ECE 3430

1
ECE 3430 Introduction to Microcomputer
SystemsUniversity of Colorado at Colorado Springs

• Lecture 11
• Agenda Today
• More on condition testing and branching
• More Bit-Wise Instructions and Some Others
• The Stack (Push, Pop) and the Stack Pointer

2
Condition Testing and Branching

• In addition to CMP and TST instructions, you can
  use BIT to test
• certain bits. The source and destination are
  ANDed together
• and the N, V, C, and Z flags updated in the
  status register. The
• Destination is not modified!
• Test bits (accompanies CMP and TST instructions)
• BIT ltsrcgt,ltdstgt
• When JMP is out-of-range, use BR as an
  alternative.
• Unlimited, unconditional jump
• BR ltdstgt

3
More Instructions

• Decrement and Increment- Often used in loops to
  adjust counters- These add or subtract 1 or 2
  from the destination (effective address)
• DEC ltdstgt x x 1 INC ltdstgt x x 1
  DECD ltdstgt x x 2 INCD ltdstgt x x 2
• Ex) COUNT EQU 10FLASH EQU 0xF800 ORG FLASH
• mov.b COUNT,R4 initialize loop counter
• LOOP ltloop bodygt
• dec.b R4 loop 10 times
  jne LOOP
• DONE jmp DONE
• END

4
More Instructions

• Arithmetic Shift- used to shift bits- note the
  characteristics of the end bits
• - preserves sign when shifting rightLEFT
• RLA ltdstgt Note LSb is filled with
  0, MSb shift into carry RIGHT
• RRA ltdstgt Note MSb is shifted into
  itself LSb is shifted into carry

C
N-1 ? 0
?
?
0
N-1? 0
?
?
C
5
More Instructions

• Rotate- used to rotate bits- note the
  characteristics of the end bits, different from
  arithmetic shiftsLEFT
• RLC ltdstgt Note full loop, MSb goes to
  carry RIGHT
• RRC ltdstgt Note full loop, carry goes
  to MSb

N-1 ? 0
?
?
C
?
N-1? 0
?
?
C
?
6
More Instructions

• Logical Shift- no explicit instructions in
  MSP430 to do this
• - to go left, use RLA
• - to go right, clear C flag (CLRC) and then use
  RRCLEFT
• Note LSb is filled with 0, MSb
  shift into carry
• RIGHT
• Note 0 is shifted into MSb
  LSb is shifted into carry
  

C
N-1 ? 0
?
?
0
N-1? 0
?
?
C
0
7
More Instructions

• ExampleWhat are the contents of R4 after each
  instruction?mov.b 10110110b, R4 R4 Carry
• rra.b R4 1101 1011 0 rrc.b R4
  0110 1101 1 rrc.b R4 1011 0110
  1What CCR bits are altered and how? N Z
  V C
• RLA
• RLC
• RRA 0 RRC
• set to one or cleared to zero depending on
  run-time circumstances

8
More Instructions

• 1s and 2s compliment
• INV ltdstgt ? 1s compliment
• INV ltdstgt
• ADD 1, ltdstgt ? 2s compliment
• Examplemov.b 11110000b, R4 ? R4 00000000
  11110000b
• inv.w R4 ? R4 11111111 00001111b
• add.w 1, R4 ? R4 11111111 00010000b

9
More Instructions

• Clearing destinationCLR ltdstgt ? zero
  out destination
• Clearing C, N, Z flags
• CLRC ? BIC 1,SR
• CLRN ? BIC 4,SR
• CLRZ ? BIC 2,SR
• Setting C, N, Z flags
• SETC ? BIS 1,SR
• SETN ? BIS 4,SR
• SETZ ? BIS 2,SR

10
More Instructions

• Swap bytes (little to big endian or vice versa)
• -gt destination must be 16-bit
• SWPB ltdstgt
• Sign extend byte to word (8-bit to 16-bit signed
  cast)
• -gt must be room for extension to 16-bits
• SXT ltdstgt
• No-operation (waste time, delay 1 cycle)
• NOP

11
The Stack

• - This is just managed RAM. The stack can live
  anywhere in RAM.
• - This is a last in, first out (LIFO) data
  structure.
• - This is a first in, last out (FILO) data
  structure.
• - Items can only be added or removed from the top
  of the stack.
• - Conceptually like a stack of cafeteria trays or
  a PEZ dispenser.
• PUSH inserting something onto the TOP of the
  STACK
• POP removing something from the TOP of the
  STACKEx) PUSH 0x0011, 0x0022, 0x0033 POP,
  we will receive 0x0033, 0x0022, 0x0011 in that
  order

TOP
Last In
This structure is useful because the order of
data is inherently kept. We dontneed to worry
about setting up dedicated memory for each data
item. This is also a necessary structure for
subroutines to work.
First In
BOTTOM
TOP
0x0033
0x0022
0x0011
12
The Stack
The size of the arguments pushed to the stack can
vary. In the MSP430, push operations can be 8 or
16-bit. Regardless, a full 16-bit value is
allocated on the stack! In other words, the
stack pointer is always even! All MSP430 pop
operations deallocate 16-bits of data into the
destination.
13
The Stack

• What the stack really is- A section of memory
  with an address pointer (stack pointer SP in
  CPU).- The SP contains the address of the top
  element of the stack.- In the MSP430, the SP is
  pre-decremented as information is PUSHED.- In
  the MSP430, the SP is post-incremented as
  information is POPPED.- We define where to
  place the stack data structure (using mov
  instruction to initialize SP).- The stack is
  variable in size and only limited by RAM
  availability.- The standard is to place the
  STACK at the end of RAM (for our MSP430,
  0x0280).
• - The MSP430 provides instructions to manipulate
  the stack.- How do we initialize the stack?
  mov.w 0280h,SP

RAM
Keep global variables close to the beginning of
RAM
0x0200 0x027F
Initialize stack at the end of RAM
0x0280
ltno RAM heregt
14
The Stack

• Stack Overflow- If we push too much information
  onto the stack, it may start overriding data
  stored in pre-defined variables.
• - Since there is no operating system running on
  our MSP430, so you have to
• be the memory manager!
• MSP430 Stack Instructions PUSH (.b or .w)
  ltsrcgt ? always adds 16-bits (only touches 8
  bits if .b)
• POP ltdstgt ? always removes 16-bits

RAM
variables
0x0200 0x027F
Stack creeps backwards through RAM as data is
pushed onto it
15
The Stack

• ExampleRemember that the SP points to the top
  word on the stack!mov.w 0280h,
  SPmov.b 0AAh, R4mov.w 0BBCCh, R5push.b
  R4push.w R5pop R4 R4 0xBBCCpop R5
  R5 0x??AA NOTE The data in RAM is not
  actually erased!

SP
0x0280
xx
0x027E0x027F
SP
AA
??
0x027C 0x027D0x027E0x027F
SP
CC
BB
AA
??
0x027E0x027F
SP
AA
??
SP
0x0280
xx
16
The Stack

• Dumb Example Program using the STACKSample port
  1 10 times as fast as you can, then sum the
  result (in 2 loops)FLASH EQU 0xF800STACK EQU 0x
  0280RAM EQU 0x0200COUNT EQU 10
  ORG RAMRESULT DS 2 allocate memory in RAM
  to hold result ORG FLASH start code at
  beginning of FLASH
• mov.w STACK,SP initialize stack point
  to end of RAM mov.b COUNT,R5 initialize
  loop counter
• clr.w RESULT initialize
  result to zero
• SAMPLE mov.b P1IN,R4 sample data on port 1
  into R4 push R4 store the sample on the
  stack dec R5 decrement loop counter
  jne SAMPLE perform this task 10 times
  continued ?

17
The Stack

• mov.b COUNT,R5 reinitialize the
  loop counter
• SUM pop R4 bring in the
  information off the top of the stack
  add.w R4,RESULT continually sum this
  with RESULT
• dec R5 decrement loop counter jne SUM
  do this 10 timesDONE jmp DONE
• END When the program finishes, the sum of
  the 10 samples is in RESULT.What is the value
  of SP after the sample loop completes?What is
  the value of SP after the sum loop
  completes? Did stack overflow occur? How
  could it have occurred?

18
The Stack

• mov.b COUNT,R5 reinitialize the
  loop counter
• SUM pop R4 bring in the
  information off the top of the stack
  add.w R4,RESULT continually sum this
  with RESULT
• dec R5 decrement loop counter jne SUM
  do this 10 timesDONE jmp DONE
• END When the program finishes, the sum of
  the 10 samples is in RESULT.What is the value
  of SP after the sample loop completes? 0x026CW
  hat is the value of SP after the sum loop
  completes? 0x0280Did stack overflow
  occur? NOHow could it have occurred? Push
  more than 63 words on the stack, the 64th
  item would have overwritten RESULT