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CPTR 220
Computer Organization
Computer Architecture
Assembly Programming
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Chapter 3
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Instructions:
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Language of the Machine
More primitive than higher level languages
e.g., no sophisticated control flow
Very restrictive
e.g., MIPS Arithmetic Instructions
We’ll be working with the MIPS instruction set architecture
– similar to other architectures developed since the 1980's
– used by NEC, Nintendo, Silicon Graphics, Sony
Design goals: maximize performance and minimize cost, reduce design time
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MIPS arithmetic
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All instructions have 3 operands
Operand order is fixed (destination first)
Example:
C code:
A = B + C
MIPS code:
add $s0, $s1, $s2
(associated with variables by compiler)
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MIPS arithmetic
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Design Principle: simplicity favors regularity.
Of course this complicates some things...
C code:
A = B + C + D;
E = F - A;
MIPS code:
add $t0, $s1, $s2
add $s0, $t0, $s3
sub $s4, $s5, $s0
Why?
Operands must be registers, only 32 registers provided
– 32 bits per register. 32bits = Word
– Names
• $s0,$s1,.. For variables
• $t0,$t1, .. For temporary variables
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Design Principle: smaller is faster.
Why?
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Registers vs. Memory
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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
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Memory Organization
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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
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Memory Organization
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Bytes are nice, but most data items use larger "words"
For MIPS, a word is 32 bits or 4 bytes.
0
4
8
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...
32 bits of data
32 bits of data
32 bits of data
Registers hold 32 bits of data
32 bits of data
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Instructions
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Load and store instructions
Example:
C code:
A[8] = h + A[8];
MIPS code:
lw $t0, 32($s3)
add $t0, $s2, $t0
sw $t0, 32($s3)
Store word has destination last
Remember arithmetic operands are registers, not memory!
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So far we’ve learned:
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MIPS
— loading words but addressing bytes
— arithmetic on registers only
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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
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Machine Language
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Instructions, like registers and words of data, are also 32 bits long
– Example: add $t0, $s1, $s2
– registers have numbers, $t0=9, $s1=17, $s2=18
– See Figure 3.6 Page 121 for mapping of registers
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Instruction Format:
000000 10001
op
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rs
10010
rt
01000
rd
00000
100000
shamt
funct
Can you guess what the field names stand for?
– Page 118 have the names for the field names
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Machine Language
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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)
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op
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8
rs
rt
32
16 bit number
Where's the compromise?
See example on page 119
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Stored Program Concept
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Instructions are bits
Programs are stored in memory
— to be read or written just like data
Processor
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Memory
memory for data, programs,
compilers, editors, etc.
Fetch & Execute Cycle
– Instructions are fetched and put into a special register
– Execute: Bits in the register "control" the subsequent actions
– Fetch the “next” instruction and continue
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Control
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Decision making instructions
– alter the control flow,
– i.e., change the "next" instruction to be executed
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MIPS conditional branch instructions:
bne $t0, $t1, Label
beq $t0, $t1, Label
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Example:
if (i==j) h = i + j;
bne $s0, $s1, Label
add $s3, $s0, $s1
Label: ....
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Control
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MIPS unconditional branch instructions:
j label
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Example:
if (i!=j)
h=i+j;
else
h=i-j;
beq $s4, $s5, Lab1
add $s3, $s4, $s5
j Lab2
Lab1: sub $s3, $s4, $s5
Lab2: ...
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Control
while (save[i] == k)
i = i + j;
Assume i = $s3; j = $s4; and k = $s5; save = $s6
Loop: add $t1,$s3,$s3
# $t1 = 2 * i
add $t1,$t1,$t1
# $t1 = 4 * i
add $t1, $t1 $s6
# $t1 = Addr(save[i])
lw $t0,0($t1)
# $t0 = save[i]
bne $t0,$s5, Exit
# go to Exit if save[I] != k
add $s3,$s3,$s4
#i=i+j
j
Loop
Exit: ….
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So far:
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Instruction
Meaning
add $s1,$s2,$s3
sub $s1,$s2,$s3
lw $s1,100($s2)
sw $s1,100($s2)
bne $s4,$s5,L
beq $s4,$s5,L
j Label
$s1 = $s2 + $s3
$s1 = $s2 – $s3
$s1 = Memory[$s2+100]
Memory[$s2+100] = $s1
Next instr. is at Label if $s4 != $s5
Next instr. is at Label if $s4 = $s5
Next instr. is at Label
Formats:
R
op
rs
rt
rd
I
op
rs
rt
16 bit address
J
op
shamt
funct
26 bit address
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Control Flow
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We have: beq, bne, what about Branch-if-less-than?
New instruction:
if $s1 < $s2 then
$t0 = 1
slt $t0, $s1, $s2
else
$t0 = 0
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Can use this instruction to build "blt $s1, $s2, Label"
— can now build general control structures
Note that the assembler needs a register to do this
Register $Zero always contains a 0
MIPS does not include branch on less than because it is too
complicated; either it would stretch the clock cycle time or this
instruction would take extra clock cycles per instruction.
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Control Flow
blt $s0,$s1,Label can be implemented as follows
slt $t0,$s0,$s1
bne $t0,$zero, Label
bgt $s0, $s1, Label can be implemented as follows
slt $t0,$s1,$s0
bne $t0,$zero, Label
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Control Flow
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Another instruction
– jr $s0 - would jump to the address stored in register $s0
Let us see the example on p129
Page 131 have all the instructions that we have learned so far.
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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
$sp
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$fp
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$ra
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Usage
the constant value 0
values for results and expression evaluation
arguments
temporaries
saved
more temporaries
global pointer
stack pointer
frame pointer
return address
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Procedures
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In order to implement procedures you must use the following
instructions:
– Jump-and-link to call a procedure
• jal ProcedureName (or address)
• jal places the return address in $ra
– jr $ra to return from the procedure
– You may need to save the contents of registers into memory by
using a stack.
• Use $sp as the stack pointer
• Do not save $t# but only $s#
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Procedure Example
int leaf_example (int g, int h, int i, int j) {
int f;
f = (g + h ) – ( i + j);
return f; }
Leaf_example:
subi $sp,$sp,4
sw $s0, 0($sp)
add $t0,$a0,$a1
add $ti,$a2,$a3
sub $s0,$t0,$t1
add $v0, $s0, $zero
lw $s0, 0($sp)
addi $sp, $sp, 4
jr $ra
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Nested Procedures
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int fact (int n)
{ if (n < 1) return (1);
else return (n * fact(n-1));
fact: subi
$sp,$sp,8
#save $a0 = n and $ra
sw
$ra, 4($sp)
sw
$a0, 0($sp)
slti
$t0,$a0,1
beq
$t0,$zero,L1
# if n >=1 go to L1
addi
$v0,$zero,1
add
$sp,$sp,8
# pop 2 items from the stack
jr
$ra
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Factorial Procedure Cont.
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L1: subi
$a0,$a0,1
# $a0 is n-1
jal
fact
# recursive call
lw
$a0, 0($sp)
# restore old value of n and
lw
$ra, 4($sp)
# return address
addi
$sp,$sp,8
mult
$v0, $a0,$v0
jr
$ra
See page 141 for summary of all the instructions so far
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Working with Strings
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void strcpy (car x [ ], char y [ ]) {
int i = 0;
while ((x[i] = y [i]) != 0 ) i++; }
strcpy:
subi
$sp,$sp,4
sw
$s0,0($sp)
add
$s0,$zero,$zero
L1:
add
$t1,$a1,$s0
lb
$t2, 0($t1)
add
$t3,$a0,$s0
sb
$t2, 0($t3)
addi
$s0,$s0,1
bne
$t2,$zero,L1
lw
$s0, 0($sp)
addi
$sp,sp,4
jr
$ra
# save $s0
# address of y[i]
# load byte
# address of x[i]
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Constants
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Small constants are used quite frequently (50% of operands)
e.g.,
A = A + 5;
B = B + 1;
C = C - 18;
Solutions?
– put 'typical constants' in memory and load them. Create hard-wired
registers (like $zero) for constants like one.
– Put the constants in the instructions
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MIPS Instructions:
addi $29, $29, 4
slti $8, $18, 10
andi $29, $29, 6
ori $29, $29, 4
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How do we make this work?
– Use instruction format op(6),rs(5),rd(5),immediate(16)
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How about larger constants?
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We'd like to be able to load a 32 bit constant into a register
Must use two instructions, new "load upper immediate" instruction
lui $t0, 1010101010101010
1010101010101010
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filled with zeros
0000000000000000
Then must get the lower order bits right, i.e.,
ori $t0, $t0, 1010101010101010
1010101010101010
0000000000000000
0000000000000000
1010101010101010
1010101010101010
1010101010101010
ori
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Assembly Language vs. Machine Language
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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
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Overview of MIPS
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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
rely on compiler to achieve performance
— what are the compiler's goals?
help compiler where we can
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Addresses in Branches and Jumps
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Instructions:
bne $t4,$t5,Label
$t5
beq $t4,$t5,Label
j Label
Next instruction is at Label if $t4 !=
Next instruction is at Label if $t4 = $t5
Next instruction is at Label
Formats:
I
op
J
op
rs
rt
16 bit address
26 bit address
Addresses are not 32 bits
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Addresses in Branches
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Instructions:
bne $t4,$t5,Label
beq $t4,$t5,Label
Formats:
I
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Next instruction is at Label if $t4 != $t5
Next instruction is at Label if $t4=$t5
op
rs
rt
16 bit address
Could specify a register (like lw and sw) and add it to address
– use Instruction Address Register (PC = program counter)
– most branches are local (principle of locality)
Since MIPS instructions are 4 bytes long, all PC-relative addressing refer to
the number of words to the next instruction instead of the number of bytes.
How many memory locations can you jump ?
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Addressing
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j
L1
– Allows to jump to an address that is contained in a 26-bit field.
– Because the number represent words so they could be stored in
28 bits
– Use the upper 4 bits of PC to complete the 32 bits needed for
addressing
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To summarize:
MIPS operands
Name
32 registers
Example
Comments
$s0-$s7, $t0-$t9, $zero, Fast locations for data. In MIPS, data must be in registers to perform
$a0-$a3, $v0-$v1, $gp,
arithmetic. MIPS register $zero always equals 0. Register $at is
$fp, $sp, $ra, $at
reserved for the assembler to handle large constants.
Memory[0],
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2
Accessed only by data transfer instructions. MIPS uses byte addresses, so
memory Memory[4], ...,
words
and spilled registers, such as those saved on procedure calls.
add
MIPS assembly language
Example
Meaning
add $s1, $s2, $s3
$s1 = $s2 + $s3
Three operands; data in registers
subtract
sub $s1, $s2, $s3
$s1 = $s2 - $s3
Three operands; data in registers
$s1 = $s2 + 100
$s1 = Memory[$s2 + 100]
Memory[$s2 + 100] = $s1
$s1 = Memory[$s2 + 100]
Memory[$s2 + 100] = $s1
Used to add constants
Category
Arithmetic
sequential words differ by 4. Memory holds data structures, such as arrays,
Memory[4294967292]
Instruction
addi $s1, $s2, 100
lw $s1, 100($s2)
sw $s1, 100($s2)
store word
lb $s1, 100($s2)
load byte
sb $s1, 100($s2)
store byte
load upper immediate lui $s1, 100
add immediate
load word
Data transfer
Conditional
branch
Unconditional jump
$s1 = 100 * 2
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Comments
Word from memory to register
Word from register to memory
Byte from memory to register
Byte from register to memory
Loads constant in upper 16 bits
branch on equal
beq
$s1, $s2, 25
if ($s1 == $s2) go to
PC + 4 + 100
Equal test; PC-relative branch
branch on not equal
bne
$s1, $s2, 25
if ($s1 != $s2) go to
PC + 4 + 100
Not equal test; PC-relative
set on less than
slt
$s1, $s2, $s3
if ($s2 < $s3) $s1 = 1;
else $s1 = 0
Compare less than; for beq, bne
set less than
immediate
slti
jump
j
jr
jal
jump register
jump and link
$s1, $s2, 100 if ($s2 < 100) $s1 = 1;
Compare less than constant
else $s1 = 0
2500
$ra
2500
Jump to target address
go to 10000
For switch, procedure return
go to $ra
$ra = PC + 4; go to 10000 For procedure call
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Addressing Mode Summary
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Register addressing
– add $s1,$s2,$s3
Base or displacement addressing
– lw $s0 8($t0)
Immediate addressing
– addi $s0, $s1, 5
PC-relative addressing
– bne $s1, $s2,Exit
Pseudodirect addressing
– j L1
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1. Immediate addressing
op
rs
rt
Immediate
2. Register addressing
op
rs
rt
rd
...
funct
Registers
Register
3. Base addressing
op
rs
rt
Memory
Address
+
Register
Byte
Halfword
Word
4. PC-relative addressing
op
rs
rt
Memory
Address
PC
+
Word
5. Pseudodirect addressing
op
Address
PC
Memory
Word
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Decoding Machine Language
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See figure 3.18
See Example on page 154
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Translation Steps
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C Program -> Compiler ->
Assembly Language Program -> Assembler
-> Object Code -> Linker ->
Executable - >Loader ->
Execution
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Procedure Swap
Swap (int v[], int k) {
int temp;
temp = v[k];
v[k] = v[k+1];
v[k+1] = temp; }
swap: add $t1,$a1,$a1
add $t1, $t1 $t1
add $t1, $a0,$t1
lw $t0, 0($t1)
lw $t2, 4($t1)
sw $t2, 0($t1)
sw $t0, 4($t1)
jr $ra
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Procedure Sort
sort (int v[], int n) {
int i,j;
for ( i = 0; i < n; i ++)
for (j = i – 1; j > = 0 && v[j] > v [ j + 1]; j--)
swap(v,j);
}
See MIPS assembly version of this sort on page 170
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Alternative Architectures
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Design alternative:
– provide more powerful operations
– goal is to reduce number of instructions executed
– danger is a slower cycle time and/or a higher CPI
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Sometimes referred to as “RISC vs. CISC”
– virtually all new instruction sets since 1982 have been RISC
– VAX: minimize code size, make assembly language easy
instructions from 1 to 54 bytes long!
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We’ll look at PowerPC and 80x86
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PowerPC
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Indexed addressing
– example:
lw $t1,$a0+$s3
#$t1=Memory[$a0+$s3]
– What do we have to do in MIPS?
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Update addressing
– update a register as part of load (for marching through arrays)
– example: lwu $t0,4($s3) #$t0=Memory[$s3+4];$s3=$s3+4
– What do we have to do in MIPS?
Others:
– load multiple/store multiple
– a special counter register “bc Loop”
decrement counter, if not 0 goto loop
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80x86
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1978: The Intel 8086 is announced (16 bit architecture)
1980: The 8087 floating point coprocessor is added
1982: The 80286 increases address space to 24 bits, +instructions
1985: The 80386 extends to 32 bits, new addressing modes
1989-1995: The 80486, Pentium, Pentium Pro add a few instructions
(mostly designed for higher performance)
1997: MMX is added
“This history illustrates the impact of the “golden handcuffs” of compatibility
“adding new features as someone might add clothing to a packed bag”
“an architecture that is difficult to explain and impossible to love”
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A dominant architecture: 80x86
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See your textbook for a more detailed description
Complexity:
– Instructions from 1 to 17 bytes long
– one operand must act as both a source and destination
– one operand can come from memory
– complex addressing modes
e.g., “base or scaled index with 8 or 32 bit displacement”
Saving grace:
– the most frequently used instructions are not too difficult to build
– compilers avoid the portions of the architecture that are slow
“what the 80x86 lacks in style is made up in quantity,
making it beautiful from the right perspective”
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Summary
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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!
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