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Section
05 Part 04 – The SP (Stack Pointer) |
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“A whole stack of memories never equal one little hope.” ~Charles M. Schulz |
Introduction
There
will be times where the 68k will need to store some important information, and there
will also be times where “you” will need to store some important information.
The
stack is designed exactly for this purpose.
The Stack
The
stack is a memory space where important information can be stored
temporarily. You can decide where in 68k
memory the stack is, and can change the address of the stack at any time.
The
address register a7 is used as the Stack Pointer
(SP), and you’ll need to decide
where in 68k memory you want the stack to be.
There are two ways of setting the stack; using a7:
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movea.l #$00100000,a7 |
Or by
using SP:
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movea.l #$00100000,sp |
Both
are acceptable, and are exactly the same.
a7 is used as the SP,
so a7 is the SP.
Storing
In
the above example, I’ve set the stack address to 68k offset 100000, however, memory is stored
inside the stack in reverse order.
Here’s an example of saving data into the stack (I’ve used SP, but you can use a7 if you prefer):
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move.w d0,-(sp) move.l d1,-(sp) |
Please
note how SP is treated as an address register,
and also note the “-“ symbol. Starting with the first instruction, a word
of d0 is saved into the stack, d0 contains FF000020, and so 0020 is saved into the stack:
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Offset |
0 |
1 |
2 |
3 |
4 |
5 |
6 |
7 |
8 |
9 |
A |
B |
C |
D |
E |
F |
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000FFFD0 |
00 |
00 |
00 |
00 |
00 |
00 |
00 |
00 |
00 |
00 |
00 |
00 |
00 |
00 |
00 |
00 |
|
000FFFE0 |
00 |
00 |
00 |
00 |
00 |
00 |
00 |
00 |
00 |
00 |
00 |
00 |
00 |
00 |
00 |
00 |
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000FFFF0 |
00 |
00 |
00 |
00 |
00 |
00 |
00 |
00 |
00 |
00 |
00 |
00 |
00 |
00 |
00 |
20 |
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00100000 |
00 |
00 |
00 |
00 |
00 |
00 |
00 |
00 |
00 |
00 |
00 |
00 |
00 |
00 |
00 |
00 |
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etc |
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As
you can see above, the word was saved into memory at offset 0FFFFE and 0FFFFF. This is because the stack is moved back
before the 0020 is saved into memory. After this, the SP will be at offset 0FFFFE.
The
next instruction moves a long-word of data from d1 into the stack, so we’ll pretend that d1 contains FEDCBA98, FEDCBA98 is saved into the stack:
|
Offset |
0 |
1 |
2 |
3 |
4 |
5 |
6 |
7 |
8 |
9 |
A |
B |
C |
D |
E |
F |
|
000FFFD0 |
00 |
00 |
00 |
00 |
00 |
00 |
00 |
00 |
00 |
00 |
00 |
00 |
00 |
00 |
00 |
00 |
|
000FFFE0 |
00 |
00 |
00 |
00 |
00 |
00 |
00 |
00 |
00 |
00 |
00 |
00 |
00 |
00 |
00 |
00 |
|
000FFFF0 |
00 |
00 |
00 |
00 |
00 |
00 |
00 |
00 |
00 |
00 |
FE |
DC |
BA |
98 |
00 |
20 |
|
00100000 |
00 |
00 |
00 |
00 |
00 |
00 |
00 |
00 |
00 |
00 |
00 |
00 |
00 |
00 |
00 |
00 |
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etc |
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Once
again, the SP is moved back from offset 0FFFFE to 0FFFFA, and FEDCBA98 is saved into memory.
Reloading
Loading
the data back out of the stack is done in reverse order:
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move.l (sp)+,d1 move.w (sp)+,d0 |
The
first instruction will load a long-word from the stack at offset 0FFFFA back into d1. FEDCBA98 is copied out of the stack and put back into d1. The SP
is then moved forwards to offset 0FFFFE.
The
next instruction loads a word from the stack at offset 0FFFFE back into d0. 0020
is copied out of the stack and put back into d0. The SP is moved forwards to the
offset 100000 once again.
Summary
It is
probably worth mentioning that when you load data out of the stack; you’re only
“copying” the data out. This means that
even though the address of the SP moves forwards, a copy of the data is still
technically inside the stack:
|
Offset |
0 |
1 |
2 |
3 |
4 |
5 |
6 |
7 |
8 |
9 |
A |
B |
C |
D |
E |
F |
|
000FFFD0 |
00 |
00 |
00 |
00 |
00 |
00 |
00 |
00 |
00 |
00 |
00 |
00 |
00 |
00 |
00 |
00 |
|
000FFFE0 |
00 |
00 |
00 |
00 |
00 |
00 |
00 |
00 |
00 |
00 |
00 |
00 |
00 |
00 |
00 |
00 |
|
000FFFF0 |
00 |
00 |
00 |
00 |
00 |
00 |
00 |
00 |
00 |
00 |
FE |
DC |
BA |
98 |
00 |
20 |
|
00100000 |
00 |
00 |
00 |
00 |
00 |
00 |
00 |
00 |
00 |
00 |
00 |
00 |
00 |
00 |
00 |
00 |
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etc |
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This
data will be overwritten when you (or the 68k) stores
something new into the stack later on.
Speaking
of the 68k storing information, the next part will teach an instruction that
causes the 68k to use the stack. This
will give you a good example of why the stack exists.