ece 265 – lecture 6 the m68hc11 basic instruction set basic arithmetic instructions 8/19/2015 1...

ECE 265 – LECTURE 6

The M68HC11 Basic Instruction Set

Basic Arithmetic Instructions

04/19/23

1ECE265

Joanne E. DeGroat, OSU

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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ECE265


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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Add instructions

Add instructions – note the addressing mode

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Add operations

Operation: 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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The overflow flag

What exactly is overflow? Does that mean you exceeded the binary range represent-able?

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Carry 0 1 A 0110 1100 108 +B 0100 0001 65 C=0 V=1 1010 1101 173 and does not represent 83 as it would in 2’s complement Carry 1 1 A 1110 0000 224 (-32) +B 0100 0000 64 C=1 V=1 0010 0000 32 2’s complement coming Carry 1 0 A 1011 1111 -65 (note: +65 is 0100 0001) +B 1100 0000 -64 C=1 V=0 0111 1111 127 and not -129


More arithemetic instructions

Decimal adjust, increment and decrement, and Two’s complement.

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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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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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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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Two’s Complement

How do you get the two’s complement? Two simple algorithms – consider 00001001(dec 9)

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Two’s Complement


1. Take the 1’s complement and add 1 Ones complement is 11110110 + 00000001 = 11110111 Which represents -9

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Two’s Complement


1. Take the 1’s complement and add 1 Ones complement is 11110110 + 00000001 = 11110111 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 0000 1110 11110010 (14 and -14)

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Two’s complement numbers

0111 7 0110 6 0101 5 0100 4 0011 3 0010 2 0001 1 0000 0

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1111 -1 1110 -2 1101 -3 1100 -4 1011 -5 1010 -6 1001 -7 1000 -8

For 4-bits


Additional Arithmetic Instr.

A few more instructions, this time for binary subtraction

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Binary subtraction

Consider the following example of 7 – 5 = 2 0111 -0101 0010 Direct and very easy to see

Now throw in a borrow - 6 – 5 = 1 0110 -0101 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) -0101 (and it is the weight when a digit is moved right) 0001

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More binary subtraction

Consider 12 – 3 = 9 1100 -0011 and we again have borrows

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1020 after 1st step of borrow -0011 but we still need a borrow

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1020 after 1st step of borrow -0011 but we still need a borrow

1012 after 2nd step of borrow -0011 1001 which is the binary for 9

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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 =1 2011 1211 - 0101 giving 1110 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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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) - 16 bit operation SBCA (opr), SBCB (opr) – subtract with carry

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Compare Instructions

Instruction to compare two values and set condition codes.

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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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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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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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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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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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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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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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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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Another multiply example

Figure 3.5

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Binary Multiplication

Multiply in the previous example $FF = 1111 1111 255 $14 = 0001 0100 20 11 1111 1100 5100 1111 1111 0000 1 0011 1110 1100 ($13EC)

Is this 5100? 4096 +512+256 +128+64+32 +8+4 4096 +768 +128+96 +12 4864 +140 +96 = 4864 + 236 = 5100

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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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Division Examples

Integer and Fractional examples

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Lecture summary

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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


Assignment

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Problems Chapter 3 page 87 Problem 13 Problem 14 Problem 17

ece 265 – lecture 6 the m68hc11 basic instruction set basic arithmetic instructions 8/19/2015 1...

Documents

s complement 8192015ece265

ece265 slide

ones complement

basic arithmetic instructions

inherent mode 8192015ece265

addressing mode 8192015ece265

decrement description

valid bcd digit