1. The M68HC11 Basic Instruction Set Basic Arithmetic Instructions
ECE265
ECE 265 – Lecture 6
The M68HC11 Basic Instruction Set
Basic Arithmetic Instructions
4/20/2017

2. Lecture Overview The M68HC11 Basic Instruction Set
The basic mathematical instructions
REF: Chapter 3 and the appendix that details the instructions.
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3. Arithmetic Instructions
These instructions are used to add, subtract, compare, increment, decrement, 2’s complement, test, and decimal adjust data.
Both 8-bit and 16-bit operations are possible.
It is possible to write code that allows data of higher precision to be operated upon. This extended precision requires the user to write code to support the higher precision.
Architecture also supports Binary Coded Decimal (BCD) operations.
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4. Add instructions Add instructions – note the addressing mode ECE265
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5. Add operations Operation: Description: 2’s complement binary addition
ABA – add the A and B accumulators – Result in A
ABX, ABY – add B to the specified index register
ADCA,ADCB – add with carry to A/B
ADDA,ADDB – add the contents of memory to A/B
ADC and these set the H CC bit to accommodate BCD arithmetic
ADDD – a 16-bit addition using memory
Description: 2’s complement binary addition
CC effects: N X V C and H as noted
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6. The overflow flag
What exactly is overflow? Does that mean you exceeded the binary range represent-able?
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7. More arithemetic instructions
Decimal adjust, increment and decrement, and
Two’s complement.
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8. Decimal Adjust DAA
Description: Adjusts the result in A to a valid BCD number.
Consider adding the BCD for $99 + $22 in Accum A
Both $99 and $22 are valid BCD numbers.
Adding these in binary gives a result of $BB with no half carry or carry. Also, $B is not a valid BCD digit.
Executing DAA will have the following effect.
For the lsd: BCD of $9 + $2= $B will give a $1 and a half carry.
For the most significant digit you have $9 + $2=$B adjusted $B is $1 + hc = $2 and a final carry out.
So here, accumulator A is adjusted to $21 and both C & H are set
CC effects: N Z and C H are corrected
Forms: DAA and only inherent mode
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9. Increment and Decrement
Description: Instructions that increment or decrement a memory location or either the A or B accumulator (Inherent mode). Note that there is no instruction to increment or decrement the D accumulator. (and C is not affected)
CC effects: N Z V
Forms: INCA INCB INC (opr)
DECA DECB DEC(opt)
Just add 1 or subtract 1 (2’s complement)
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10. Two complement operations
Description: Replace the contents of the location with its two’s complement value
CC effects: N Z V and C
The C bit will be set in all cases except when the contents are $00
Forms: NEGA NEGB NEG (opr)
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11. Two’s Complement How do you get the two’s complement?
Two simple algorithms – consider (dec 9)
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12. Two’s Complement How do you get the two’s complement?
Two simple algorithms – consider (dec 9)
1. Take the 1’s complement and add 1
Ones complement is =
Which represents -9
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13. Two’s Complement How do you get the two’s complement?
Two simple algorithms – consider (dec 9)
1. Take the 1’s complement and add 1
Ones complement is =
Which represents -9
2. Starting with the lsb keep all binary digits until you come to the 1st 1. Keep that 1. Invert all the more significant binary digits.
Easy to see on the above example.
Now consider  (14 and -14)
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14. Two’s complement numbers
For 4-bits
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15. Additional Arithmetic Instr.
A few more instructions, this time for binary subtraction
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16. Binary subtraction Consider the following example of 7 – 5 = 2
0111
-0101
Direct and very easy to see
Now throw in a borrow – 5 = 1
0110
but right off see that we need to subtract 1 from 0
so borrow from the next binary position and get
0102 (and yes 2 is not a binary digit – but this is for illustration)
(and it is the weight when a digit is moved right)
0001
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17. More binary subtraction
Consider 12 – 3 = 9
1100
and we again have borrows
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18. More binary subtraction
Consider 12 – 3 = 9
1100
and we again have borrows
1020 after 1st step of borrow
but we still need a borrow
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19. More binary subtraction
Consider 12 – 3 = 9
1100
and we again have borrows
1020 after 1st step of borrow
but we still need a borrow
1012 after 2nd step of borrow
-0011
1001 which is the binary for 9
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20. How about a 2’s complement result
An example to get a 2’s complement result, i.e., a negative number result.
Consider 3 – 5
0011
- 0101
Borrow in =  1211
giving which is the 2’s
complement for -2
And here would also have a CC carry bit of 1 to indicate a borrow.
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21. The subtract instruction
Description: Subtract the contents of two locations
CC effects: N Z V and C
C is set if contents of 2nd operand is larger
Forms: SBA – subtract accumulator B from A  A
SUBA (opr) – subtract memory location
SUBB (opr) contents from accumulator
SUBD (opr) bit operation
SBCA (opr), SBCB (opr) – subtract with
carry
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22. Compare Instructions
Instruction to compare two values and set condition codes.
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23. Compare Instructions
Description: compare the data at two locations and set the CC bits. The data is not altered. (Compare instructions perform a subtraction to update the bits of the CC register.)
CC effects: N V C Z
Forms: CBA
CPD (opr) CMPA (opr) and CMPB (opr)
Addressing modes: Note the difference in the mnemonic for register compare of D, A, or B to memory.
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24. Compare instruction example
Comparison of a subtract versus a compare. A subtract instruction does alter one of the operands which is the destination.
Note that only the affected
bit of the CCR is shown.
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25. Compare example problem
PROBLEM: What programming steps are needed to compare the data at address $1031 with a set point value of $50 using accumulator A. Note that this address is one of the A/D result registers, i.e., where the result of an A/D conversion is stored.
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26. Compare example problem
PROBLEM: What programming steps are needed to compare the data at address $1031 with a set point value of $50 using accumulator A. Note that this address is one of the A/D result registers, i.e., where the result of an A/D conversion is stored.
LDDA #$50 Load the set point into Reg A
CMPA $1031 Compare A to memory (A-M)
Results will indicate if A=M : Z=1
A>M : N=0 Z=0
A<M : N=1 Z=0
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27. Another example
If the data at address $1031 in the previous example was $45 which bits in the CC register are set or cleared?
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28. Test Instruction The test instructions
Allows testing of values for positive, negative, or zero values.
The instruction subtracts $00 from the location, setting the CC register bits. The contents of the location are not modified.
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29. Multiply and Divide instructions
The processor can perform and 8-bit by 8-bit multiply and two forms of divide. The forms for divide are integer and fractional.
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30. The multiply
Example 3.12 from text: What programming steps are needed to multiply the data at location $D500 with the data at location $D510. The result is to be stored at $D520 and $D521.
LDAA $D500 Load value #1 into A accum
LDAB $D510 Load value #2 into B accum
MUL A x B -> D
STD $D520 Store result (16 bits)
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31. Another multiply example
Figure 3.5
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32. Binary Multiplication
Multiply in the previous example
$FF =
$14 =
($13EC)
Is this 5100?
= = 
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33. The Divide instruction
2 divide instructions
IDIV Binary integer division
Used when D is larger than X
Divide D/X -> X with the remainder going into D
FDIV Binary fractional division
Used when X is larger than D
Divide D/X -> X with the remainder (or continuation of the fractional part going into D
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34. Division Examples
Integer and Fractional examples
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35. Lecture summary Have covered
Cover data transfer instructions in Lecture 5
In this lecture went over the basic arithmetic instructions
Add
Subtract
Increment and Decrement
Testing data
Multiply and Divide
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36. Assignment Problems Chapter 3 page 87 Problem 13 Problem 14 Problem 17
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