ECE 265 – Lecture 8

ECE 265 – LECTURE 8

The M68HC11 Basic Instruction Set

The remaining instructions

04/22/23

1ECE265

Joanne E. DeGroat, OSU

Lecture Overview

The M68HC11 Basic Instruction Set Stack and index register instructions The remaining instructions classes are for program flow

control Branch Instructions Jump Instructions Subroutine Calls No Operation Stop

REF: Chapter 3 and the appendix that details the instructions.

04/22/23

2

ECE265


Stack &Index register instructions

These instructions allow manipulation of the index registers, X and Y, and exchange with the D accumulator. Additionally, they provide the ability to push and pop X and Y to/from the stack.

This class of instructions can be further subdivided into sub-classes of the type of operation.

04/22/23ECE265

3


Modify value index reg or SP

To modify the value of the index register Add B accumulator to X or Y: ABX ABY Decrement the register: DES (stack pointer) DEX DEY Increment the register: INS (stack pointer) INX INY

INX and INY (DEX,DEY) are only ones that affect CC register and then only the Z bit.

04/22/23ECE265

4


Data manipulation of contents

Add the B accumulator to and index register ABX ABY Does not affect the CC register

Data testing of contents of index register CPX CPY There are compares to a 16-bit value in memory

04/22/23ECE265

5


Move the X and Y index registers

Push them on the stack – copy the value in the register to the top of the stack. The register stays the same value. These are 2 byte push and pop operations. PSHX PSHY

Pull the top 2 bytes from the stack into X or Y PULX PULY

Store the X or Y index register or the SP STX STY STS

Load the X or Y index register or the SP LDX LDY LDS

04/22/23ECE265

6


Register interchange instructions

Exchanges D with X: XGDX D with Y: XGDY

Transfers Stack pointer to X or Y: TSX TSY X or Y to the Stack Pointer: TXS TYS On a transfer what is the value of the register

transferred from after the operation? What values is transfrred?

04/22/23ECE265

7


Register interchange instructions

Values before execution SP -- $C100 X -- $000C

Execute a TSX instruction Values now

SP -- $C100 X -- $C101

X is loaded with 1 plus the contents of the stack pointer so that it points to the last value that was stored on the stack.

04/22/23ECE265

8


Ended here on Fri Jan 27, 12

04/22/23ECE265

9


Program flow control instructions

When a task is sequential simply increment through the steps of tasks will accomplish it.

However, in computer programming for most applications, and especially embedded control applications, the programs typically have actions based on the status of the various inputs or internal counters and timers.

These instructions control what part of the program will execute next. They control the flow of execution and are often referred to as program flow control instructions.

04/22/23ECE265

10


Conditional Branches

These are instructions to the control the flow of execution based on a value in the condition code register.

Their general execution is: If the condition tested is false execution continues with

the next instruction. If the condition tested is true then execution continues

at the relative location indicated in the argument of the instruction.

04/22/23ECE265

11


The effect of a branch

A branch controls the flow of execution

04/22/23ECE265

12


A table of the HC11 branches

The Instructions Note the mode is relative

04/22/23ECE265

13


Relative addressing mode

Relative means the address is related to something. It is related to the current instruction address which

is in the program counter (PC). The instruction argument is the offset to be taken if the condition is TRUE.

The offset can be positive of negative. It is an 8 bit quantity that is treated as a 2’s complement value and allows offsets of -128 to +127.

04/22/23ECE265

14


An example

This is a timing delay loop DECB takes 1usec BNE takes 1.5 usec LDAB immediate mode takes 1us

Coding LDAB #$count DELAY: DECB BNE DELAY

Time delay = 1us+(1us+1.5us)*count

04/22/23ECE265

15


More calculation

Most times you will know to time delay that you need and must calculate what count will create a loop with the correct delay. Time Delay – LDAB time Count = DECB time + BNE time

Taking into account the MCUs clock frequency which then tells us the time for the instructions. Count = (Time Delay – 1us) / (2.5 us)

04/22/23ECE265

16


Simple Branches summary

Branch on Carry Set of Carry Clear BCS BCC

Branch on Zero Equal to zero – BEQ Not Equal to zero – BNE

Branch on Minus – BMI Branch on Plus – BPL Branch on Overflow clear or set

BVC BVS

04/22/23ECE265

17


Unsigned Binary Branches

Branch if higher BHI CMPA $D380 compares A to the contents of $D380 BHI will branch if A > $D380 i.e. A > M

Branch on higher or same BHS BHI is > BHS is a test

Branch if lower BLO < Branch if lower or equal BLS

04/22/23ECE265

18


Signed Branches

These branches treat the data as 2’s complement Branch if greater than BGT Branch is greater than or equal BGE Branch is less than BLT Branch is less than or equal BLE

04/22/23ECE265

19


Jump and Branch always

Sometimes called unconditional branches Transfers control to the location specified in the

argument of the instruction. Jump, JMP, uses direct, extended, or indexed

addressing for the target address Branch always – BRA – us a relative transfer

04/22/23ECE265

20


Subroutines

High level languages and good programming practice use subroutines.

The 68HC11 has two branches to subroutine and one return from subroutine The BSR is a relative branch to subroutine

04/22/23ECE265

21


The effect of the JSR,BSR & RTS

Instruction and their effect

JSR shows the effect of the addressing mode

04/22/23ECE265

22


Lecture summary

04/22/23ECE265

23

Have covered the remaining instructions

Now knowing the instruction set, the processor may be used in applications

To do that you need a program. How do you design a program? How do you design a well documented,

understandable and maintainable program?


Assignment

04/22/23ECE265

24

Problems Chapter 3 page 87

Problem 28

ECE 265 – Lecture 8

Documents

Transcript of ECE 265 – Lecture 8