Stack-Interrupts

Stack in LC3 &

Interrupt Processing

(Chapters 10)

Subroutines in LC3

• we covered TRAP routines

• System calls to process I/O (or other system tasks)

• Written by system, called by user

ØResides as part of system code

• Steps: Call, Process, Return

• Subroutines – i.e., functions

• Written by user

• Called by user program

• Steps: Call, Process, Return

Finally. . last concept at Machine Level. . The

Stack: An abstract data type (ADT)

• An important abstraction you will encounter in many

applications

• Abstract Data Type

• Defined by behavior not implementation

• Stack

• Last in/First Out

• Many ways to implement

• Many uses in CS

Ø Interrupt drive I/O, Saving state of Processor, function calls, etc.

• Push/Pop

Stack: An Abstract Data Type

•An important abstraction that you will encounter

in many applications.

•Example: three uses --

•Interrupt-Driven I/O

• The rest of the story…

•Evaluating arithmetic expressions

• Store intermediate results on stack instead of in registers

Ø Do this in HW 5

•Data type conversion

• 2’s comp binary to ASCII strings

• You will read this on your own and use it in HW5 and Project 2

Stacks

•Simplest form of storage ?

•A LIFO (last-in first-out) storage structure.

• The first thing you put in is the last thing you take out.

• The last thing you put in is the first thing you take out.

•This means of access is what defines a stack,

not the specific implementation.

•Two main operations:

PUSH: add an item to the stack

• POP: remove an item from the stack

A Physical Stack

•Coin rest in the arm of an automobile

•First quarter out is the last quarter in.

1995 1996

1998

1982

1995

1998

1982

1995

Initial State After

One Push

After Three

More Pushes

After

One Pop

A Hardware Implementation

•Data items move between registers to memory

•

/ / / / / /

YesEmpty:

TOP

#18

/ / / / / /

NoEmpty:

TOP

#12

#31

#18

/ / / / / /

NoEmpty:

TOP

#31

#18

/ / / / / /

NoEmpty:

TOP

Initial State After

One Push

After Three

More Pushes

After

Two Pops

Problems….

• Implement a stack as an array.

• Push will involve moving each element over 1:

• a(n+1) = a(n)

• a(0) = new_element

• Etc.

• So what’s the solution….

A Software Implementation

•Data items don't move in memory, just our idea about where

the TOP of the stack is.

• Note: stack “grows” towards lower numbered addresses

/ / / / / /

/ / / / / / TOP

/ / / / / /

#18

/ / / / / /

TOP

#12

#31

#18

/ / / / / /

TOP

#12

#31

#18

/ / / / / /

TOP

Initial State After

One Push

After Three

More Pushes

After

Two Pops

x4000 x3FFF x3FFC x3FFE

R6 R6 R6 R6

By convention, in LC3 R6 holds the Top of Stack (TOS) pointer.

every ISA specifies such a register

Memory addresses increase ‘downward’; bottom of stack has highest address

Question…State of stack

• R6 points to top of stack

•What happens after a push ?

•What happens after a pop ?

Basic Push and Pop Code

•For our implementation, stack grows downward

in address space

• (when item added, TOS moves closer to 0)

•Push: value in register R0 into the stack

• ADD R6, R6, #-1 ; decrement stack ptr

STR R0, R6, #0 ; store data (R0)

•Pop: value at top of stack into register R0

• LDR R0, R6, #0 ; load data from TOS

ADD R6, R6, #1 ; decrement stack ptr

Anything missing…?

•Detect underflow

• Trying to pop from empty stack

•Detect overflow

• Trying to push into a “full” stack

•Need to send a “signal” back to calling program indicating

success or failure of pop or push operations

Pop subroutine with Underflow Detection

•If we try to pop too many items,an underflow condition occurs.

• Check for underflow by checking TOS before removing data.

• Return status code in R5 (0 for success, 1 for underflow)

• Return value at TOS into R0

•POP LD R1, EMPTY ; EMPTY = -x4000

ADD R2, R6, R1 ; Compare stack pointer

BRz FAIL ;

LDR R0, R6, #0

ADD R6, R6, #1

AND R5, R5, #0 ; SUCCESS: R5 = 0

RET

FAIL AND R5, R5, #0 ; FAIL: R5 = 1

ADD R5, R5, #1

RET

EMPTY .FILL xC000

Push Subroutine with Overflow Detection

•If we try to push too many items,an overflow condition occurs.

• Check for underflow by checking TOS before adding data.

• Return status code in R5 (0 for success, 1 for overflow)

• Push value in R0 onto the stack

•PUSH LD R1, MAX ; MAX = -x3FFB

ADD R2, R6, R1 ; Compare stack pointer

BRz FAIL ;

ADD R6, R6, #-1

STR R0, R6, #0

AND R5, R5, #0 ; SUCCESS: R5 = 0

RET

FAIL AND R5, R5, #0 ; FAIL: R5 = 1

ADD R5, R5, #1

RET

MAX .FILL xC005

More Stack Use examples

•Simple calculator program

• Also illustrates use of subroutines

•Converting ASCII to Binary

• System always reads ASCII characters from terminal/input

• System operates on binary numbers

• Who does the conversion ?

ØNeed a subroutine

•Interrupts

• Interrupt driven I/O

Will need to understand stack concept and use it later to

implement high level language program execution

HW: Calculator using Stack

•Simulate a calculator using LC3

•You MUST read the Example in Chapter 10

• You will be required to implement it in Team homework

• various portions of the source code have been provided, but you have

to put them together and get them to work

ØThere are some integration issues you need to worry about – i.e.,

ignoring them will lead to bugs

• Source code can help with your Project 4 (and Homework 5)

ØYou can use any of the code provided

ØAvailable in electronic reserves on Blackboard

Arithmetic Using a Stack

•Instead of registers, some ISA's used a stack for

source and destination operations: a zero-address machine.

• Example:

ADD instruction pops two numbers from the stack,

adds them, and pushes the result to the stack.

•Evaluating (A+B)·(C+D) using a stack:

•(A,B,+)(C,D,+)*

• (1) push A

(2) push B

(3) ADD

(4) push C

(5) push D

(6) ADD

(7) MULTIPLY

(8) pop result

Why use a stack?

• Limited registers.

• Convenient calling convention

for subroutines.

• Algorithm naturally expressed

using FIFO data structure.

Historical note: HP calculators used reverse polish notation…similar to this

Example: So how does Add operation (OpAdd)

work on this calculator ?

•POP two values, ADD, then PUSH result.

START

POP POP

OK? OK?

ADD

Range

OK?

PUSH

RETURN

Put back bothPut back first

Yes

No No

Yes

Example: OpAdd

•

OpAdd JSR POP ; Get first operand.

ADD R5,R5,#0 ; Check for POP success.

BRp Exit ; If error, bail.

ADD R1,R0,#0 ; Make room for second.

JSR POP ; Get second operand.

ADD R5,R5,#0 ; Check for POP success.

BRp Restore1 ; If err, restore & bail.

ADD R0,R0,R1 ; Compute sum.

JSR RangeCheck ; Check size.

BRp Restore2 ; If err, restore & bail.

JSR PUSH ; Push sum onto stack.

RET

Restore2 ADD R6,R6,#-1 ; Decr stack ptr (undo POP)

Restore1 ADD R6,R6,#-1 ; Decr stack ptr

Exit RET

Data Type Conversion

• LC3 keyboard input routines read ASCII characters,

not binary values.

•Similarly, LC3 output routines write ASCII.

•You’ve seen this program:

• IN ; input from keybd into R0

ADD R1, R0, #0 ; move to R1

IN ; read next input from keybd

ADD R2, R0, #0 ; move to R2 add two inputs

ADD R0, R1,R2 ; add the two inputs

OUT ; print result

•What happened ?

• Why? ASCII '2' (x32) + ASCII '3' (x33) = ASCII 'e' (x65)

ASCII to Binary

•Useful to deal with mult-digit decimal numbers

•Assume we've read three ASCII digits (e.g., "259")

into a memory buffer.

•How do we convert this to a number

we can use?

• Convert first character to digit (subtract x30)

and multiply by 100.

• Convert second character to digit and

multiply by 10.

• Convert third character to digit.

• Add the three digits together.

x32

x35

x39

'2'

'5'

'9'

Curious ? Look up atoi function in C…

Multiplication via a Lookup Table

•How can we multiply a number by 100?

• One approach:

Add number to itself 100 times.

• Another approach:

Add 100 to itself <number> times. (Better if number < 100.)

•Since we have a small range of numbers (0-9),

use number as an index into a lookup table.

• Entry 0: 0 x 100 = 0

Entry 1: 1 x 100 = 100

Entry 2: 2 x 100 = 200

Entry 3: 3 x 100 = 300

etc.

Binary to ASCII Conversion

•Converting a 2's complement binary value to

a three-digit decimal number

• Resulting characters can be output using OUT

•Instead of multiplying, we need to divide by 100

to get hundreds digit.

• Why wouldn't we use a lookup table for this problem?

• Subtract 100 repeatedly from number to divide.

•First, check whether number is negative.

• Write sign character (+ or -) to buffer and make positive.

More Stack uses…

•Interrupt processing

•Will need to understand stack concept and use it later to

implement high level language program execution

Recall: Interrupt-Driven I/O

• Two ways to process I/O

• Polling: CPU waits for input device

• Interrupt driven: device signals when it is ready

• Why?

• Polling consumes a lot of cycles,

especially for rare events – these cycles could be used to do useful

computations.

• Example: Process previous input while waiting for network/disk.

Interrupt Processing

External device can:

(1) Force currently executing program to stop;

(2) Have the processor satisfy the device’s needs; and

(3) Resume the stopped program as if nothing happened.

Interrupt-Driven I/O

•To implement an interrupt mechanism, we need:

• A way for the I/O device to signal the CPU that an

interesting event has occurred.

• A way for the CPU to test whether the interrupt signal is set

and whether its priority is higher than the current program.

•Generating Signal

• Software sets "interrupt enable" bit in device register.

• When ready bit is set and IE bit is set, interrupt is signaled.

KBSR

15 14 0

ready bit

interrupt enable bit

interrupt signal

to processor

Priority

•Every instruction executes at a stated level of urgency.

•LC-3: 8 priority levels (PL0-PL7)

• Example:

Ø email program runs at PL0.

Ø Nuclear power correction program runs at PL6.

• It’s OK for PL6 device to interrupt PL0 program,

but not the other way around.

•Priority encoder selects highest-priority device,

compares to current processor priority level,

and generates interrupt signal if appropriate.

Testing for Interrupt Signal

•CPU looks at signal between STORE and FETCH phases.

•If not set, continues with next instruction.

•If set, transfers control to interrupt service routine.

interrupt

signal?

Transfer to

ISR

YES

More details in Chapter 10.

Full Implementation of LC-3 Memory-Mapped I/O

Because of interrupt enable bits, status registers (KBSR/DSR)

must be written, as well as read.

Interrupt-Driven I/O

•Interrupts processing:

1. External device signals need to be serviced.

2. Processor saves state and starts service routine.

3. When finished, processor restores state and resumes program.

•How do steps (2) and (3) occur,

involves a stack.

Interrupt is an

unscripted subroutine call

triggered by an external event.

Processor State

•What state is needed to completely capture the

state of a running process?

•Processor Status Register

• Privilege [15], Priority Level [10:8], Condition Codes [2:0]

•Program Counter

• Pointer to next instruction to be executed.

•Registers

• All temporary state of the process that’s not stored in memory.

Ø Can be saved by Interrupt processing routine (similar to TRAP routines)

Where to Save Processor State?

•Can’t use registers.

• Programmer doesn’t know when interrupt might occur,

so she can’t prepare by saving critical registers.

• When resuming, need to restore state exactly as it was.

•Memory allocated by service routine?

• Must save state before invoking routine,

so we wouldn’t know where.

• Also, interrupts may be nested –

that is, an interrupt service routine might also get interrupted!

•Use a stack!

• Location of stack “hard-wired”.

• Push state to save, pop to restore.

Supervisor Stack

•A special region of memory used as the stack

for interrupt service routines.

• Initial Supervisor Stack Pointer (SSP) stored in Saved.SSP.

• Another register for storing User Stack Pointer (USP):

Saved.USP.

•Want to use R6 as stack pointer.

• So that our PUSH/POP routines still work.

•When switching from User mode to Supervisor mode (as

result of interrupt), save R6 to Saved.USP.

Invoking the Service Routine – The Details

1. If Priv = 1 (user),

Saved.USP = R6, then R6 = Saved.SSP.

2. Push PSR and PC to Supervisor Stack.

3. Set PSR[15] = 0 (supervisor mode).

4. Set PSR[10:8] = priority of interrupt being serviced.

5. Set PSR[2:0] = 0.

6. Set MAR = x01vv, where vv = 8-bit interrupt vector

provided by interrupting device (e.g., keyboard = x80).

7. Load memory location (M[x01vv]) into MDR.

8. Set PC = MDR; now first instruction of ISR will be fetched.

Note: This all happens between

the STORE RESULT of the last user instruction and

the FETCH of the first ISR instruction.

Returning from Interrupt

• Special instruction – RTI – that restores state.

1. Pop PC from supervisor stack. (PC = M[R6]; R6 = R6 + 1)

2. Pop PSR from supervisor stack. (PSR = M[R6]; R6 = R6 + 1)

3. If PSR[15] = 1, R6 = Saved.USP.

(If going back to user mode, need to restore User Stack Pointer.)

• RTI is a privileged instruction.

• Can only be executed in Supervisor Mode.

• If executed in User Mode, causes an exception.

(More about that later.)

Example (1)

/ / / / / /

x3006

Program A

ADD

x3006

Executing ADD at location x3006 when Device B interrupts.

Saved.SSP

Example (2)

/ / / / / /

x3007

PSR for A

/ / / / / /

x6200

Program A

ADD

x3006

Saved.USP = R6. R6 = Saved.SSP.

Push PSR and PC onto stack, then transfer to

Device B service routine (at x6200).

x6200

ISR for

Device B

x6210

RTI

Example (3)

/ / / / / /

x3007

PSR for A

/ / / / / /

x6203

Program A

ADD

x3006

Executing AND at x6202 when Device C interrupts.

x6200

ISR for

Device B

AND

x6202

x6210

RTI

Example (4)

/ / / / / /

x3007

PSR for A

x6203

PSR for B

x6300

Program A

ADD

x3006

x6200

ISR for

Device B

AND

x6202

ISR for

Device C

Push PSR and PC onto stack, then transfer to

Device C service routine (at x6300).

x6300

x6315

RTI

x6210

RTI

Example (5)

/ / / / / /

x3007

PSR for A

x6203

PSR for B

x6203

Program A

ADD

x3006

x6200

ISR for

Device B

AND

x6202

ISR for

Device C

Execute RTI at x6315; pop PC and PSR from stack.

x6300

x6315

RTI

x6210

RTI

Example (6)

/ / / / / /

x3007

PSR for A

x6203

PSR for B

x3007

Program A

ADD

x3006

x6200

ISR for

Device B

AND

x6202

ISR for

Device C

Execute RTI at x6210; pop PSR and PC from stack.

Restore R6. Continue Program A as if nothing happened.

x6300

x6315

RTI

x6210

RTI

Saved.SSP

Exception: Internal Interrupt

•When something unexpected happens

inside the processor, it may cause an exception.

•Examples:

• Privileged operation (e.g., RTI in user mode)

• Executing an illegal opcode

• Divide by zero

• Accessing an illegal address (e.g., protected system memory)

•Handled just like an interrupt

• Vector is determined internally by type of exception

• Priority is the same as running program

What Next ?

Problem Transformation

- levels of abstraction

Natural Language

Algorithm

Program

Machine Architecture

Devices

Micro-architecture

Logic Circuits

Interaction and

Translation from

High level to ISA

How do user programs

get executed?

What to study next..

•How are high level programs executed on the processor ?

• Review some aspects of C

ØHow are operators implemented?

• Function calls and recursion

ØHow does the system support this?

– Any guesses?

• Memory

ØConcept of virtual memory

•Performance

• How “efficient is your program” ?