Embedded Systems ARM Computer Architecture

Embedded Systems ARM Computer Architecture including Instruction Set

1. What is “Computer Architecture” ?
INSTRUCTION SET ARCHITECTURE
Operating
System
Processor Architecture I/O System
Digital Design
VLSI Circuit Design
Application
Compiler
L
e
v
e
ls

o
f
A
b
s
t
r
a
c
t
io
n
low
high
Key: Instruction Set Architecture (ISA)
Different levels of abstraction

2. Internal Organisation
Major components of Typical Computer System
Data is mostly stored in the computer memory separate from the
Processor, however registers in the processor datapath can also
store small amounts of data
Processor
Computer
Control
Datapath
Memory Devices:
Input
Output
Processor aka CPU (Central Processing Unit)

3. A Very Simple Processor
Based on von Neumann model
Stored program and data in same
memory
Central Processing Unit (CPU)
contains:
Arithmetic/Logic Unit (ALU)
Control Unit
Registers: fast memory, local to the
CPU
CPU
Memory
I/O
The point of philosophy is to start with something so simple as not to seem worth
stating, and to end with something so paradoxical that no one will believe it."
Bertrand Russell

4. ARM ’s CPU
•ARM ’s CPU
•ALU
•16 General Purpose registers
(R0 to R15)
•PC register (R15)
•Instruction decoder
CPU
PC
ALU
registers
R1
R0
R13 (SP)
R2
…
R14 (LR)
Instruction Register
Instruction decoder
CPSR:ITHSVN CZ
R15 (PC)

5. Main memory organisation
Main memory is used to store programs, data, intermediate
results
Two main organisations: Harvard & von Neumann
Harvard architecture.
In A Harvard architecture CPU programs are stored in a separate
memory (possibly with a different width) from the data memory. This
has the added benefit that instructions can be fetched at the same
time as data, simplifying & speeding up the hardware.
In practice, the convenience of being able to read and write programs
just like normal data makes this less usual
still popular for fixed program microcontrollers.
CPU
Data
Memory
Instruction
Memory

6. Von Neumann memory architecture
Von Neumann architecture
Programs and data occupy a single memory.
Think of main memory as being an array of words, the array
index being the memory address. Each word (array location)
has data which can be separately written or read.
Usually instructions are one word in length – but can be
either more or less
CPU
Data bus
Data &
Instruction
Memory
Address bus
Control bus
memory bus

7. Memory in detail
Memory locations store instructions data and
each have unique numeric addresses
Usually addresses range from 0 up to some
maximum value.
Memory space is the unique range of possible
memory addresses in a computer system
We talk about “the address of a memory
location”.
Each memory location stores a fixed number
of bits of data, normally 8, 16, 32 or 64
We write mem
8[100], mem
16[100] to indicate
the value of the 8 or 16 bits with memory
address 100 etc
0 02E
machine
code
2 02F
1 030
7 000
--
0AA0
0110
0BB0
--
--
000
001
002
003
004
005
006
02E
02F
030
.
.
.

8. Byte addresses for words
Most computer systems now use little-endian byte addressing, in
which the least-significant byte has the lower address.
It is inconvenient to have completely separate byte and word
addresses, so word addressing usually follows byte addressing.
The word address of a word is the byte address of its lowest
numbered byte. This means that consecutive words have addresses
separated by 2 (16 bit words) or 4 (32 bit words) etc.
… …
7 6
5 4
3 2
1 0
8:
6:
4:
2:
0:
Word
address
MSB
Little-endian
LSB
16 bit memory
with consecutive
word addresses
separated by 2
4:
3:
2:
1:
0:
Word
number
Not used

9. Internal Registers & Memory
Internal registers (e.g. A, R0) are
same length as memory word
Word READ:
A := Mem
16
[addr]
Word WRITE:
Mem
16[addr] := A
Byte READ:
A := 00000000 Mem
8
[addr]
Byte WRITE:
Mem
8[addr] := A(7:0) (bottom 8 bits)
16 bits
8 bits8 bits
bottom
8
Top
8
A
Memory
16 bits

10. What are memory locations used for?
Read-write memory (RAM) is used for
data and programs. It loses its contents
on power-down.
Read-only memory (ROM) typically used
to hold programs that do not change
Flash ROM allows data to be changed by
programming (but not by memory write).
Memory-mapped I/O. Some locations
(addresses) in memory allow
communication with peripheral devices.
For example, a memory write to the data
register of a serial communication
controller might output a byte on a serial
port of a PC.
In practice, all I/O in modern systems is
memory-mapped
RAM
ROM
I/O
E007 0000:
0:
7 FFFF:
400 0000:
E000 0000:
LPC2138 microcontroller
On-chip memory map
400 7FFF:
512K
32K
28 X 16K

11. The ARM Instruction Set
Load-Store architecture
Fixed-length (32-bit) instructions
3-operand instruction format (2 source operand regs, 1
result operand reg): ALU operations very powerful (can
include shifts)
Conditional execution of ALL instructions (v. clever
idea!)
Load-Store multiple registers in one instruction
A single-cycle n-bit shift with ALU operation
“Combines the best of RISC with the best of CISC”

12. ARM Programmer’s Model
16 X 32 bit registers
R15 is equal to the PC
Its value is the current PC value
Writing to it causes a branch!
R0-R14 are general purpose
R13, R14 have additional functions,
described later
Current Processor Status Register (CPSR)
Holds condition codes AKA status bits
r0
r1
r2
r3
r4
r5
r6
r7
r8
r9
r10
r11
r12
r13 (stack pointer)
r14 (link register)
r15PC
CVNZ Iunused modeFT
31 29 7 6 5 4 0CPSR

13. ARM Programmer's Model (con't)
CPSR is a special register, it cannot be read or written like
other registers
The result of any data processing instruction can modify status bits (flags)
These flags are read to determine branch conditions etc
Main status bits (AKA condition codes):
N (result was negative)
Z (result was zero)
C (result involved a carry-out)
V (result overflowed as signed number)
Other fields described later

14. ARM's memory organization
Byte addressed memory
Maximum 2
32
bytes of memory
A word = 32-bits, half-word = 16 bits
Words aligned on 4-byte boundaries
NB - Lowest byte
address = LSB of
word
“Little-endian”
Word addresses
follow LSB byte
address
20
16
12
8
4
0

15. CPU
15

16. 1. Levels of representation in
computers
High Level Language
Program
Assembly Language
Program
Machine Language
Program
Control Signal
Specification
Compiler
Assembler
Machine
Interpretation
temp := v[k];
v[k] := v[k+1];
v[k+1] := temp;
lw $15,0($2)
lw $16,4($2)
sw $16,0($2)
sw $15,4($2)
0000 1001 1100 0110 1010 1111 0101 1000
1010 1111 0101 1000 0000 1001 1100 0110
1100 0110 1010 1111 0101 1000 0000 1001
0101 1000 0000 1001 1100 0110 1010 1111

17. Instruction Description
ADD Rd, Rn,Op2 *ADD Rn to Op2 and place the result in Rd
ADC Rd, Rn,Op2 ADD Rn to Op2 with Carry and place the result in Rd
AND Rd, Rn,Op2 AND Rn with Op2 and place the result in Rd
BIC Rd, Rn,Op2 AND Rn with NOT of Op2 and place the result in Rd
CMP Rn,Op2 Compare Rn with Op2 and set the status bits of CPSR**
CMN Rn,Op2 Compare Rn with negative of Op2 and set the status bits
EOR Rd, Rn,Op2 Exclusive OR Rn with Op2 and place the result in Rd
MVN Rd,Op2 Store the negative of Op2 in Rd
MOV Rd,Op2 Move (Copy) Op2 to Rd
ORR Rd, Rn,Op2 OR Rn with Op2 and place the result in Rd
RSB Rd, Rn,Op2 Subtract Rn from Op2 and place the result in Rd
RSC Rd, Rn,Op2 Subtract Rn from Op2 with carry and place the result in Rd
SBC Rd, Rn,Op2 Subtract Op2 from Rn with carry and place the result in Rd
SUB Rd, Rn,Op2 Subtract Op2 from Rn and place the result in Rd
TEQ Rn,Op2 Exclusive-OR Rn with Op2 and set the status bits of CPSR
TST Rn,Op2 AND Rn with Op2 and set the status bits of CPSR
* Op2 can be an immediate 8-bit value #K which can be 0–255 in decimal, (00–FF in hex).
Op2 can also be a register Rm. Rd, Rn and Rm are any of the general purpose registers
** CPSR is discussed later in this chapter

18. ARM Assembly Quick Introduction
MOV ra, rb
MOV ra, #n
ra := rb
ra := n
n decimal in range -128 to 127
(other values possible, see later)
ADD ra, rb, rc
ADD ra, rb, #n
ra := rb + rc
ra := rb + n
SUB => – instead of +
CMP ra, rb
CMP ra, #n
set status bits on ra-rb
set status bits on ra-n
CMP is like SUB but has no destination
register ans sets status bits
B label branch to label BL label is branch & link
BEQ label
BNE label
BMI label
BPL label
branch to label if zero
branch if not zero
branch if negative
branch if zero or plus
Branch conditions apply to the
result of the last instruction to set
status bits
(ADDS/SUBS/MOVS/CMP etc).
LDR ra, label
STR ra, label
ADR ra, label
LDR ra, [rb]
STR ra, [rb]
ra := mem[label]
mem[label] := ra
ra :=address of label
ra := mem[rb]
mem[rb] := ra
LDRB/STRB => byte transfer
Other address modes:
[rb,#n] => mem[rb+n]
[rb,#n]! => mem[rb+n], rb := rb+n
[rb],#n => mem[rb], rb:=rb+n
[rb+ri] => mem[rb+ri]

19. Introduction to ARM data processing
a := b+c-d
ADD R0, R1, R2
SUB R0, R0, R3
a: R0
b: R1
c: R2
d:R3
Machine Instructions:
ADD Rx,Ry,Rz ;Rx := Ry + Rz
SUB Rx,Ry,Rz ;Rx := Ry - Rz
ARM has 16 registers R0-R15
If a,b,c,d are in registers:
LDR R1, B
LDR R2, C
LDR R3, D
ADD R0, R1, R2
SUB R0, R0, R3
STR R0, A
a
b
c
d
mem[A]
mem[B]
mem[C]
mem[D]
LOAD data to reg
from memory
STORE result to
memory from reg

20. AREA Example, CODE ;name a code block
TABSIZEEQU 10 ;defines a numeric constant
X DCW 3 ; X (initialised to 3)
Y DCW 11 ; Y (initialised to 11)
Z % 4 ; 4 bytes (1 word) space for Z, uninitialised
ENTRY ;mark start
LDR r0, X ;load multiplier from mem[X]
LDR r1, Y ;load number to be multiplied from mem[Y]
MOV r2, #0 ;initialise sum
LOOP
ADD R2, R2, R1 ;add Y to sum
SUB r0, r0, #1 ;decrement count
CMP r0, #0 ;compare & set codes on R0
BNE LOOP ;loop back if not finished (R0 ≠ 0)
STR r2, Z ;store product in mem[Z]
END
An ARM assembly module
opcode operands
comments
symbols
module
header
and end

21. CMP instruction & condition codes
CMP R0, #n
computes x = R0 - n
x = 0 <=> Z = 1
z(x) < 0 <=> N = 1
C is carry from addition
V is two's complement overflow
BNE ;branch if Z=0 (x ≠ 0)
BEQ;branch if Z=1 (x = 0)
BMI;branch if N=1 (z(x) < 0)
BPL;branch if N=0 (z(x) ≥ 0)
CMP R0, #0 ; set condition codes
BNE LOOP; branch if Z=0
N
Z
C
V
condition codes
AKA status bits
Negative
Zero
Carry
oVerflow
(signed)
z(x) two complement
interpretation of bits x

22. Some simple instructions
1. MOV (MOVE)
•MOV Rd, #k
• Rd = k
•k is an 8-bit value
•Example:
•MOV R5,#53
•R5 = 53
•MOV R9,#0x27
•R9 = 0x27
•MOV R3,#2_11101100

MOV Rd, Rs

Rd = Rs

Example:

MOV R5,R2

R5 = R2

MOV R9,R7

R9 = R7

23. LDR pseudo-instruction (loading 32-bit values)
•LDR Rd, =k
• Rd = k
•k is an 32-bit value
•Example:
•LDR R5,=5543
•R5 = 5543
•LDR R9,=0x123456
•R9 = 0x123456
•LDR R4,=2_10110110011011001

24. Some simple instructions
2. Arithmetic calculation
•Opcode destination, source1, source2
•Opcodes: ADD, SUB, AND, etc.
•Examples:
•ADD R5,R2,R1
•R5 = R2 + R1
•SUB R5, R9,#23
•R5 = R9 - 23

25. A simple program
•Write a program that calculates 19 + 95
MOV R6, #19 ;R6 = 19
MOV R2, #95 ;R2 = 95
ADD R6, R6, R2 ;R6 = R6 + R2

26. A simple program
•Write a program that calculates 19 + 95 - 5
MOV R1, #19 ;R6 = 19
MOV R2, #95 ;R2 = 95
MOV R3, #5 ;R21 = 5
ADD R6, R1,R2 ;R6 = R1 + R2
SUB R6, R6,R3 ;R6 = R6 - R3
MOV R1, #19 ;R6 = 19
MOV R2, #95 ;R2 = 95
ADD R6, R1,R2 ;R6 = R1 + R2
MOV R2, #5 ;R21 = 5
SUB R6, R6,R2 ;R6 = R6 - R2

27. What is subtraction in binary?
In a microprocessor
Subtract generates correct two's complement
answer for two's complement operands.
Subtract = negate followed by add: a - b = a + (-b)
Example: 4 - 1
two's comp negate is
invert bits & add 1:
0001 => 1110=> 1111
0100
0001 -
0100
1111 +
10011
No overflow because:
c
n=1
c
n-1=1

28. Status Register (CPSR)
CPSR:
Thumb
InterruptoVerflow
carryZero
Negative
Example: Show the status of the C and Z flags after the addition of
0x38 and 0x2F in the following instructions:
MOV R6, #0x38 ;R6 = 0x38
MOVR7, #0x2F ;R17 = 0x2F
ADDSR6, R6,R7 ;add R7 to R6
Solution:
38 00000000 00000000 00000000 0011 1000
+ 2F 00000000 00000000 00000000 0010 1111
67 00000000 00000000 00000000 01100111
R6 = 0x67
C = 0 because there is no carry beyond the D31 bit.
Z = 0 because the R6 (the result) has a value other than 0 after the addition.
Example: Show the status of the C and Z flags after the addition of
0x0000009C and 0xFFFFFF64 in the following instructions:
LDR R0,=0x9C
LDR R1,=0xFFFFFF64
ADDS R0,R0,R1 ;add R1 to R0
Solution:
0000009C 00000000 00000000 00000000 10011100
+ FFFFFF64 11111111 11111111 11111111 01100100
1 00000000 1 00000000 00000000 00000000 00000000
R0 = 00000000
C = 1 because there is a carry beyond the D7 bit.
Z = 1 because R0 (the result) has a value 0 in it after the addition.
Example: Show the status of the Z flag after the subtraction of 0x9C
from 0x9C in the following instructions:
LDR R0,=0x9C
LDR R1,=0x9C
SUBS R0,R0,R1 ;subtract R21 from R20
Solution:
9C 1001 1100
- 9C 1001 1100
00 0000 0000 R0 = $00
Z = 1 because the R20 is zero after the subtraction.
C = 1 because R21 is not bigger than R20 and there is no borrow from D32 bit.
M0M1M2M3M4TFIReservedVCZN
D28D29D30D31 D0D1D2D3D4D5D6D7
……….
Example: Show the status of the Z flag after the subtraction of 0x73
from 0x52 in the following instructions:
LDR R0,=0x52
LDR R1,=0x73
SUBS R0,R0,R1 ;subtract R1 from R0
Solution:
52 0101 0010
- 730111 0011
DF 1101 1111 R0 = 0xDF
Z = 0 because the R20 has a value other than zero after the subtraction.
C = 0 because R1 is bigger than R0 and there is a borrow from D32 bit.
Example: Show the status of the Z flag after the subtraction of 0x23
from 0xA5 in the following instructions:
LDR R0,=0xA5
LDR R1,=0x23
SUBS R0,R0,R1 ;subtract R1 from R0
Solution:
0xA5 1010 0101
- 0x230010 0011
0x82 1000 0010 R0 = 0x82
Z = 0 because the R20 has a value other than 0 after the subtraction.
C = 1 because R1 is not bigger than R0 and there is no borrow from D32 bit.