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 • 1K=1024=210, 1M=1024K=220, 1G=1024M=230

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 3 2 1 0 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 3 2 1 0 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 (@, #, $, ^, &, ~, _, |, \) • 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): • -(2^27+2^26+2^24+2^21) • 8D280000 as floating-point (in decimal): • 1000110100101000 0000 0000 0000 0000 • (-1) 2^{26-127}*(1+0.025+0.0625)= • 8D280000 as instruction: • 1000 110100101000 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 $a2=6 # $a1 = 61 00 64 63 lbu $a1, -7($a2) # assume $a2=12 # $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 • e=1 to 254: (-1)S *2e-127 * 1.M • Special cases: • Exponent=0, mantissa=0: 0 • Exponent=0, mantissa0: denormalized (-1)S *2-127 * 0.M • Exponent=255, mantissa=0: Infty • Exponent=255, 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. PC=PC+4, go to 2 • If there is branch instruction, PC is modified

19. Exercise • Count the number of 1’s in $a0 and store result in $s0 or $s0, $zero, $zero # $s0=0 addi $t0, $zero, 1 # $t0=00000001 loop: beq $t0, $zero, exit # if $t0=0, 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 (cont’d) • Base (or displacement) addressing • Data address is base (register content) plus displacement • For example, lw $a1, 12($a0)

22. Five Addressing Mode (cont’d) • The last two modes are for branch only. Address is instruction address. • PC-relative addressing • Branch address is PC+4+offset*4 • 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 (a[i] == k) i = i + 1; • MIPS assembly code loop: sll $t1, $s3, 2 # $t1 = i*4 add $t1, $t1, $s6 # $t1 = addr of a[i] lw $t0, 0($t1) # $t0 = a[i] bne $t0, $s5, exit # if a[i] != k, exit addi $s3, $s3, 1 # i = i+1 j loop # go to loop

24. Addressing Mode Example loop: sll $t1, $s3, 2 # $t1 = i*4 add $t1, $t1, $s6 # $t1 = addr of a[i] lw $t0, 0($t1) # $t0 = a[i] bne $t0, $s5, exit # if a[i] != k, exit addi $s3, $s3, 1 # i = i+1 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 (i=0; i<=100; i=i+1) a[i] = b[i] +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