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DAY 5. Data Manipulation

novice intermediate advanced expert
On Day 3, you developed an elementary understanding of registers and memory, and how to integrate the two. We will now do something more interesting than merely moving data around and actually operate on it, then finish things off by introducing two very simple, but very useful data stuctures: the array and the structure.

Extension

On Day 3 I said that two's complement integers have a fixed length. There is also a limit, to some extent, on the size of unsigned numbers. What if you want to increase the number of bits, yet want to retain the value? This problem is solved with zero extension and sign extension.

To do a sign extension, consider the numbers 64 and -64. Let's derive these numbers for various bit sizes and see if anything interesting happens.
Size 64 -64
8-bit $40 $C0
16-bit $0040 $FFC0
24-bit $000040 $FFFFC0
32-bit $00000040 $FFFFFFC0
From this, we can deduce that to perform a sign extension, you copy the sign bit into every additional bit, and a zero extension is just a special case of sign extensioning where you consider the sign bit to always be zero (regardless of whether it is or not).

Zero extension on the Z80 is easy: 

    ;Zero extend DE
    LD    D, 0
Sign extension is tougher, you need to decide whether to store $00 or $FF. The instructions to do this haven't been learned yet, and I don't want to introduce them out of their context, so...

Fun With Data

Okay! Let's play around with some data manipulation instructions.

Adding and Subtracting

Just a whole slew of instructions:
INC { reg8 | reg16 | (HL) } Adds one to the operand.
S affected
Z affected
P/V detects overflow
C not affected
DEC { reg8 | reg16 | (HL) } Subtracts one from the operand.
S affected
Z affected
P/V detects overflow
C not affected
ADD A, { reg8 | imm8 | (HL) } Adds to the accumulator.
ADD HL, reg16 Adds to HL.
S affected
Z affected
P/V detects overflow
C affected
SUB { reg8 | imm8 | (HL) } Subtracts from the accumulator.
S affected
Z affected
P/V detects overflow
C affected
SBC HL, reg16 Subtracts reg16 and the carry flag from HL.
S affected
Z affected
P/V detects overflow
C affected
Examples:
Before Instruction After
A = 45 INC A A = 46
DE = 12116 INC DE DE = 12117
B = 19 DEC B B = 18
A = 5
L = 21		 ADD A, L A = 26
A = 95 SUB 90 A = 5
HL = 5516
DE = 1102
CY = 1		 SBC HL, DE HL = 4413
The last instruction incorporates the carry flag into the calculation, which implies that there would be situations where this would be desirable. In fact, these instructions exist for a significant technique, but you won't get to know it for a long time. So why do I bring SBC up now? Because SBC is the only 16-bit subtraction instruction!

One thing that needs to be pointed out about the instructions that allow two 16-bit operands, is that the registers HL and IX are mutually exclusive. What that means is that if the first operand is HL, the second can be any other 16-bit register except IX (and, of course, AF). Similarly for IX. Also, IX can never be an operand for SBC. Anyway, if you're ever confused, just look in the Z80 Instruction Set Reference.

If you want to subtract a constant number x from HL, you should use ADD and load into the other operand the negative of x.

Example: 

    LD     HL, 46243
    LD     BC, -1000
    ADD    HL, BC
; HL now equals 45243

However, if the number is already in a register from a previous calculation, you have to use SBC. This becomes quite a sticky situation, because you might not know what the carry flag's value is, thus giving an erroneous result 50% of the time. The solution is to ensure that the carry is reset before doing the subtraction. How to do that? 

    SCF    ; Force carry = 1
    CCF    ; Flip carry so it is 0
    SBC    HL, BC
This is actually the most idiotic way to force the carry to zero, since it can be done in just one instruction. Problem is, that instruction doesn't just reset the carry flag, and it belongs to a family of instructions that do similar operations, and the whole thing would be just too much and too messy for one day.

Finally, before I forget, what if you wanted to do the above, but with IX? Since SBC won't accept an index register, you must use ADD, and manually negate the second register. 

    LD     A, B
    CPL
    LD     B, A
    LD     A, C
    CPL
    LD     C, A
; We have now found the one's complement of BC so, by definition of
; the two's complement:
    INC    BC
    ADD    IX, BC

Multiplying

If the number you want to multiply happens to be a power of two, then it's a cake walk, because you only need a sequence of ADD instructions.

    LD     HL, 10
    ADD    HL, HL        ; 10 * 21 = 20
    ADD    HL, HL        ; 10 * 22 = 40
    ADD    HL, HL        ; 10 * 23 = 80
    ; et cetera
If the number is not a power of two, but can be expressed as the sum or difference of two powers of two, then its still pretty easy, just a little less efficient.

; Calculate HL * 40 as (HL * 32) + (HL * 8)
    ADD    HL, HL
    ADD    HL, HL
    ADD    HL, HL        ; HL * 8
    LD     D, H
    LD     E, L          ; Save it for later
    ADD    HL, HL
    ADD    HL, HL        ; HL * 32
    ADD    HL, DE        ; HL * 32 + HL * 8

; Calculate HL * 15 as (HL * 16) - (HL * 1)
    LD     D, H
    LD     E, L          ; Save HL * 1 for later
    ADD    HL, HL
    ADD    HL, HL
    ADD    HL, HL
    ADD    HL, HL        ; HL * 16
    SCF
    CCF
    SBC    HL, DE        ; HL * 16 - HL * 1
What if it is an awkward number like 13? In this case, it might be better to follow this general-purpose algorithm:
  1. If the multiplier is even, divide it by 2 and type ADD HL, HL.
  2. If the multiplier is odd, subtract 1 and type ADD HL, DE.
  3. Go to step one until you have one. This time, type each instruction above the preceding one.
  4. Load HL into DE. 
;Calculate HL * 13
    LD     D, H
    LD     E, L
    ADD    HL, HL    ; HL * 2
    ADD    HL, DE    ; HL * 3
    ADD    HL, HL    ; HL * 6
    ADD    HL, HL    ; HL * 12
    ADD    HL, DE    ; HL * 13

Dividing

Dividing is trickier still. The best way to do this is to take a page out of your math text book and multiply by a reciprocal.
So now the question on everyone's minds is how to generate a reciprocal when all you've got are integers. The answer to that, as with everything else in life, is to cheat.

  1. Determine the number you want to divide by (the divisor). Figure out 256 divided by this number and round. This is the number to multiply by.
  2. Put the dividend into HL.
  3. Look in H for the quotient.
Example: 127 ÷ 52 (expect 2). 
    LD    HL, 127
    LD    D, H
    LD    E, L

; 256 ÷ 52 = 5, find 127 × 5
    ADD    HL, HL    ; HL = $00FE
    ADD    HL, HL    ; HL = $01FC
    ADD    HL, DE    ; HL = $027B
Please note that this this method gives only a very rough approximation for the quotient. Later on, I will show you a way to divide a number perfectly, and even get the remainder!

Overflow

When a register's value is increased beyond the largest value it can contain, it's value will start over at the smallest value and continue incrementing: 
    LD     A, 203
    ADD    A, 119
If we add 119 to 203, we would get 322, but this does not fit in eight bits, so we have to wrap around. If we look at the binary value of 322, which is %101000010, then eliminating all but the rightmost eight bits will give us the value A will hold. The end result is that A holds 66, but the carry flag is set to hold that ninth bit of the result. This affect applies equally if we consider A to be signed (in this case, the largest and smallest possible values are 127 and -128). There is a similar phenomenon when subtracting.

Registers and RAM

Suppose you type out an instruction like 
    LD    HL, $D361
Which puts $D361 into HL. No big suprise there, but since 16-bit registers are just two 8-bit registers taken together, what happens to H and L?
[H <- D3] [L <- 61]Two hex digits mean one byte, so $D3 is one byte and $61 is one byte. Since $D3 and H are first from the left, it makes sense that $D3 is stored into H. Similarily, $61 would be stored into L.

Now take this instruction 

    LD    ($2315), HL
Since H comes before L, You'd figure that register H would be stored in byte $2315 and L would be stored into byte $2316. I mean, it just makes sense, right?
$2314 $2315 $2316 $2317 $2318 $2319
  D3 61      
  Λ
H 		Λ
L 		×
Wrong. Because the Z80 is what's called a "little-endian processor", when you store HL to memory, the number gets "twisted around": The byte in register L is loaded first, then the byte in H (the number is stored "little-end" first). When you store from RAM to HL, the first byte goes into L and the next byte goes into H.
$2314 $2315 $2316 $2317 $2318 $2319
  61 D3      
  Λ
L 		Λ
H 		=)
You should stop and think about these points until it's second-nature to you. It is amazing how many people get confused about little-endian. It is very important to remember when dealing with memory.

Arrays

An array is an collection of data structures, each exactly the same. The data structure you use can be as simple as a single byte or as complex as a set of records, as long as you are consistent with regards to handling the array. Each of these data structures is otherwise unique and is referred to as an array element, distinguished from each other with an index, which will range from zero to some number. The elements should be contiguous in memory (though this is by no means required, just more efficient).

1-D Arrays

A one-dimensional array can be though of as a list of elements. To access an element, you need a function to convert an element's index into that element's address:
element_address = base_address + (index × element_size)
Where base_address is the address of the array's first element, index is the element you want to get (starting from zero), and element_size is the size (in bytes) of each element.

Example, using this array, and considering each element to be an 8-bit number...

Base Address: $8000
Element Size: One Byte
Address $8000 $8001 $8002 $8003 $8004 $8005 $8006 $8007 $8008 $8009 $800A $800B
Element No 0 1 2 3 4 5 6 7 8 9 10 11
Value 232 37 131 103 187 11 86 254 51 204 243 56
...access the 5th element and store it into C: 

array_base      .EQU    $8000
element_size    .EQU    3
    LD     A, (array_base+(4*element_size))
    LD     C, A
If the index is in a register, you have a bit more work to do. 
    LD     A, 3              ; Put index in A
    LD     B, A              ; Multiply by element size
    ADD    A, A
    ADD    A, B

    LD     D, 0
    LD     E, A              ; Put A in DE

    LD     HL, array_base    ; Add index to base
    ADD    HL, DE

    LD     C, (HL)

2-D Arrays

Whereas a 1-D array can be thought of as a list, a 2-D array is probably best thought of as a matrix or table. Instead of one index, you have two, which, for the sake of comprehension, are called the row index and the column index. However, the array's elements are still stored sequentially in memory. Which brings up an important question: how do you represent an n×m array? Well, you have your choice of two options.

With row-major ordering, you fill up each row from left to right, then move down to the next row when you have exhausted a row.

Column
0 1 2 3
R
o
w		 0 232 37 131 103
1 187 11 86 254
2 51 204 243 56
Or you could have column-major ordering, where each column is filled up top to bottom before moving to the next one:
Column
0 1 2 3
R
o
w		 0 232 103 86 204
1 37 187 234 243
2 131 11 51 56
As you might've guessed, you need a different function to match an index to an array.

Row major:
address = base_address + (col_index × row_size + row_index) × element_size
Column major:
address = base_address + (row_index × col_size + col_index) × element_size
Example, let's say we have a 4×6 row-major array of words with a base address of $8000, and we have row index in B and a column index in C. What we would like to do is get the indexed element into HL. 
array_base    .EQU    $8000
row_size      .EQU    4
col_size      .EQU    3

    LD     HL, array_base
    LD     A, C       ; Multiply by row size
    ADD    A, A
    ADD    A, A
    ADD    A, B       ; Add in row index
    LD     D, 0
    LD     E, A
    ADD    HL, DE

    LD     A, (HL)
    INC    HL
    LD     H, (HL)
    LD     L, A

Structures

Whereas an array's elements are all the same type, a structure's elements can vary. The whole purpose of a structure is to encapsulate data that is different but logically connected. If you were managing a CD database, you might use this hypothetical example: 
struct CD {
    byte  title[32];   // Name of the CD
    byte  band[32];    // The guys what made it
    word  release;     // Year of release
    byte  tracks;      // Number of songs
    word  length;      // Total disc length in seconds
    byte  rating;      // How am I reflecting upon having thrown
}                      // my hard-earned cash at the RIAA today? (/10)
The structure's elements are allocated one after another in memory, just like an array is. To access an element of the structure, you need to know the offset from the beginning of the structure to the first byte of that element. Continuing with the example, we might define some manifest constants to help us: 
CD.title    .EQU  0
CD.band     .EQU  32
CD.release  .EQU  64
CD.tracks   .EQU  66
CD.length   .EQU  67
CD.rating   .EQU  69
These equates will help enormously in maintaining readability. To access an element, you can put the structure base address into HL, then add the offset. Alternatively, you might use IX and use the equated displacement. Slow, but easy to follow.

Example, given this instance of our CD: 

    CD.title   =  "P�u�l�s�e"
    CD.band    =  "Pink Floyd"
    CD.release =  1995
    CD.tracks  =  23
    CD.length  =  8863
    CD.rating  =  10   // Watch the video, it ownz.
And say we wanted to set the length element to its proper value: 
    LD     HL, disc01 + length
    LD     (HL), $9F
    INC    HL
    LD     (HL), $22

    LD     IX, disc01
    LD     (IX + CD.length), $9F
    LD     (IX + CD.length + 1), $22

disc01:
.DB   "Pulse"
.DB   "Pink Floyd"
.DW   1995
.DB   23
.DW   6502
.DB   10

Arrays of Structures

Oh, to be sure, you can have an array of structures. I mean, a database would be pretty useless if all you keep track of was one measly CD. To access a structure element, just index the array and go for it. E.g. Suppose we have an array for 4 sprites in a game located at AppBackupScreen, and each element has this structure: 
struct sprite {
    byte  x;      // x-position
    byte  y;      // y-position
    byte  dx;     // delta-x each frame
    byte  dy;     // delta-y each frame
    byte  hp;     // hit points
    byte  frame;  // animation frame
}
And suppose we wanted to add the dx byte to the x byte, and the dy byte to the y byte of each element. This could be done 
x       .EQU  0
y       .EQU  1
dx      .EQU  2
dy      .EQU  3
hp      .EQU  4
frame   .EQU  5
sizeof  .EQU  6     ; Size of each element

    LD    IX, AppBackupScreen     ; Get array base
    LD    DE, sizeof              ; Use this to update IX

    LD    A, (IX + x)
    ADD   A, (IX + dx)
    LD    (IX + x), A
    ADD   IX, DE

    LD    A, (IX + x)
    ADD   A, (IX + dx)
    LD    (IX + x), A
    ADD   IX, DE

    LD    A, (IX + x)
    ADD   A, (IX + dx)
    LD    (IX + x), A
    ADD   IX, DE

    LD    A, (IX + x)
    ADD   A, (IX + dx)
    LD    (IX + x), A
PREVIOUS - Day 4: Flags

NEXT - Day 6: Stacks

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This is part of Learn TI-83 Plus Assembly In 28 Days
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