Chapter 2
Instructions: Language of the Computer
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Outline
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2.1 Introduction
2.2 Operations of the Computer Hardware
2.3 Operands of the Computer Hardware
2.4 Representing Instructions in the Computer
2.5 Logical Operations
2.6 Instructions for Making Decisions
2.7 Supporting Procedures in Computer Hardware
2.8 Communicating with People
2.9 MIPS Addressing for 32-Bits Immediates and Addresses
2.10 Translating and Starting a Program
2.16 Real Stuff: IA-32 Instructions
Summary
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2.1 Introduction
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Four design principles guide
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Simplicity favors regularity
Smaller is faster
Make the common case fast
Good design demands good
compromises
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Instructions:
•
•
Language of the Machine
We’ll be working with the MIPS instruction set architecture
– similar to other architectures developed since the 1980's
– Almost 100 million MIPS processors manufactured in 2002
– used by NEC, Nintendo, Cisco, Silicon Graphics, Sony, …
1400
1200
Other
SPARC
Hitachi SH
1100
PowerPC
1300
1000
Motorola 68K
900
MIPS
IA-32
800
ARM
700
600
500
400
300
200
100
0
1998
1999
2000
2001
2002
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Instruction set: The vocabulary of
commands understood by a given
architecture.
Stored-Program Concept : The idea that
instructions and data of many types can be
stored in memory as numbers, leading to
the stored program computer.
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2.2 Operations of the Computer Hardware
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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 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”
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Four design principles guide
•
•
•
•
Simplicity favors regularity
Smaller is faster
Make the common case fast
Good design demands good
compromises
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FIGURE 2.1 MIPS architecture revealed in section 2.2.
MIPS assembly language
Category
Instruction
Example
Meaning
Comments
Arithmetic
add
add a, b, c
a=b+c
Always three operands
subtract
sub a, b, c
a=b-c
Always three operands
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f = (g+h) - (i+j);
add t0, g, h
add t1, I, j
sub f, t0, t1
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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;
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.
Why?
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f = (g+h) - (i+j);
$s0, $s1, $s2, $s3, $s4
add t0, $s1, $s2
add t1, $s3, $s4
sub $s0, t0, t1
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2.3 Operands of the Computer Hardware
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Data transfer instruction: a command
that moves data between memory and
registers.
Address: A value used to delineate the
location of a specific data element
within a memory arry.
Alignment restriction : A requirement
that data be aligned in memory on
natural boundaries.
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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
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Four design principles guide
•
•
•
•
Simplicity favors regularity
Smaller is faster
Make the common case fast
Good design demands good
compromises
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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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•
•
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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 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?
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FIGURE 2.2 Memory addresses and contents of
memory at those locations.
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FIGURE 2.3 Actual MIPS memory addresses and
contents of memory for those words.
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g = h + A[8]
lw $t0, 8($s3)
add $s1, $s2, $t0
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Instructions
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•
•
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Load and store instructions
Example:
C code:
A[12] = h + A[8];
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)
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Our First Example
•
Can we figure out the code?
swap(int v[], int k);
{ int temp;
temp = v[k]
v[k] = v[k+1];
v[k+1] = temp;
swap:
}
muli $2, $5, 4
add $2, $4, $2
lw $15, 0($2)
lw $16, 4($2)
sw $16, 0($2)
sw $15, 4($2)
jr $31
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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
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addi $s3, $s4, 4
# $s3 = $s3 +4
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Four design principles guide
•
•
•
•
Simplicity favors regularity
Smaller is faster
Make the common case fast
Good design demands good
compromises
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FIGURE 2.4 MIPS architecture revealed through
section 2.3.
MIPS operands
Name
32 registers
230memory
words
Example
Comments
$s0, $s1, ……
Fast locations for data. In MIPS, data must be in registers to perform arithmetic.
$t 0, $t 1, ……
Memory[0],
Accessed only by data transfer instructions in MIPS. MIPS uses byte addresses, so
Memory[4], …,
sequential word addresses differ by 4. Memory holds data structures arrays, and
Memory[4294967292]
spilled registers.
MIPS assembly language
Category
Arithmetic
Instruction
Example
Meaning
Comments
add
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
add immediate
addi $s1, $s2, 100
$s1 = $s2 + 100
Used to add constants
load word
lw $s1, 100($s2)
$s1 = Memory [$s2 + 100]
Data from memory to register
store word
sw $s1, 100 ($s2)
Memory [$s2 + 100] = $s1
Data from register to memory
Data transfer
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2.4 Representing Instructions in the
Computer
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Binary digit Also called binary bit. One of
the two numbers in base 2, 0 or 1, that are
the components of information.
Machine language: Binary representation
used for communication within a computer
system.
Instruction format: A form of representation
of an instruction composed of fields of
binary numbers.
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Machine Language
•
Instructions, like registers and words of data, are also 32 bits long
– Example: add $t1, $s1, $s2
– registers have numbers, $t1=9, $s1=17, $s2=18
•
Instruction Format:
000000 10001
op
•
rs
10010
rt
01000
rd
00000
100000
shamt
funct
Can you guess what the field names stand for?
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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)
35
18
9
op
rs
rt
32
16 bit number
Where's the compromise?
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op: Opcode
rs: the first register source operand.
rt: the second register source operand.
rd: the destination operand.
shamt: Shift amount.
funct: Function (function code).
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Four design principles guide
•
•
•
•
Simplicity favors regularity
Smaller is faster
Make the common case fast
Good design demands good
compromises
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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
•
Memory
memory for data, programs,
compilers, editors, etc.
Fetch & Execute Cycle
– Instructions are fetched and put into a special register
– 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: ...
Can you build a simple for loop?
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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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FIGURE 2.7 MIPS architecture revealed through
section 2.4
MIPS operands
Name
32 registers
230memory
words
Example
Comments
$s0, $s1, …, $s7
Fast locations for data. In MIPS, data must be in registers to perform arithmetic.
$t 0, $t 1, …, $t 7
Registers $s0 - $s7 map to 16-23 and $t0 - $t7 map to 8-15.
Memory[0],
Accessed only by data transfer instructions in MIPS. MIPS uses byte addresses, so
Memory[4], …,
sequential word addresses differ by 4. Memory holds data structures arrays, and
Memory[4294967292]
spilled registers.
MIPS assembly language
Category
Arithmetic
Instruction
Example
Meaning
Comments
add
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
load word
lw $s1, 100($s2)
$s1 = Memory [$s2 + 100]
Data from memory to register
store word
sw $s1, 100 ($s2)
Memory [$s2 + 100] = $s1
Data from register to memory
Data transfer
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FIGURE 2.7 MIPS architecture revealed through
section 2.4
MIPS machine language
Name
Format
Example
Comments
add
R
0
18
19
17
0
32
add $s1, $s2, $s3
sub
R
0
18
19
17
0
34
sub $s1, $s2, $s3
addi
I
8
18
17
100
addi $s1, $s2, 100
lw
I
35
18
17
100
lw $s1, 100($s2)
sw
I
43
18
17
100
sw $s1, 100 ($s2)
6 bits
5 bits
5 bits
5 bits
5 bits
6 bits
ALL MIPS instructions 32 bits
rd
shamt
funct
Arithmetic instruction format
Field size
R-format
R
op
rs
rt
I-format
I
op
rs
rt
address
Data transfer format
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FIGURE 2.8 The stored-program concept.
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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
•
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,
— there are policy of use conventions for registers
•
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2.5
Logical Operations
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FIGURE 2.9 C and JAVA logical operators and their
corresponding MIPS instructions.
Logical operations
C operators
JAVA operators
MIPS instructions
Shift left
<<
<<
sll
Shift right
>>
>>>
srl
Bit-by-bit AND
&
&
and, andi
Bit-by-bit OR
|
|
or, ori
Bit-by-bit NOT
~
~
nor
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FIGURE 2.10 MIPS architecture revealed thus far.
MIPS operands
Name
Example
Comments
32 registers
$s0, $s1, …, $s7
$t 0, $t 1, …, $t 7
Fast locations for data. In MIPS, data must be in registers to perform arithmetic.
Registers $s0 - $s7 map to 16-23 and $t0 - $t7 map to 8-15.
Memory[0],
Memory[4], …,
Memory[4294967292]
Accessed only by data transfer instructions in MIPS. MIPS uses byte addresses, so
sequential word addresses differ by 4. Memory holds data structures arrays, and
spilled registers.
230memory
words
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MIPS assembly language
Category
Arithmetic
Logical
Data
transfer
Instruction
Example
Meaning
Comments
add
add $s1, $s2, $s3
$s1 = $s2 + $s3
Three operands; Overflow detected
subtract
sub $s1, $s2, $s3
$s1 = $s2 - $s3
Three operands; Overflow detected
add immediate
addi $s1, $s2, 100
$s1 = $s2 + 100
+constant; overflow detected
and
add $s1, $s2, $s3
$s1 = $s2 & $s3
Three reg. operands; bit-by-bit AND
or
or
$s1, $s2, $s3
$s1 = $s2 | $s3
Three reg. operands; bit-by-bit OR
nor
nor
$s1, $s2, $s3
$s1 = ~($s2 | $s3)
Three reg. operands; bit-by-bit NOR
and immediate
andi $s1, $s2, 100
$s1 = $s2 & 100
Bit-by-bit AND reg with constant
or immediate
ori
$s1, $s2, 100
$s1 = $s2 | 100
Bit-by-bit OR reg with constant
shift left logical
sll
$s1, $s2, 10
$s1 = $s2 << 10
Shift left by constant
shift right
logical
srl
$$s1, $s2, 10
$s1 = $s2 >> 10
Shift right by constant
load word
lw
$s1, 100($s2)
$s1 = Memory [$s2 + 100]
Word from memory to register
store word
sw
$s1, 100 ($s2)
Memory [$s2 + 100] = $s1
Word from register to memory
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2.6 Instructions for Making Decisions
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Conditional branch: An Instruction that
requires the comparison of two values and
that allows for a subsequent transfer of
control to a new address in the program
based on the outcome of the comparison.
Basic block: A sequence of instructions
without branches and without branch
targets or branch labels.
Jump address table also called jump table.
A table of addresses of alternative
instruction sequences.
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FIGURE 2.11 Illustration of the options in the IF
statement above.
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Compiling if-then-else into Conditional Branches
if (i==j) f=g+h; else f=g-h;
bne $s3, $s4, Else
add $s0, $s1, $s2
j Exit
Else: sub $s0, $s1, $s2
Exit:
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Compiling a while loop in C
while (save[i]==k)
i+=1;
Loop: sll $t1, $s3,2
add $t1, $t1, $s6
lw $t0, 0($t1)
bne $t0, $s5, Exit
add $s3, $s3, 1
j Loop
Exit:
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The MIPS instruction is called set on less than
slt $t0, $s3, $s4
means that register $t0 is set to 1 if the
value in register $s3 is less than the value
in register $s4; otherwise, register $t0 is set
to o.
slti $t0, $s2, 10
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FIGURE 2.12 MIPS architecture revealed through
section 2.6
MIPS operands
Name
Example
Comments
32 registers
$s0, $s1, …, $s7
$t 0, $t 1, …, $t 7
$zero
Fast locations for data. In MIPS, data must be in registers to perform arithmetic.
Registers $s0 - $s7 map to 16-23 and $t0 - $t7 map to 8-15. MIPS register $zero
always equals 0.
Memory[0],
Memory[4], …,
Memory[4294967292]
Accessed only by data transfer instructions in MIPS. MIPS uses byte addresses, so
sequential word addresses differ by 4. Memory holds data structures arrays, and
spilled registers.
230memory
words
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FIGURE 2.13 MIPS machine language revealed
through section 2.6.
MIPS machine language
Name
Format
Example
Comments
add
R
0
18
19
17
0
32
add $s1, $s2, $s3
sub
R
0
18
19
17
0
34
sub $s1, $s2, $s3
lw
I
35
18
17
100
lw
$s1, 100($s2)
sw
I
43
18
17
100
sw
$s1, 100 ($s2)
and
R
0
18
19
17
0
36
and $s1, $s2, $s3
or
R
0
18
19
17
0
37
or
$s1, $s2, $s3
nor
R
0
18
19
17
0
39
nor
$s1, $s2, $s3
andi
I
12
18
17
100
andi $s1, $s2, 100
ori
I
13
18
17
100
ori
$s1, $s2, 100
sll
R
0
0
18
17
10
0
sll
$s1, $s2, 10
srl
R
0
0
18
17
10
2
srl
$s1, $s2, 10
beq
I
4
17
18
25
beq $s1, $s2, 100
bne
I
5
17
18
25
bne $s1, $s2, 100
slt
R
0
18
19
j
J
2
Field size
17
0
42
2500
slt
$s1, $s2, $s3
j 10000 (see Section 2.9)
6 bits
5 bits
5 bits
5 bits
5 bits
6 bits
ALL MIPS instructions 32 bits
rd
shamt
funct
Arithmetic instruction format
R-format
R
op
rs
rt
I-format
I
op
rs
rt
address
Data transfer, branch format
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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
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
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Constants
•
•
•
Small constants are used quite frequently (50% of operands)
e.g.,
A = A + 5;
B = B + 1;
C = C - 18;
Solutions? Why not?
– put 'typical constants' in memory and load them.
– create hard-wired registers (like $zero) for constants like one.
MIPS Instructions:
addi $29, $29, 4
slti $8, $18, 10
andi $29, $29, 6
ori $29, $29, 4
•
Design Principle: Make the common case fast.
Which format?
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How about larger constants?
•
•
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
•
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
•
•
•
•
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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Other Issues
•
Discussed in your assembly language programming lab:
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’ll talk about compiler optimizations when we hit chapter 4.
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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
rely on compiler to achieve performance
— what are the compiler's goals?
help compiler where we can
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FIGURE 2.13 MIPS machine language revealed
through section 2.6.
MIPS machine language
Name
Format
Example
Comments
add
R
0
18
19
17
0
32
add $s1, $s2, $s3
sub
R
0
18
19
17
0
34
sub $s1, $s2, $s3
lw
I
35
18
17
100
lw
$s1, 100($s2)
sw
I
43
18
17
100
sw
$s1, 100 ($s2)
and
R
0
18
19
17
0
36
and $s1, $s2, $s3
or
R
0
18
19
17
0
37
or
$s1, $s2, $s3
nor
R
0
18
19
17
0
39
nor
$s1, $s2, $s3
andi
I
12
18
17
100
andi $s1, $s2, 100
ori
I
13
18
17
100
ori
$s1, $s2, 100
sll
R
0
0
18
17
10
0
sll
$s1, $s2, 10
srl
R
0
0
18
17
10
2
srl
$s1, $s2, 10
beq
I
4
17
18
25
beq $s1, $s2, 100
bne
I
5
17
18
25
bne $s1, $s2, 100
slt
R
0
18
19
j
J
2
Field size
17
0
42
2500
slt
$s1, $s2, $s3
j 10000 (see Section 2.9)
6 bits
5 bits
5 bits
5 bits
5 bits
6 bits
ALL MIPS instructions 32 bits
rd
shamt
funct
Arithmetic instruction format
R-format
R
op
rs
rt
I-format
I
op
rs
rt
address
Data transfer, branch format
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2.7 Supporting Procedures in Computer
Hardware
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Procedure: A stored subroutine that performs a
specific task based on the parameters with which it
is provided.
Jump-and-link instruction: An instruction that
jumps to an address and simultaneously saves the
address of the following instruction in a register
($ra in MIPS)
Return address: A link to the calling site that allows
a procedure to return to the proper address; in
MIPS it is stored in register $ra.
Program Counter (PC): The register containing the
address of the instruction in the program being
executed.
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Caller: The program that instigates a procedure and
provides the necessary parameter values.
Callee: A procedure that executes a series of stored
instructions based on parameters provided by the
caller and then returns control to the caller.
Stack: A data structure for spilling registers
organized as a last-in-first-out queue.
Stack pointer: A value denoting the most recently
allocated address in a stack that shows where
registers should be spilled or where old register
values can be found.
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Addresses in Branches and Jumps
•
•
Instructions:
bne $t4,$t5,Label
beq $t4,$t5,Label
$t5
j Label
Formats:
op
I
J
•
op
rs
Next instruction is at Label if $t4 ° $t5
Next instruction is at Label if $t4 =
Next instruction is at Label
rt
16 bit address
26 bit address
Addresses are not 32 bits
— How do we handle this with load and store instructions?
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Addresses in Branches
•
•
Instructions:
bne $t4,$t5,Label
beq $t4,$t5,Label
Formats:
I
•
•
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)
Jump instructions just use high order bits of PC
– address boundaries of 256 MB
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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],
30
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
16
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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The six steps of the execution of a procedure
1. Place parameter in a place where the procedure
can access them.
2. Transfer control to the procedure
3. Acquire the storage resources needed for the
procedure.
4. perform the desired task.
5. place the result value in a place where the calling
program can access it.
6. Return control to the point of origin, since a
procedure can be called from several points in a
program.
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Registers for procedure calling:
• $a0-$a3: four argument registers in which to
pass parameters
• $v0-$v1: two value registers in which to
return values
• $ra: one return address register to return to
the point of origin
• $t0-$t9: 10 temporary registers that are not
preserved by the callee on a procedure call
• $s0-$s7: 8 saved registers that must be
preserved on a procedure call
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Compiling a C procedure(Doesn’t Call Another)
Int leaf_example (int g, int h, int i, int j)
{ int f:
f=(g+h)-(i+j);
return f;}
leaf_example:
addi $sp, $sp, -12
sw $t1, 8($sp)
sw $t0, 4($sp)
sw $s0, 0($sp)
add $t0, $a0, $a1
add $t1, $a2, $a3
sub $s0, $t0, $t1
add $v0, S0, $zero
lw $s0, 0($sp)
lw $t0, 4($sp)
lw $t1, 8($sp)
addi $sp, $sp, 12
jr $ra
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72
Compiling a Recursive C Procedure
Int fact (int n)
{
if (n<1) return (1);
else return (n*fact(n-1));}
fact:
addi $sp, $sp, -8
sw $ra, 4($sp)
sw $a0, 0($sp)
slti $t0, $a0, 1
beq $t0, $zero, L1
addi $v0, $zero, 1
addi $sp, $sp, 8
jr $ra
L1: addi $a0, $a0, -1
jalfact
lw $a0, 0($sp)
lw $ra, 4($sp)
addi $sp, $sp, 8
mul $v0, $a0, $v0
jr $ra
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FIGURE 2.14 The values if the stack pointer and the stack
(a) before, (b) during, and (c) after the procedure call
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FIGURE 2.15 What is and what is not preserved across a
procedure call.
Preserved
Not preserved
Saved registers: $s0 - $s7
Temporary registers: $t 0-$t 9
Stack pointer register: $sp
Argument registers: $a0 - $a3
Return address register: $ra
Return value registers: $v0 - $v1
Stack above the stack pointer
Stack below the stack pointer
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FIGURE 2.16 Illustration of the stack allocation (a)
before, (b) during, and (c) after the procedure call.
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FIGURE 2.17 The MIPS memory allocation for program
and data.
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FIGURE 2.18 MIPS register conventions.
Name
Preserved on
call?
Register number Usage
$zero
0
the constant value 0
n.a.
$v0-$v1
2-3
values for results and expression evaluation
no
$a0-$a3
4-7
arguments
no
$t0-$t7
8-15
temporaries
no
$s0-$s7
16-23
saved
yes
$t8-$t9
24-25
more temporaries
no
$gp
28
global pointer
yes
$sp
29
stack pointer
yes
$fp
30
frame pointer
yes
$ra
31
return address
yes
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FIGURE 2.19 MIPS architecture revealed through
section 2.7.
MIPS operands
Name
Example
Comments
32 registers
$s0-$s7, $t 0- $t 9,
$zero, $a0-a3, $v0-$v1,
$gp, $fp, $sp, $ra
Fast locations for data. In MIPS, data must be in registers to perform arithmetic.
MIPS register $zero always equals 0. $gp (28) is the global pointer, $sp (29) is the
stack pointer, $fp (30) is the frame pointer, and $ra (31) is the return address.
30memory
2
words
Memory[0],
Memory[4], …,
Memory[4294967292]
Accessed only by data transfer instructions in MIPS. MIPS uses byte addresses, so
sequential word addresses differ by 4. Memory holds data structures, arrays, and
spilled registers, such as those saved on procedure calls.
2004 Morgan Kaufmann Publishers
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MIPS assembly language
Category
Instruction
Example
Meaning
Comments
add
add $s1, $s2, $s3
$s1 = $s2 + $s3
Three operands; Overflow detected
subtract
sub $s1, $s2, $s3
$s1 = $s2 - $s3
Three operands; Overflow detected
load word
lw
$s1, 100($s2)
$s1 = Memory [$s2 + 100]
Data from memory to register
store word
sw
$s1, 100 ($s2)
Memory [$s2 + 100] = $s1
Data from register to memory
and
add $s1, $s2, $s3
$s1 = $s2 & $s3
Three reg. operands; bit-by-bit AND
or
or
$s1, $s2, $s3
$s1 = $s2 | $s3
Three reg. operands; bit-by-bit OR
nor
nor
$s1, $s2, $s3
$s1 = ~($s2 | $s3)
Three reg. operands; bit-by-bit NOR
and immediate
andi $s1, $s2, 100
$s1 = $s2 & 100
Bit-by-bit AND reg with constant
or immediate
ori
$s1, $s2, 100
$s1 = $s2 | 100
Bit-by-bit OR reg with constant
shift left logical
sll
$s1, $s2, 10
$s1 = $s2 << 10
Shift left by constant
shift right logical
srl
$$s1, $s2, 10
$s1 = $s2 >> 10
Shift right by constant
branch on equal
beq $s1, $s2, L
if ($s1 == $s2) go to L
Equal test and branch
branch on not
equal
bne $s1, $s2, L
if ($s1 != $s2) go to L
Not equal test and branch
set on less than
slt
$s1, $s2, $s3
if ($s2 < $s3) $s1 = 1;
else $s1 = 0
Compare less than; used with beq,
bne
set on less than
immediate
slt
$s1, $s2, 100
if ($s2 < 100) $s1 = 1;
else $s1 = 0
Compare less than immediate; used
with beq, bne
jump
j
L
go to L
Jump to target address
jump register
jr
$ra
go to $ra
For procedure return
jump and link
jal
L
$ra = PC + 4; go to L
For procedure
callKaufmann Publishers
2004 Morgan
Arithmetic
Data transfer
Logical
Condition
al branch
Unconditi
onal jump
80
FIGURE 2.20 MIPS machine language revealed
through section 2.7.
Name
Format
Example
Comments
add
R
0
18
19
17
0
32
add $s1, $s2, $s3
sub
R
0
18
19
17
0
34
sub $s1, $s2, $s3
lw
I
35
18
17
100
lw
$s1, 100($s2)
sw
I
43
18
17
100
sw
$s1, 100 ($s2)
and
R
0
18
19
17
0
36
and $s1, $s2, $s3
or
R
0
18
19
17
0
37
or
$s1, $s2, $s3
nor
R
0
18
19
17
0
39
nor
$s1, $s2, $s3
andi
I
12
18
17
100
andi $s1, $s2, 100
ori
I
13
18
17
100
ori
$s1, $s2, 100
sll
R
0
0
18
17
10
0
sll
$s1, $s2, 10
srl
R
0
0
18
17
10
2
srl
$s1, $s2, 10
beq
I
4
17
18
25
beq $s1, $s2, 100
bne
I
5
17
18
25
bne $s1, $s2, 100
slt
R
0
18
19
j
J
2
jr
R
0
jal
J
3
Field size
17
0
42
2500
31
0
0
slt
$s1, $s2, $s3
j 10000 (see Section 2.9)
0
8
2500
jr
$ra
jal
10000 (see section 2.9)
6 bits
5 bits
5 bits
5 bits
5 bits
6 bits
ALL MIPS instructions 32 bits
rd
shamt
funct
Arithmetic instruction format
R-format
R
op
rs
rt
I-format
I
op
rs
rt
address
Data transfer,
branch
formatPublishers
2004
Morgan Kaufmann
81
2.8
Communicating with People
2004 Morgan Kaufmann Publishers
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Instructions to move bytes
lb $t0, 0($sp); Loads a bytes from
memory, placing it in the rightmost 8
bits of a register.
sb $t0, 0($gp); Stores a bytes from the
rightmost 8 bits of a register, placing it
to memory.
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FIGURE 2.21 ASCII representation of characters.
ASCII
value
Character
ASCII
value
Character
ASCII
value
Character
ASCII
value
Character
ASCII
value
Character
ASCII
value
Character
32
space
48
0
64
@
80
P
96
`
112
p
33
!
49
1
65
A
81
Q
97
a
113
q
34
“
50
2
66
B
82
R
98
b
114
r
35
#
51
3
67
C
83
S
99
c
115
s
36
$
52
4
68
D
84
T
100
d
116
t
37
%
53
5
69
E
85
U
101
e
117
u
38
&
54
6
70
F
86
V
102
f
118
v
39
‘
55
7
71
G
87
W
103
g
119
w
40
(
56
8
72
H
88
X
104
h
120
x
41
)
57
9
73
I
89
Y
105
I
121
y
42
*
58
:
74
J
90
Z
106
j
122
z
43
+
59
;
75
K
91
[
107
k
123
{
44
,
60
<
76
L
92
\
108
l
124
|
45
-
61
=
77
M
93
]
109
m
125
}
46
.
62
>
78
N
94
^
110
n
126
~
47
/
63
?
79
O
95
_
111
o
127
DEL
2004 Morgan Kaufmann Publishers
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2.9 MIPS Addressing for 32-Bits Immediates
and Addresses
2004 Morgan Kaufmann Publishers
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FIGURE 2.23 The effect of the lui instruction.
The machine language version of lui $t0, 255
001111
00000
01000
# $t0 is register 8:
0000000011111111
Contents of register $t0 after executing lui $t0, 255:
0000 0000 1111 1111
0000000011111111
2004 Morgan Kaufmann Publishers
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Loading a 32-bit Constant
Load 0000 0000 0011 1101 0000 1001 0000
0000 to register $s0
lui $s0, 61 #61 decimal = 0000 0000 0011 1101
ori $s0, $s0, 2304 #2304 decimal = 0000 1001
0000 0000
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MIPS Addressing Mode Summary
1. Register addressing,
2. Base or displacement addressing,
3. Immediate addressing,
4. PC-relative addressing,
5. Pseudodirect addressing,
2004 Morgan Kaufmann Publishers
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FIGURE 2.24 Illustration of the five MIPS addressing modes.
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
2004 Morgan Kaufmann Publishers
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FIGURE 2.25 MIPS instruction encoding
Op (31:26)
28-26
31-29
0(000)
1(001)
2(010)
0(000) R-format
Bltz/gez
jump
1(001) add
immediate
addiu
2(010) TLB
FLPt
3(011)
5(101)
6(110)
branch
eq
branch
blez
bgtz
set less
sltiu
than imm.
andi
ori
xori
load upper
imm
lbu
lhu
lwr
Jump &
link
4(100)
7(111)
3(011)
4(100) load byte
load half
lw1
load word
5(101) store byte
store half
sw1
store word
6(110)
lwc0
lwc1
7(111)
swc0
swc1
swr
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Op (31:26)=010000 (TLB), rs(25:21)
23-21
25-24
0(00)
0(000)
mfc0
1(001)
2(010)
cfc0
3(011)
4(100)
mtc0
5(101)
6(110)
7(111)
ctc0
1(01)
2(10)
3(11)
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Op (31:26)=000000 (R-format), funct(5:0)
2-0
5–3
0(000)
1(001)
0(000) shift left
logical
2(010)
shift right
logical
3(011)
sra
1(001) jump reg.
jalr
2(010) mfhi
mthi
mflo
mtlo
3(011)
mult
multu
div
divu
4(100) add
addu
subtract
subu
set l . t .
sltu
5(101)
4(100)
5(101)
sllv
syscall
break
and
or
6(110)
7(111)
srlv
srav
xor
not or (nor)
6(110)
7(111)
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FIGURE 2.26 MIPS instruction formats in Chapter 2.
Name
Field size
Fields
6 bits 5 bits 5 bits 5 bits
Comments
5 bits
6 bits
ALL MIPS instructions 32 bits
shamt
funct
Arithmetic instruction format
R-format
op
rs
rt
rd
I-format
op
rs
rt
address/immediate
J-format
op
target address
Transfer, brance, imm. format
jump instruction format
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FIGURE 2.27 MIPS assembly language revealed in
Chapter 2.
MIPS operands
Name
Example
Comments
32 registers
$s0-$s7, $t 0- $t 9,
$zero, $a0-a3, $v0-$v1,
$gp, $fp, $sp, $ra, $at
Fast locations for data. In MIPS, data must be in registers to perform arithmetic.
MIPS register $zero always equals 0. Register $at is reserved for the assembler to
handle large constants.
memory
words
Memory[0],
Memory[4], …,
Memory[4294967292]
Accessed only by data transfer instructions in MIPS. MIPS uses byte addresses, so
sequential word addresses differ by 4. Memory holds data structures, arrays, and
spilled registers, such as those saved on procedure calls.
2004 Morgan Kaufmann Publishers
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MIPS assembly language
Category
Arithmetic
Data transfer
Logical
Instruction
Example
Meaning
Comments
add
add $s1, $s2, $s3
$s1 = $s2 + $s3
Three operands; Overflow detected
subtract
sub $s1, $s2, $s3
$s1 = $s2 - $s3
Three operands; Overflow detected
add immediate
addi $s1, $s2, 100
$s1 = $s2+ 100
Used to add constants
load word
lw
$s1, 100($s2)
$s1 = Memory [$s2 + 100]
Word from memory to register
store word
sw
$s1, 100 ($s2)
Memory [$s2 + 100] = $s1
Word from register to memory
load half
lh
$s1, 100($s2)
$s1 = Memory [$s2 + 100]
Halfword memory to register
store half
sh
$s1, 100 ($s2)
Memory [$s2 + 100] = $s1
Halfword register to memory
load byte
lb
$s1, 100($s2)
$s1 = Memory [$s2 + 100]
Byte from memory to register
store byte
sb
$s1, 100 ($s2)
Memory [$s2 + 100] = $s1
Byte from register to memory
load upper immed.
lui
$s1, 100
$s1 = 100 * 2^16
Loads constant in upper 16 bits
and
add $s1, $s2, $s3
$s1 = $s2 & $s3
Three reg. operands; bit-by-bit AND
or
or
$s1, $s2, $s3
$s1 = $s2 | $s3
Three reg. operands; bit-by-bit OR
nor
nor
$s1, $s2, $s3
$s1 = ~($s2 | $s3)
Three reg. operands; bit-by-bit NOR
and immediate
andi $s1, $s2, 100
$s1 = $s2 & 100
Bit-by-bit AND reg with constant
or immediate
ori
$s1, $s2, 100
$s1 = $s2 | 100
Bit-by-bit OR reg with constant
shift left logical
sll
$s1, $s2, 10
$s1 = $s2 << 10
Shift left by constant
shift right logical
srl
$$s1, $s2, 10
$s1 = $s2 >> 10
Shift right by constant
2004 Morgan Kaufmann Publishers
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Continue..
Condition
al branch
Unconditi
onal jump
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 L
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 on less than
immediate
slt
$s1, $s2, 100
if ($s2 < 100) $s1 = 1;
else $s1 = 0
Compare less than constant
jump
j
2500
go to 10000
Jump to target address
jump register
jr
$ra
go to $ra
For procedure return
jump and link
jal
2500
$ra = PC + 4; go to 10000
For procedure call
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2.10 Translating and Starting a Program
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• Pseudoinstruction: A common variation of assembly
language instructions often treated as if it were an instruction
in its own right.
• Symbol table: A table that matches names of labels to the
addresses of the memory words that instructions occupy.
• Linker Also called link editor: A systems program that
combines independtly assembled machine language
programs and resolves all undefined labels into an
executable file.
• Executable file: A functional program in the format of an
object file that contains no unresolved references, relocation
information, symbol table, or debugging information.
• Loader: A systems program that places an object program in
main memory so that it is ready to execute.
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FIGURE 2.28 A translation hierarchy for C.
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FIGURE 2.29 Dynamically linked library via lazy
procedure linkage.
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FIGURE 2.30 A translation hierarchy for JAVA.
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2.16 Real Stuff: IA-32 Instructions
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Alternative Architectures
•
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
–“The path toward operation complexity is thus fraught with peril.
To avoid these problems, designers have moved toward simpler
instructions”
•
Let’s look (briefly) at IA-32
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IA - 32
•
•
•
•
•
•
•
•
•
•
•
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: 57 new “MMX” instructions are added, Pentium II
1999: The Pentium III added another 70 instructions (SSE)
2001: Another 144 instructions (SSE2)
2003: AMD extends the architecture to increase address space to 64 bits,
widens all registers to 64 bits and other changes (AMD64)
2004: Intel capitulates and embraces AMD64 (calls it EM64T) and adds
more media extensions
“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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IA-32 Overview
•
•
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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IA-32 Registers and Data Addressing
•
Registers in the 32-bit subset that originated with 80386
Name
Use
31
0
EAX
GPR 0
ECX
GPR 1
EDX
GPR 2
EBX
GPR 3
ESP
GPR 4
EBP
GPR 5
ESI
GPR 6
EDI
GPR 7
EIP
EFLAGS
CS
Code segment pointer
SS
Stack segment pointer (top of stack)
DS
Data segment pointer 0
ES
Data segment pointer 1
FS
Data segment pointer 2
GS
Data segment pointer 3
Instruction pointer (PC)
Condition codes
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IA-32 Register Restrictions
•
Registers are not “general purpose” – note the restrictions below
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IA-32 Typical Instructions
•
Four major types of integer instructions:
– Data movement including move, push, pop
– Arithmetic and logical (destination register or memory)
– Control flow (use of condition codes / flags )
– String instructions, including string move and string compare
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IA-32 instruction Formats
•
Typical formats: (notice the different lengths)
a. JE EIP + displacement
4
4
8
CondiDisplacement
tion
JE
b. CALL
8
32
CALL
Offset
c. MOV
6
MOV
EBX, [EDI + 45]
1 1
8
d w
r/m
Postbyte
8
Displacement
d. PUSH ESI
5
3
PUSH
Reg
e. ADD EAX, #6765
4
3 1
32
ADD Reg w
f. TEST EDX, #42
7
1
TEST
w
Immediate
8
32
Postbyte
Immediate
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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!
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