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CSE 230 - Final Study Guide

by: RianMartins

CSE 230 - Final Study Guide CSE 230

GPA 4.0
Computer Org/Assemb Lang Prog

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This material contain all the notes and study guides of the previous exams and an extra material to help you to prepare better for this final exam.
Computer Org/Assemb Lang Prog
Study Guide
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This page Study Guide was uploaded by RianMartins on Wednesday December 2, 2015. The Study Guide belongs to CSE 230 at Arizona State University taught by Nakamura in Fall 2015. Since its upload, it has received 88 views. For similar materials see Computer Org/Assemb Lang Prog in Computer Science and Engineering at Arizona State University.

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Date Created: 12/02/15
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Ls fax 10 5P 4 ng EPAr a Camp mks 53 3quot Ema Lad 1h Q it i move 32 V 39 3 Wk I 0W4 4m 5amp0 MA Wo bgf f6 122 E p39 n 113 363395 1m so aquot w eas L51quot 439 163 56 9 036 5ackn WWW v 3915qu W mgrr 39 r r5quot ck 55p 3 r Ex male 351 5P 392 7 VQ uG i i j a HWY Handy r Aw 331 2 359 MA ism 5610 m1 7 LS 3 H MM 352 3012 9amp3 35g Hd sub 30351 51 227 524 a 152 v mng f 3 1390 59 2 3W a 4 1n 4 M 50 W539 Qua 33 4 5 M 532 O 359 a M13 53 I 1 3 6quot 5W 7 Why CSE 230 STUDY GUIDE EXAM 1 CHAPTER 1 0 Classic Components of Computer I Datapath Manipulates the data coming through the processor It also provides a small amount of temporary data storage I Control Generates control signals that direct the operation of memory and the datapath I Memory Holds instructions and most of the data for currently executing programs I Input External devices such as keyboard mice disks and networks that provide input to the processor I Output External devices such as display printers disks and networks that receive data from the processor CHAPTER 2 0 Number conversion Binary Decimal and Hex representation I To convert numbers to different bases we need to look at each number separately I xyz gt xbase2 ybase1 zbase0 I Converting Binary base 2 0 Binary Decimal o 1101gt12312202112 840113 I Hex System base 16 O From 0 9 the hex system is the same as the decimal system but after 9 we have 0 A 10 O B 11 O C 12 O D 13 O E 14 O F 15 0 Ex Converting 4002AC to binary with 24 bytes 39 4 0100 0 0000 0 0000 2 0010 A 1010 C 1100 39 4002AChex 0100 000 0000 0010 1010 1100two 0 Ex Converting 4002AC to decimal 39 44000022A10C12 I 4002AChex 4105 0104 0103 2102 10101 1210o 400312 0 See some more examples below Obs If you are not familiar with number conversions check the back of your green sheet 9 M0111 193quot 515 6105 quot DGC W OX Nomi GB 2 37 quot ESQCUS ilol en leo A quo Buo Numbers 2 0 lxz x1 DJ 1x1 thgt i o 13 6x 39 43gi 5 i t5gtlt39 O Registers O Registers can be numbered or named zero Used to represent zero v0 v1 Used in return values in functions and also to print and receive data a0 a3 Used to parse arguments to functions t0 t7 Temporary registers 50 57 Saved registers correspond to variables in C program t8 t9 Temporary registers sp Stack Pointer Sra Return Address MIPS Assembly Instructions add Description Adds two registers and stores the result in a register Operation Syntax Encoding Descriptio Operation Syntax Encoding d s t advancepc 4 add dst 0000 0088 ssst tttt dddd 1000 0010 0000 sub 11 Subtracts two registers and stores the result in a register d s t advancepc 4 sub d s t 0000 0088 ssst tttt 01010101 01000 0010 0010 I addi Adds a register and a sign extended immediate value and D Z O escnp on stores the result in a register Operation t s imm advancepc 4 Syntax addi t s imm Encoding 0010 0088 ssst tttt 1111 1111 1111 1111 I lw Description A word is loaded into a register from the specified address Operation t MEMs offset advancepc 4 Syntax lw t offsets Encoding 1000 1188 ssst tttt 1111 1111 1111 1111 39 SW Description The contents of t is stored at the specified address Operation MEMs offset t advancepc 4 Syntax sw t offsets Encoding 1010 1188 ssst tttt 1111 1111 1111 1111 I and Description Bitwise ands two registers and stores the result in a register Operation d s amp t advancepc 4 Syntax and d s t Encoding 0000 0088 ssst tttt 01010101 01000 0010 0100 I or Bitwise logical ors two registers and stores the result in a Description register Operation d s t advancepc 4 Syntax or d s t Encoding 0000 0088 ssst tttt 01010101 01000 0010 0101 39 nor Bitwise logical not ors two registers and stores the result in Description a register Operation d s t advancepc 4 Syntax or d s t Encoding 0000 0088 ssst tttt oldolol d000 0010 0101 I andi Description Bitwise ands a register and an immediate value and stores the result in a register Operation t s amp imm advancepc 4 Syntax andi t s imm Encoding 0011 0088 ssst tttt 1111 1111 1111 1111 I ori Description Bitwise ors a register and an immediate value and stores the result in a register Operation t s imm advancepc 4 Syntax ori t s imm Encoding 0011 0188 ssst tttt iiii iiii iiii iiii I sll Shifts a register value left by the shift amount listed in the instruction and Descnp on places the result in a third register Zeroes are shifted in Operation d t ltlt h advancepc 4 Syntax sll d t h Encoding 0000 0033 ssst tttt dddd dhhh 111100 0000 I srl Shifts a register value right by the shift amount shamt and places the value Descnp on in the destination register Zeroes are shifted in Operation d t gtgt h advancepc 4 Syntax srl d t h Encoding 0000 00 t tttt dddd dhhh 111100 0010 I bne Description Branches if the two registers are not equal Operation if s t advancepc offset ltlt 2 else advancepc 4 Syntax bne s t offset Encoding 0001 0188 ssst tttt 1111 1111 1111 1111 I beq Description Branches if the two registers are equal Operation if s t advancepc offset ltlt 2 else advancepc 4 Syntax beq s t offset Encoding 0001 0088 ssst tttt 1111 1111 1111 1111 I slt Description If s is less than t d is set to one It gets zero otherwise Operation if s lt t d l advancepc 4 else d O advancepc 4 Syntax slt d s t Encoding 0000 0088 ssst tttt 01010101 01000 0010 1010 I slti Description If s is less than immediate t is set to one It gets zero otherwise Operation if s lt imm t l advancepc 4 else t O advancepc 4 Syntax slti t s imm Encoding 0010 1088 ssst tttt 1111 1111 1111 1111 39 i Description Jumps to the calculated address Operation PC nPC nPC PC amp OxfOOOOOOO target ltlt 2 Syntax j target Encoding 0000 1011 1111 1111 1111 1111 1111 1111 I jal Description Jumps to the calculated address and stores the return address in 31 Operation 31 PC 8 or nPC 4 PC nPC nPC PC amp OxfOOOOOOO target ltlt 2 Syntax jal target Encoding 0000 1111 1111 1111 1111 1111 1111 1111 l jr Description Jump to the address contained in register s Operation PC nPC nPC s Syntax jrs Encoding 0000 0088 8880 0000 0000 0000 0000 1000 I mult Description Multiplies s by t and stores the result in LO Operation LO s t advancepc 4 Syntax mult s t Encoding 0000 0088 ssst tttt 0000 0000 0001 1000 I div Description DiVides s by t and stores the quotient in LO and the remainder in HI Operation LO s t HI s t advancepc 4 Syntax diV s t Encoding 0000 0088 ssst tttt 0000 0000 0001 1010 I mflo Description The contents of register LO are moved to the specified register Operation d LO advancepc 4 Syntax m o d Encoding 0000 0000 0000 0000 01010101 01000 0001 0010 I mfhi lDescription lThe contents of register HI are moved to the specified register lOperation ld HI advancepc 4 lSyntaX lm ii d lEncoding 0000 0000 0000 0000 01010101 01000 0001 0000 I Now Let s see some examples and notes for those commands o Encoding and Decoding Machine Language gtDecoma Mac iune chch MEL i SJ TUC llon is a a A 539 g 0 I a a 00 U f kw Oil H H O oto o o Jf c 3 al39n Q J m or 9 J IS 1639 0 gt90er Ts 20W Ti 4 S WJC Jam 11133 opCoAE LS 0 um be R 5499 4149 5931 0 1 5 t 9 A 0 Conditions Statements and Loops I For IFELSE in MIPS we use bne Branch of Not Equal or beq Branch of Equals I Loops must be written using thej jump to defined labels gt Onraj D 61550n Making 6999 368 LOJ Suki LOJDGQ 7 i fb 3 L105 ii 1an 9 quotI D bne site Sh M562 4 1 0 ii w i riox quad 83 b Cl 1 if 2 33 MIP55 lane 35o 34 LoJoGJZ h t A id MA 353 no 354 Law 39 o Arithmetic Operations amp Use of array of integers I Check quiz 3 solution 0 Spilling Register CHAPTER3 o Two s Complement I To convert back to decimal number we need to do the same procedure 1 Complement each bit 2 Add 1 3 Convert to decimal 0 Ex Convert the binary 1011 in two s complement to the equivalent decimal number 1 0100 complement each bit 2 0101 add 1 3 0101 5 convert to decimal and get the magnitude So we can affirm that 1011 represent negative 5 5 CSE 230 STUDY GUIDE EXAM 2 Chapter 3 Arithmetic for Computers Two s complement How to detect overflow for addition and subtraction Binary integer MultiplicationDivision Multiplication Control WE uso m tip ioamd register Multipiiier o multiplior rogibtorj and product rogistori Product registeer initial Halo is 1 Multipiiier 1 M ul tipliior i5 the right most bit f39th mtl tipiiori 321i MUCHiiiioom ii ii ili liii iii ii i i Mi ii Hr i i rm iiwii ii i ii Min 4 32 repetitions Exampie muitiipiy 011D and 0011 Assume 4bit numbers instead of 32bit numbers liberation Stop irritiipiliioond iiiuitiipilior Product Register value Register value Register value 0 iirri iiiail yoiuoo 0110 i101 1i 0 1539 iteration 11o Prod 21 1 100 i101 1 1 02101 1D Prodliilirrltiiplioondi 215W i uitipiioand by 11 sril iiiuitiiplior by 1 2nd imamamquot 11o Prod 21 01 1000 no 01 1iD 1l1l 2 Prodliilirrltiiplioondi 0100110 oil Huitpiioand by 11 srii multiplier by 1 oii i uitpiioand by 1 G11 0000 0 0100110 sril multiplier by 1 3m it i it39ii ii39li w M w 41h iteration oii i uitpiioand by 11 011 i 0000 01 0010 3 sri multiplier by1 Final Product Control for Division A V v 1 smummtomm Mimmmdphmim Remainder register 15 initialized to Minnquot I I new the D1v1dend value 39 Y Divisor is shifted to left so that the new o gemMrlt 0 left most digit is aligned with quot 39 mamquot 39 the left most digit of the dividend 39 39l 23 wuwmwwwm 2h Rmnlwahltdvdwbywim To decide whether divisor can MWWWW WWNWW39 e su rac e rom remain er mm1mmm we need to subtract and check mmmwnmminow MHWW N100 it the result IS negative or posmve If we cannot subtract V v we need to add the subtracted value I a sun om mam sq back to the remainder 39 NO39 lt 3mm m Yes 33 Was V Example Divide 0000 0111 by 0010 Step Quotient Dillisur Remainder ll ll ll 0 Initial valus man 0010 anon anon am all tn he 8 hit l ividsnd l Rsms smDiv 3300 03100003 1110 111 2b Remso Rem2 im sllll c1 nasal 3300 Minimum man will 3 srll Bill 3300 0331 mm new will 2 17Rsms smlliiv man 0331 com 1111 0111 Zia Rem3 Remaining sill c1 Ginsu man mm mm mm Mill 3 srll Bill man man mm mm am 3 l RsmsRsmDiv man man 1003 1111 1111 2h Rem3 Rem Dilla sill t1 man mm 00001000 mm am 3 srll Bill 3300 000001100 man will 4 l RsmsRsmDiv man man 01100 new 0011 2a Rams20 all a 2021 3301 man mm mm 331 ll 3 srll Bill 0001 00000010 mm 0011 5 1 RsmsRsmDiv 0001 00000010 new 0001 2a Ramaan all a 0021 331 ll 00000010 new 0001 3 sril Bill 0011quot mun 0001 mm 0001 3quot Binary fraction floating point numbers EEE 754 floating point single precision representation double precision representation Biased exponent Binary floating point additionmultiplication Guard digits rounding Section 16 Performance CPI cycles per instruction lt quot if po f M of I r I r J x 1 quot2quot J J r 5 39 V y A x f lt I 2 JJ 4 l 1 ll t r Q 39 l 2 gt r x j I x r I r l l 7 1 39 I f f 1 x 1 malt C1le 12 I may f 4 I A ril o y 7 Chapter 4 Processor Single Cycle Datapath amp Control lw sw Rtype add sub and or slt bne j which datapath rs rt rd and shamt data can take Op and funct code are used for controls Rtypeformat Op 3126 rs 2521 rt 2016 rd 1511 shamt 106 funct 50 typeformat Op 3126 rs 2521 rt 2016 immediate 150 Jtypeformat Op 3126 address 250 be able to add datapaths and controls to the single cycle datapath for a given instruction Contro Signals RegDst RegWrite ALUSrc PCSrc MemRead MemWrite MemtoReg Branch ALUOp ALUOp1 and ALUOp0 We already have some good examples gave by the professor here is some more examples of other datapaths that can be asked implementation on the exam jal Jump And Link Instruction um Jun mo 310 26Vjkm I mamas may auto hubris PM 031 Hr quotIvIt il In Inuucuon 13126 JI IIW IIC Ir Ill 1 JDL r Jquot I n InslmsuonRS 21 K I mg awr l quotand has 36639 Luannnu LN IBI a duo l u rog lov 2 Aw uu W5 and f ALU Lil I a m a 2 0 will Jade ham L V m I u ns39mcunnl 1511 39 Wm 39 I quot39ll Dd Ll il mm 2 39 wmo quot I am InsznmnnHS Jl E0 9 3 2 ouow Aw I ennllm DE 30 f H Insuuwm 15 0l ALUSrcX RegWrite1 MemRead0 MemWriteO Branch0 ALUOp01XX Jump1 j Jump That is what needs to be added dumps 1 str F IEJIJ mp I M M ll l I I RegWrite 1 RegDst X ALUSrc X Branch X MemWrite 0 ALUOpO X ALUOp1X Jump 1 CSE 230 STUDY GUIDE EXAM 3 5 exaspjrion Inerrulofion Edenfs o Hmr H39Iam branches 0quot UMPS H39WJ change Hme novmo Q Wow we Conan m grom Wham IO Device Edema ln6rrap Request An meiic ln l39ernoQ exce mLion Ovevj ow J 9Ari Hlme LC OVer fQow E CGP I On I 11 3 r j I x 15 035 T l54ch39l390 ng EMJL where I here 5 09 oVCVflow lF Husk signal 0 28 con nes in IF Fro39m Brunok F osbi aignql Jro 26m C0nro 3 m ID L9 0 Udi Xe M sisml ram Wc Hazard QDGJFC L DN on Ea f usb 1 abnal o zero CoM39ron a EX b68931 15 6 MISS 120ij His ra ILe oat cache aces when f 16 mgueajrecl word 5 n64 Qun i my we Miss Pena ky Tme requin 4 6 640k 91 f0quot ch nearqu 0 cache 39 gtAeroaa Access pmr Tune O CC CkG Q races L14 17m 0 u v Manama MasPem y P14551136 Am Axes U rime Black 5525 mesa 34am use 6er Eonk 16mm 4w Me 10103 10 Cmth re ux llecu 608 r606 Jr Wm who Lr 6 MQSG g39 15 e x e QC39Dm memory J Cache m4 is W4 Mia Wml lo I 4 f 2 1 3 O may quot10quot 7 ma i 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