Review Chpt 2 and 3

1
Review Chpt 2 and 3

• Five components of a computer
• Input
• Output
• Memory
• Arithmetic Logic Unit (ALU)
• Control
• Two types of information
• Data and instruction

2
Think of a Calculator

• Components in a calculator
• Input
• Output
• Memory
• ALU
• One type of information
• Data

3
Memory in Calculator

• One register R, which is implicit
• Its content is shown at the LCD panel
• Arithmetic operations are performed on R
• One memory M, which is explicit
• Its content is not shown.
• Arithmetic operation on M is limited
• M is additional storage in case R is not enough
• Copy data from R to M is done by MS
• Copy data from M to R is done by MR

4
Memory in Computer (MIPS)

• More registers
• 32 integer registers 0 to 31
• 32 floating-point registers f0 to f31
• More memory
• 4 Giga Byte (GB) memory, organized as 1 Gita Word
• A word is 32 bits (b), a Byte (B) is 8 bits
• A unique address is designated for each memory
  byte
• 1K1024210, 1M1024K220, 1G1024M230

5
Examples

• Which one can store more information a 128MB
  memory card or a half Gb memory chip?
• How many times more data can a 40GB hard disk
  store than a 256KB memory?

6
Memory Address
0
1
2
3
00000000
00000004
00000008
...
FFFFFFF8
FFFFFFFC
7
Example Memory content

• chara .word 0x00000061 assume starting mem 0
• msg1 .asciiz ab01cd
• msg2 .ascii ab
• value .word 15

0
1
2
3
00000000
00000004
00000008
0000000C
00000010
...
8
Data and Instruction

• Memory contains data and instruction
• The content itself does not tell if it is data or
  instruction
• Each instruction takes one word, and instruction
  address must be a word address

9
Data Encoding

• ASCII code (American Standard Code for
  Information Interchange)
• Total 128 characters, including
• digits (0-9)
• alphabets (a-z, A-Z)
• math (, -, , /, ,
• punctuations (!, ,, ., ?, , , , , , ,)
• special (_at_, , , , , , _, , \)
• control (back space, tab, line return, null)

10
Data Encoding

• Unicode (universal encoding)
• Capable of expressing most languages in the world

11
Data Encoding

• Fixed point integer
• Single precision floating point
• Double precision floating point
• Question Which format is most efficient in
  expressing a number, say 12345678? How about
  1234.5678?
• ASCII character string
• Fixed point
• Floating point

12
Example on Numbers

• What is the following hexadecimal?
• 8D280000 as binary
• 1000 1101 0010 1000 0000 0000 0000 0000
• 8D280000 as integer (in decimal)
• -(227226224221)
• 8D280000 as floating-point (in decimal)
• 1000 1101 0010 1000 0000 0000 0000 0000
• (-1) 226-127(10.0250.0625)
• 8D280000 as instruction
• 1000 1101 0010 1000 0000 0000 0000 0000
• lw 8 0(9)

13
Word v.s. Byte

• Memory can be addressed either as words, each 32
  bits, or as bytes, each 8 bits
• Whether an address is word address or byte
  address depends on the instruction
• lw a1, 0(zero) a1 00 00 00 61
• lw a1, 2(a2) assume a26
• a1 61 00 64 63
• lbu a1, -7(a2) assume a212
• a1 00 00 00 62

14
ALU

• In a calculator, ALU performs operation between R
  and the new number typed on key pad
• 12 enter 12 into R and operation is
• 3 enter 3, and perform addition with R
• In a computer, ALU performs more complex
  operations between registers

15
Example

• Convert temperature from Fahrenheit in f0 to
  Celsius C(5/9)(F-32)
• Assume 5.0 is in f16, 9.0 is in f17, and 32.0
  is in f18
• div.s f16, f16, f17 .s is single
  precision
• sub.s f0, f0, f18
• mul.s f0, f0, f16
• Puzzle How to swap values in a1 and a2 without
  using additional register?

16
Fixed Point Mult/Div

• Assume F temperature is in a0, store C
  temperature result in a1
• addi a1, a0, -32 a1 F-32
• addi t0, zero, 5 t0 5
• mult a1, t0 mult first reduces error
• mflo a1 a1 (F-32)5
• addi t0, zero, 9 t0 9
• div a1, t0
• mflo a1

17
IEEE 754 Standard Exception

• Normal case
• e1 to 254 (-1)S 2e-127 1.M
• Special cases
• Exponent0, mantissa0 0
• Exponent0, mantissa?0 denormalized
• (-1)S 2-127 0.M
• Exponent255, mantissa0 ?Infty
• Exponent255, mantissa ?0 NaN

18
Control

• In a calculator, human controls the operation at
  each step
• In a computer, control follows from the
  instruction
• 1. Program Counter (PC) points to the first
  instruction to be executed
• 2. Execute the instruction pointed by PC
• 3. PCPC4, go to 2
• If there is branch instruction, PC is modified

19
Exercise

• Count the number of 1s in a0 and store result
  in s0
• or s0, zero, zero s00
• addi t0, zero, 1 t000000001
• loop beq t0, zero, exit if t00, exit
• and t1, a0, t0 check one bit
• sll t0, t0, 1 shift t0 left 1 bit
• beq t1, zero, loop
• addi s0, s0, 1 found a 1
• j loop
• exit

20
Five Address Modes (p. 101)

• Address Mode tells where to find the data or
  branch address
• Immediate addressing
• It is rather immediate than addressing since
  the data is included in the instruction
• For example, addi a1, a1, 100
• Register addressing
• Data address is register number
• For example, add a1, a1, a0

21
Five Addressing Mode (contd)

• Base (or displacement) addressing
• Data address is base (register content) plus
  displacement
• For example, lw a1, 12(a0)

22
Five Addressing Mode (contd)

• The last two modes are for branch only. Address
  is instruction address.
• PC-relative addressing
• Branch address is PC4offset4
• For example, beq a1, zero, 100
• Pseudo-direct addressing
• Branch address is 26 bits in instruction
  concatenated with top 4 bits of PC
• For example, j 10000

23
Addressing Mode Example

• C program, assume i and k correspond to s3 and
  s5, and base address of array a is in s6
• while (ai k)
• i i 1
• MIPS assembly code
• loop sll t1, s3, 2 t1 i4
• add t1, t1, s6 t1 addr of ai
• lw t0, 0(t1) t0 ai
• bne t0, s5, exit if ai ! k, exit
• addi s3, s3, 1 i i1
• j loop go to loop

24
Addressing Mode Example

• loop sll t1, s3, 2 t1 i4
• add t1, t1, s6 t1 addr of ai
• lw t0, 0(t1) t0 ai
• bne t0, s5, exit if ai ! k, exit
• addi s3, s3, 1 i i1
• j loop go to loop
• exit

FFFF0004
FFFF0008
FFFF000C
FFFF0010
FFFF0014
FFFF0018
25
Homework 2, due Friday

• 2.21, 2.36
• 3.27, 3.42

26
Exercise 2.21

• Write a subroutine convert to convert an ASCII
  decimal string to an integer. You can expect
  register a0 to hold the address of a
  null-terminated string containing some
  combination of the digits 0 through 9. Your
  program should compute the integer value
  equivalent to this string of digits, then place
  the number in register v0. If a non-digit
  character, including - or ., appears anywhere
  in the string, your program should stop with the
  value 1 in register v0 .
• For example, if register a0 points to a sequence
  of three character 50ten , 52ten , 0ten (the
  null-terminated string 24), then when the
  program stops, register v0 should contain the
  value 24ten .

27
Exercise 2.36

• Consider the following fragment of C code
• for (i0 i
• ai bi c
• Assume that a and b are arrays of words and the
  base address of a is in a0 and the base address
  of b is in a1 . Register t0 is associated with
  variable i and register s0 with c.
• Write the code for MIPS. How many instructions
  are executed during the running of this code? How
  many memory data references will be made during
  execution?

28
Chapter 4 Performance

• What is performance?
• Response time The time between start and
  completion of the task. (unit is time, say sec)
• Throughput the total amount of work done at a
  given time. (unit is jobs/sec)
• Do the following changes to a computer system
  increase/decrease performance?
• Replacing the processor by a faster one
• Adding additional processors to a multi-processor
  system, such as google search engine