONT SIZE=+1>0</FONT></CENTER>
|
||
|
|
</TD>
|
||
|
|
</TR>
|
||
|
|
<TR>
|
||
|
|
<TD COLSPAN="16">
|
||
|
|
<CENTER><FONT SIZE=+1>W</FONT></CENTER>
|
||
|
|
</TD>
|
||
|
|
</TR>
|
||
|
|
<TR>
|
||
|
|
<TD WIDTH="6%">
|
||
|
|
<CENTER><FONT SIZE=+1>S</FONT></CENTER>
|
||
|
|
</TD>
|
||
|
|
<TD COLSPAN="12" WIDTH="75%">
|
||
|
|
<CENTER><FONT SIZE=+1>M</FONT></CENTER>
|
||
|
|
</TD>
|
||
|
|
<TD COLSPAN="3" WIDTH="19%">
|
||
|
|
<CENTER><FONT SIZE=+1>X</FONT></CENTER>
|
||
|
|
</TD>
|
||
|
|
</TR>
|
||
|
|
<TR>
|
||
|
|
<TD COLSPAN="11" WIDTH="69%">
|
||
|
|
<CENTER> </CENTER>
|
||
|
|
</TD>
|
||
|
|
<TD COLSPAN="5" WIDTH="31%">
|
||
|
|
<CENTER><FONT SIZE=+1>A</FONT></CENTER>
|
||
|
|
</TD>
|
||
|
|
</TR>
|
||
|
|
<TR>
|
||
|
|
<TD COLSPAN="13" WIDTH="81%">
|
||
|
|
<CENTER> </CENTER>
|
||
|
|
</TD>
|
||
|
|
<TD WIDTH="6%">
|
||
|
|
<CENTER><FONT SIZE=+1>XS</FONT></CENTER>
|
||
|
|
</TD>
|
||
|
|
<TD COLSPAN="2" WIDTH="13%">
|
||
|
|
<CENTER><FONT SIZE=+1>B</FONT></CENTER>
|
||
|
|
</TD>
|
||
|
|
</TR>
|
||
|
|
</TABLE></CENTER>
|
||
|
|
<BR>
|
||
|
|
A, B, C, and D are structured this way. A[A] means field A of A while D[S] means field S of D.
|
||
|
|
C[P] specifies the nibble pointed to by P. I.e. if P=15 then C[P] is equal to C[S]. A=A+A WP
|
||
|
|
doubles the values in register A from nibbles 0 to P.
|
||
|
|
|
||
|
|
<H2> Reading data </H2>
|
||
|
|
To point D0 at the screen for example, use the supported entry =D0->Row1. Now you can read from the screen and/or
|
||
|
|
write to the screen:
|
||
|
|
reading 5 nibbles:
|
||
|
|
<PRE>
|
||
|
|
GOSBVL =D0->Row1 point D0 to top-left corner of currently displayed grob
|
||
|
|
C=DAT0 A read 5 nibbles
|
||
|
|
</PRE>
|
||
|
|
writing 5 nibbles:
|
||
|
|
<PRE>
|
||
|
|
GOSBVL =D0->Row1
|
||
|
|
LC(5) #FFFFF load C with 5*4 pixels (#Fh = #1111b)
|
||
|
|
DAT0=C A
|
||
|
|
</PRE>
|
||
|
|
Observe that by doing D0=D0+ 1 you advance the pointer in this case by four pixels, a nibble. Since the screen is
|
||
|
|
34 nibbles wide, you can move down one row by executing:
|
||
|
|
<PRE>
|
||
|
|
D0=D0+ 16
|
||
|
|
D0=D0+ 16
|
||
|
|
D0=D0+ 2
|
||
|
|
</PRE>
|
||
|
|
Say for example, that you want to turn on the pixel at coordinate { # 65d # 32d }, then you could go about it like this:
|
||
|
|
<PRE>
|
||
|
|
|
||
|
|
CODE GOSBVL =SAVPTR save rpl pointers
|
||
|
|
GOSBVL =D0->Row1 point D0 to top-left corner of display
|
||
|
|
LC(5) 34*32+16 32 rows down, 16*4 pixels to the right
|
||
|
|
AD0EX swap A[A] with D0 (actually you don't need to swap here, see with DB)
|
||
|
|
A=A+C A add offset
|
||
|
|
AD0EX swap back
|
||
|
|
LC(5) #2 #2h = #0010b
|
||
|
|
DAT0=C 1 write one nibble
|
||
|
|
* note: since data is written "backwards", the second pixel in the nibble will be lit: 0*00 (0-off, *-lit)
|
||
|
|
GOVLNG =GETPTRLOOP restore rpl pointers and continue rpl
|
||
|
|
ENDCODE
|
||
|
|
</PRE>
|
||
|
|
Assemble it, put it on stack level 2 and run <TT> << CLLCD EVAL 7 FREEZE >> </TT>
|
||
|
|
|
||
|
|
<H2>Accessing the stack</H2>
|
||
|
|
Dropping an object:
|
||
|
|
<PRE>
|
||
|
|
D1=D1+ 5 advance stack pointer to next level
|
||
|
|
D=D+1 A increment available memory
|
||
|
|
</PRE>
|
||
|
|
Dropping is easy and can be done quite fast. Dropping several objects:
|
||
|
|
<PRE>
|
||
|
|
CODE
|
||
|
|
P= 16-5 drop five objects
|
||
|
|
loop D1=D1+ 5 drop it
|
||
|
|
D=D+1 A
|
||
|
|
P=P+1 add one to the counter
|
||
|
|
GONC loop loop until done
|
||
|
|
D0=D0+ 5
|
||
|
|
A=DAT0 A
|
||
|
|
PC=(A)
|
||
|
|
ENDCODE
|
||
|
|
</PRE>
|
||
|
|
I let P=11 and then count up to 16. When P=15 and I add once more, P wraps around to 0 and the carry flag is set, the
|
||
|
|
loop is done.
|
||
|
|
<P>
|
||
|
|
Reading another pointer, D[A], amount of available addresses:
|
||
|
|
<PRE>
|
||
|
|
CODE
|
||
|
|
GOSBVL =SAVPTR save pointers
|
||
|
|
C=D A multiply available memory by five
|
||
|
|
C=C+C A
|
||
|
|
C=C+C A
|
||
|
|
C=C+D A
|
||
|
|
CSRB.F A divide number by two
|
||
|
|
R0=C.F A prepare to push number
|
||
|
|
GOSBVL =PUSH# push it to the stack, restore pointers
|
||
|
|
LC(5) =UNCOERCE exit, converting the "bint" to a float
|
||
|
|
A=C A
|
||
|
|
PC=(A)
|
||
|
|
ENDCODE
|
||
|
|
</PRE>
|
||
|
|
Since an address is five nibbles we multiply by five, and we want the answer in bytes, thus we
|
||
|
|
divide by two. CSRB.F A shifts register C field A right one bit, effectively dividing its contents
|
||
|
|
by two.
|
||
|
|
<H2> The program counter (PC) </H2>
|
||
|
|
This counter contains the address of the current instruction being executed. The
|
||
|
|
simple example here calculates the address of itself and puts itself on the stack.
|
||
|
|
Remember the stack is only a pile of pointers, not the objects themself. See the glossary
|
||
|
|
for help in understanding how objects are structured.
|
||
|
|
<PRE>
|
||
|
|
CODE
|
||
|
|
A=PC read the program counter into A[A]
|
||
|
|
LC(5) 14 load C[A] with the amount to subtract
|
||
|
|
A=A-C A A[A] now contains the prolog of the code object
|
||
|
|
GOVLNG =PUSHA push it to the stack
|
||
|
|
ENDCODE
|
||
|
|
</PRE>
|
||
|
|
Example programs greatly appreciated!
|
||
|
|
</BODY>
|
||
|
|
</HTML>
|