Download Chapter 2

ECE369
Chapter 2
ECE369
1
Instruction Set Architecture
•
A very important abstraction
– interface between hardware and low-level software
– standardizes instructions, machine language bit patterns, etc.
– advantage: different implementations of the same architecture
– disadvantage: sometimes prevents using new innovations
•
Modern instruction set architectures:
– IA-32, PowerPC, MIPS, SPARC, ARM, and others
ECE369
2
The MIPS Instruction Set
•
•
•
•
Used as the example throughout the book
Stanford MIPS commercialized by MIPS Technologies (www.mips.com)
Large share of embedded core market
– Applications in consumer electronics, network/storage equipment,
cameras, printers, …
Typical of many modern ISAs
– See MIPS Reference Data tear-out card, and Appendixes B and E
ECE369
3
MIPS arithmetic
•
•
All instructions have 3 operands
Operand order is fixed (destination first)
Example:
C code:
a = b + c
MIPS ‘code’:
add a, b, c
(we’ll talk about registers in a bit)
“The natural number of operands for an operation like addition is
three…requiring every instruction to have exactly three operands, no
more and no less, conforms to the philosophy of keeping the
hardware simple”
ECE369
4
MIPS arithmetic
•
•
Design Principle: simplicity favors regularity.
Of course this complicates some things...
C code:
a = b + c + d;
MIPS code:
add a, b, c
add a, a, d
•
•
Operands must be registers, only 32 registers provided
Each register contains 32 bits
•
Design Principle: smaller is faster.
ECE369
Why?
5
Registers vs. Memory
•
•
•
Arithmetic instructions operands must be registers,
— only 32 registers provided
Compiler associates variables with registers
What about programs with lots of variables
Control
Input
Memory
Datapath
Processor
Output
I/O
ECE369
6
Memory Organization
•
•
•
Viewed as a large, single-dimension array, with an address.
A memory address is an index into the array
"Byte addressing" means that the index points to a byte of memory.
0
1
2
3
4
5
6
...
8 bits of data
8 bits of data
8 bits of data
8 bits of data
8 bits of data
8 bits of data
8 bits of data
ECE369
7
Memory Organization
•
•
•
•
•
Bytes are nice, but most data items use larger "words"
For MIPS, a word is 32 bits or 4 bytes.
0 32 bits of data
4 32 bits of data
Registers hold 32 bits of data
32
bits
of
data
8
12 32 bits of data
...
232 bytes with byte addresses from 0 to 232-1
230 words with byte addresses 0, 4, 8, ... 232-4
Words are aligned
i.e., what are the least 2 significant bits of a word address?
ECE369
8
Instructions
•
•
Load and store instructions
Example:
C code: A[12] = h + A[8];
# $s3 stores base address of A and $s2 stores h
MIPS code:
lw $t0, 32($s3)
add $t0, $s2, $t0
sw $t0, 48($s3)
•
•
•
Can refer to registers by name (e.g., $s2, $t2) instead of number
Store word has destination last
Remember arithmetic operands are registers, not memory!
Can’t write:
add 48($s3), $s2, 32($s3)
ECE369
9
Instructions
•
Example:
C code:
g = h + A[i];
# $s3 stores base address of A and
# g,h and i in $s1,$s2 and $s4
Add $t1,$s4,$s4
t1 = 2*i
Add $t1,$t1,$t1
t1 = 4*i
Add $t1,$t1,$s3
t1 = 4*i + s3
Lw $t0,0($t1)
t0 = A[i]
Add $s1,$s2,$t0
g = h + A[i]
ECE369
10
So far we’ve learned:
•
MIPS
— loading words but addressing bytes
— arithmetic on registers only
•
Instruction
Meaning
add $s1, $s2, $s3
sub $s1, $s2, $s3
lw $s1, 100($s2)
sw $s1, 100($s2)
$s1 = $s2 + $s3
$s1 = $s2 – $s3
$s1 = Memory[$s2+100]
Memory[$s2+100] = $s1
ECE369
11
Policy of Use Conventions
Name Register number
$zero
0
$v0-$v1
2-3
$a0-$a3
4-7
$t0-$t7
8-15
$s0-$s7
16-23
$t8-$t9
24-25
$gp
28
$sp
29
$fp
30
$ra
31
Usage
the constant value 0
values for results and expression evaluation
arguments
temporaries
saved
more temporaries
global pointer
stack pointer
frame pointer
return address
Register 1 ($at) reserved for assembler, 26-27 for operating system
ECE369
12
MIPS Format
•
Instructions, like registers and words
– are also 32 bits long
–
–
•
add $t1, $s1, $s2
Registers: $t1=9, $s1=17, $s2=18
Instruction Format:
000000
op
10001
rs
10010
rt
01001
rd
00000
shamt
ECE369
100000
funct
13
Machine Language
•
•
•
•
Consider the load-word and store-word instructions,
– What would the regularity principle have us do?
– New principle: Good design demands a compromise
Introduce a new type of instruction format
– I-type for data transfer instructions
– other format was R-type for register
Example: lw $t0, 32($s2)
35
18
9
op
rs
rt
32
16 bit number
Where's the compromise?
ECE369
14
Summary
Name Register number
$zero
0
$v0-$v1
2-3
$a0-$a3
4-7
$t0-$t7
8-15
$s0-$s7
16-23
$t8-$t9
24-25
$gp
28
$sp
29
$fp
30
$ra
31
Usage
the constant value 0
values for results and expression evaluation
arguments
temporaries
instruction format op
saved
add
R 0
more temporaries
global pointer
sub
R 0
stack pointer
lw
I 35
frame pointer
sw
I 43
return address
A[300]=h+A[300]
Lw $t0,1200($t1)
Add $t0, $s2, $t0
Sw $t0, 1200($t1)
rs
reg
reg
reg
reg
rt
reg
reg
reg
reg
rd shamt funct address
reg 0 32 na
reg 0 34 na
na na na address
na na na address
# $t1 = base address of A, $s2 stores h
# use $t0 for temporary register
Op rs,rt,address
Op,rs,rt,rd,shamt,funct
Op,rs,rt,address
ECE369
35,9,8,1200
0,18,8,8,0,32
43,9,8,1200
15
Summary of Instructions We Have Seen So Far
ECE369
16
Summary of New Instructions
ECE369
17
Example
swap(int* v, int k);
{ int temp;
temp = v[k]
v[k] = v[k+1];
v[k+1] = temp;
}
swap:
sll $t0, $a1, 4
add $t0, $t0, $a0
lw $t1, 0($t0)
lw $t2, 4($t0)
sw $t2, 0($t0)
sw $t1, 4($t0)
jr $31
ECE369
18
Control Instructions
ECE369
19
Using If-Else
$s0 = f
$s1 = g
$s2 = h
$s3 = i
$s4 = j
$s5 = k
Where is 0,1,2,3 stored?
ECE369
20
Addresses in Branches
•
•
Instructions:
bne $t4,$t5,Label
beq $t4,$t5,Label
Next instruction is at Label if $t4≠$t5
Next instruction is at Label if $t4=$t5
Formats:
I
op
rs
rt
16 bit address
•What if the “Label” is too far away (16 bit address is not enough)
ECE369
21
Addresses in Branches and Jumps
• Instructions:
bne $t4,$t5,Label
beq $t4,$t5,Label
j Label
if $t4 != $t5
if $t4 = $t5
Next instruction is at Label
• Formats:
I
op
J
op
rs
rt
16 bit address
26 bit address
•
ECE369
22
Control Flow
• We have: beq, bne, what about Branch-if-less-than?
If (a<b) # a in $s0, b in $s1
slt
$t0, $s0, $s1
bne $t0, $zero, Less
# t0 gets 1 if a<b
# go to Less if $t0 is not 0
Combination of slt and bne implements branch on less than.
ECE369
23
While Loop
While (save[i] == k)
i = i+j;
# i, j and k correspond to registers
# $s3, $s4 and $s5
# array base address at $s6
Loop: add $t1, $s3, $s3
add $t1, $t1, $t1
add $t1, $t1, $s6
lw $t0, 0($t1)
bne $t0, $s5, Exit
add $s3, $s3, $s4
j
loop
Exit:
ECE369
24
What does this code do?
ECE369
25
Overview of MIPS
•
•
•
simple instructions all 32 bits wide
very structured, no unnecessary baggage
only three instruction formats
R
op
rs
rt
rd
I
op
rs
rt
16 bit address
J
op
shamt
funct
26 bit address
ECE369
26
Arrays vs. Pointers
clear1( int array[ ], int size)
{
int i;
for (i=0; i<size; i++)
array[i]=0;
}
clear2(int* array, int size)
{
int* p;
for( p=&array[0]; p<&array[size]; p++)
*p=0;
}
CPI for arithmetic, data transfer, branch type of instructions are 1, 2,
and 1 correspondingly. Which code is faster?
ECE369
27
Clear1
array in $a0
size in $a1
i in $t0
clear1( int array[ ], int size)
{
int i;
for (i=0; i<size; i++)
array[i]=0;
}
add
loop1: add
$t0,$zero,$zero # i=0, register $t0=0
$t1,$t0,$t0
# $t1=i*2
add
add
sw
addi
slt
$t1,$t1,$t1
$t2,$a0,$t1
$zero, 0($t2)
$t0,$t0,1
$t3,$t0,$a1
# $t1=i*4
# $t2=address of array[i]
# array[i]=0
# i=i+1
# $t3=(i<size)
bne
$t3,$zero,loop1 # if (i<size) go to loop1
ECE369
28
Clear2, Version 2
clear2(int* array, int size)
{
int* p;
for( p=&array[0]; p<&array[size]; p++)
*p=0;
}
loop2:
Array and size
to registers $a0 and $a1
add
$t0,$a0,$zero
# p = address of array[0]
add
add
add
sw
addi
slt
bne
$t1,$a1,$a1
$t1,$t1,$t1
$t2,$a0,$t1
$zero,0($t0)
$t0,$t0,4
$t3,$t0,$t2
$t3,zero,loop2
# $t1 = size*2
# $t1 = size*4 Distance of last element
# $t2 = address of array[size]
# memory[p]=0
# p = p+4
# $t3=(p<&array[size])
# if (p<&array[size]) go to loop2
29
ECE369
Array vs. Pointer
loop1:
add
add
add
add
sw
addi
slt
bne
add
loop2:
add
add
add
sw
addi
slt
bne
$t0,$zero,$zero
$t1,$t0,$t0
$t1,$t1,$t1
$t2,$a0,$t1
$zero, 0($t2)
$t0,$t0,1
$t3,$t0,$a1
$t3,$zero,loop1
# i=0, register $t0=0
# $t1=i*2
# $t1=i*4
# $t2=address of array[i]
# array[i]=0
# i=i+1
# $t3=(i<size)
# if (i<size) go to loop1
$t0,$a0,$zero
7 instructions
inside loop
# p = address of array[0]
$t1,$a1,$a1
$t1,$t1,$t1
$t2,$a0,$t1
$zero,0($t0)
# $t1 = size*2
# $t1 = size*4
4 instructions
# $t2 = address of array[size] inside loop
# memory[p]=0
$t0,$t0,$4
$t3,$t0,$t2
$t3,zero,loop2
# p = p+4
# $t3=(p<&array[size])
# if (p<&array[size]) go to loop2
ECE369
30
Summary
ECE369
31
Other Issues
•
More reading:
support for procedures
linkers, loaders, memory layout
stacks, frames, recursion
manipulating strings and pointers
interrupts and exceptions
system calls and conventions
•
Some of these we'll talk more about later
•
We have already talked about compiler optimizations
ECE369
32
Elaboration
Name Register number
$zero
0
$v0-$v1
2-3
$a0-$a3
4-7
$t0-$t7
8-15
$s0-$s7
16-23
$t8-$t9
24-25
$gp
28
$sp
29
$fp
30
$ra
31
Usage
the constant value 0
values for results and expression evaluation
arguments
temporaries
saved
more temporaries
global pointer
stack pointer
frame pointer
return address
What if there are more than 4 parameters for a function call?
Addressable via frame pointer
References to variables in the stack have the same offset
ECE369
33
What is the Use of Frame Pointer?
Variables local to procedure do not fit in registers !!!
ECE369
34
Nested Procedures,
function_main(){
function_a(var_x);
:
return;
}
function_a(int size){
function_b(var_y);
:
return;
}
function_b(int count){
:
return;
}
/* passes argument using $a0 */
/* function is called with “jal” instruction */
/* passes argument using $a0 */
/* function is called with “jal” instruction */
Resource Conflicts ???
ECE369
35
Stack
• Last-in-first-out queue
• Register # 29 reserved as stack
pointer
• Points to most recently
allocated address
• Grows from higher to lower
address
• Subtracting $sp
• Adding data – Push
• Removing data – Pop
ECE369
36
Function Call and Stack Pointer
jr
ECE369
37
Recursive Procedures Invoke Clones !!!
int fact (int n) {
if (n < 1 )
return ( 1 );
else
return ( n * fact ( n-1 ) );
}
Registers $a0 and $ra
“n” corresponds to $a0
Program starts with the label of the procedure “fact”
How many registers do we need to save on the stack?
ECE369
38
Factorial Code
200 fact:addi
204
sw
sw
L1:
236
240
$sp, $sp, -8 #adjust stack for 2 items
$ra, 4($sp) #save return address
$a0, 0($sp) #save argument n
slti
beq
$t0, $a0, 1
# is n<1?
$t0, $zero, L1 # if not  go to L1
addi
addi
jr
$v0, $zero, 1
$sp, $sp, 8
$ra
addi
jal
$a0, $a0, -1
fact
lw
lw
addi
mult
jr
:
100 fact(3)
104 add ….
#return result
#pop items off stack
#return to calling proc.
#decrement n
# call fact(n-1)
$a0, 0($sp) # restore “n”
$ra, 4($sp) # restore address
$sp, $sp,8
# pop 2 items
$v0,$a0,$v0 # return n*fact(n-1)
$ra
# return to caller
ECE369
ra = 104
a0= 3
sp= 40
vo=
int fact (int n) {
if (n < 1 )
return ( 1 );
else
return ( n * fact ( n-1 ) );
}
39
Assembly to Hardware Example
k is stored in $t0;
3 stored in $t2;
int i;
int k = 0;
add
for (i=0; i<3; i++){
add
k = k + i + 1;
addi
}
loop: add
k = k/2;
addi
addi
slt
bne
srl
i is stored in $t1
$t3 used as temp
$t0,$zero,$zero
$t1,$zero,$zero
$t2,$zero,3
$t0,$t0,$t1
$t0,$t0,1
$t1,$t1,1
$t3,$t1,$t2
$t3,$zero,loop
$t0,$t0,1
ECE369
# k=0, register $t0=0
# i=0, register $t1=0
# $t2=3
#k=k+i
#k=k+1
# i=i+1
# $t3= (i<3)
# if (i<3) go to loop
#k=k/2
40
Assembly to Hardware Example
add
add
addi
loop: add
addi
addi
slt
bne
$t0,$zero,$zero R-Type
$t1,$zero,$zero R-Type
$t2,$zero,3
I-Type
$t0,$t0,$t1
R-Type
$t0,$t0,1
I-Type
$t1,$t1,1
I-Type
$t3,$t1,$t2
R-Type
$t3,$zero,loop I-Type
srl
$t0,$t0,1
Instruction Types?
R-Type
R
op
rs
rt
rd
I
op
rs
rt
16 bit address
ECE369
shamt
funct
41
How do we represent in machine language?
op
rs
rt
rd
shamt
funct
0:
add
$t0,$zero,$zero 000000_00000_00000_01000_00000_100000
4:
add
$t1,$zero,$zero 000000_00000_00000_01001_00000_100000
8:
addi
$t2,$zero,3
001000_00000_01010_0000000000000011
12: loop:
add
$t0,$t0,$t1
000000_01000_01001_01000_00000_100000
16:
addi
$t0,$t0,1
001000_01000_01000_0000000000000001
20:
addi
$t1,$t1,1
001000_01001_01001_0000000000000001
24:
slt
$t3,$t1,$t2
000000_01001_01010_01011_00000_101010
28:
bne
$t3,$zero,loop
000101_00000_01011_1111111111111011
32:
srl
$t0,$t0,2
000000_00000_01000_01000_00001_000010
6 bits
R
I
5 bits
5 bits
5 bits
5 bits
6 bits
shamt
funct
op
rs
rt
rd
op
rs
rt
16 bit address
ECE369
PC+4+BR Addr
- 5
$t0 is reg 8
$t1 is reg 9
$t2 is reg 10
$t3 is reg 11
42
How do we represent in machine language?
loop:
add
$t0,$zero,$zero
add
$t1,$zero,$zero
addi
$t2,$zero,3
op
rs
rt
rd
shamt
funct
Instruction Memory
0
000000_00000_00000_01000_00000_100000
4
000000_00000_00000_01001_00000_100000
8
001000_00000_01010_0000000000000011
add
$t0,$t0,$t1
addi
$t0,$t0,1
12
000000_01000_01001_01000_00000_100000
addi
$t1,$t1,1
16
001000_01000_01000_0000000000000001
20
001000_01001_01001_0000000000000001
24
000000_01001_01010_01011_00000_101010
28
000101_00000_01011_1111111111111011
32
000000_00000_01000_01000_00001_000010
slt
$t3,$t1,$t2
bne
$t3,$zero,loop
srl
$t0,$t0,2
ECE369
43
Representation in MIPS Datapath
op
rs
rt
rd
shamt
Instruction Memory
0
000000_00000_00000_01000_00000_100000
4
000000_00000_00000_01001_00000_100000
8
001000_00000_01010_0000000000000011
12
000000_01000_01001_01000_00000_100000
16
001000_01000_01000_0000000000000001
20
001000_01001_01001_0000000000000001
24
000000_01001_01010_01011_00000_101010
28
000101_00000_01011_1111111111101100
32
000000_00000_01000_01000_00001_000010
funct
Name Register number
$zero
0
$v0-$v1
2-3
$a0-$a3
4-7
$t0-$t7
8-15
$s0-$s7
16-23
$t8-$t9
24-25
$gp
28
$sp
29
$fp
30
$ra
31
ECE369
Usage
the constant value 0
values for results and expression evaluation
arguments
temporaries
saved
more temporaries
global pointer
stack pointer
frame pointer
return address
44
Big Picture
ECE369
45
Compiler
ECE369
46
Addressing Modes
ECE369
47
Our Goal
add $t1, $s1, $s2 ($t1=9, $s1=17, $s2=18)
– 000000 10001 10010 01001 00000
op
rs
rt
rd
shamt
ECE369
100000
funct
48
Assembly Language vs. Machine Language
•
•
•
•
Assembly provides convenient symbolic representation
– much easier than writing down numbers
– e.g., destination first
Machine language is the underlying reality
– e.g., destination is no longer first
Assembly can provide 'pseudoinstructions'
– e.g., “move $t0, $t1” exists only in Assembly
– would be implemented using “add $t0,$t1,$zero”
When considering performance you should count real instructions
ECE369
49
Summary
•
•
•
Instruction complexity is only one variable
– lower instruction count vs. higher CPI / lower clock rate
Design Principles:
– simplicity favors regularity
– smaller is faster
– good design demands compromise
– make the common case fast
Instruction set architecture
– a very important abstraction indeed!
ECE369
50