CS ECE 252 INTRODUCTION TO COMPUTER ENGINEERING COMPUTER SCIENCES DEPARTMENT UNIVERSITY OF WISCONSIN MADISON Prof David A Wood TAs Spyros Blanas Priyananda Shenoy Shengnan Wang Midterm Examination 3 In Class 50 minutes Monday April 21 2008 Weight 15 CLOSED BOOK NOTE CALCULATOR PHONE COMPUTER The exam is two sided and has 10 pages including two blank pages and a copy of the LC 3 Instruction Set handout on the final page please feel free to detach this final page but insert it into your exam when you turn it in You are required to present a valid UW Madison student ID card or other government issued photo ID to one of the teaching assistants who are proctoring this exam before leaving the room If you fail to do so we cannot grade your exam Plan your time carefully since some problems are longer than others NAME ID 1 Problem Number Maximum Points 1 2 3 4 5 6 7 12 9 15 24 20 10 10 Total 100 Points Awarded Problem 1 12 points The following LC 3 program has been loaded into memory Address Instruction x5000 1001 0010 0011 1111 x5001 0001 0010 0110 0001 x5002 0001 0010 0100 0010 x5003 0000 0110 0000 0110 a The contents of the registers are shown below R0 is 4 R1 is 3 R2 is 0 R3 is 5 Which condition codes are set and what is the value of the PC after instruction at x5003 is executed b The contents of the registers are shown below R0 is 3 R1 is 0 R2 is 5 R3 is 4 Which condition codes are set and what is the value of the PC after instruction at x5003 is executed 3 Problem 2 9 points The contents of the memory and the registers are shown below Register Value Memory address Value R0 x3012 x3010 x3012 R1 x002A x3011 x3010 R2 x300B x3012 xBEEF We start executing from address x3000 What s the value in R1 after each instruction is executed a If x3000 is 0010 0010 0001 0000 LD R1 PTR b If x3000 is 0110 0010 1000 0111 LDR R1 7 R2 c If x3000 is 1010 0010 0001 0000 LDI R1 PTR Problem 3 15 points The program below sums a sequence of numbers The address to the start of the sequence is stored in R2 and the number of elements in that sequence is stored in R3 Insert the missing LC 3 machine language instructions Adding comments to each machine language instruction will assist in awarding partial credit Address Instruction x5000 0101 1111 1110 0000 AND R7 R7 0 x5001 0110 0010 1000 0000 LDR R1 R2 0 x5002 0001 1111 1100 0001 ADD R7 R7 R1 x5003 x5004 0001 0110 1111 1111 ADD R3 R3 1 x5005 x5006 1111 0000 0010 0101 HALT 5 Problem 4 24 points There is something wrong with the following code This code is supposed to count the number of positive numbers in a sequence save this number in R7 and then exit The sequence always has 10 elements and the address of the start of the sequence has been stored in register R2 Address Instruction x3005 0101 0000 0010 0000 AND R0 R0 0 x3006 0001 0000 0010 1010 ADD R0 R0 10 x3007 0101 1111 1110 0000 AND R7 R7 0 x3008 0110 0010 1000 0000 LDR R1 R2 0 x3009 0000 0010 0000 0001 BRp 1 x300A 0001 1111 1110 0001 ADD R7 R7 1 x300B 0001 0100 1010 0001 ADD R2 R2 1 x300C 0001 0000 0011 1111 ADD R0 R0 1 x300D 0000 0011 1111 1010 BRp 6 x300E 1111 0000 0010 0101 HALT Explain what happens when we try to execute this code Comments are provided to save you the effort of decoding the machine language If you had a debugger briefly describe how you would use it to debug this program Problem 5 20 points The current state of the memory is given below Memory Address x3006 xABCD x2FFF x1220 xABDB x30F3 x200E x3258 x300E Memory Contents xABCD x1220 x4567 x9876 x0001 x0020 x3258 x0000 x92FE x3005 x200E We load and execute the following program Address Instruction x3000 1010 0000 0000 0101 x3001 0110 0110 0000 0000 LDR R3 R0 x0 x3002 1110 0010 1111 0000 x3003 0010 0101 1111 1110 LD R2 x1FE x3004 1111 0000 0010 0101 What will be the final contents of registers R0 R3 when we reach the HALT instruction Write your answers in hexadecimal format Register R0 R1 R2 R3 Initial contents x200E x200E x3001 x3001 Final contents 7 Problem 6 10 points If the value stored in R0 is 1 at the end of the execution of the following instructions what can be inferred about R3 Address Instruction x4000 1110 0011 1111 1111 x4001 0101 0000 0010 0000 AND R0 R0 x0 x4002 0001 0100 0100 0001 x4003 0101 0110 1000 0011 x4004 0000 0100 0000 0001 x4005 0001 0000 0010 0001 ADD R0 R0 x1 a R3 is negative b R3 is positive c R3 is equal to 0 d R3 is zero or positive Problem 7 10 points What does the following instruction sequence do Address Instruction x4000 1110 0000 0001 0000 x4001 1100 0000 0000 0000 Can you do the same using a single instruction only If yes which one If no why Scratch Sheet 1 in case you need additional space for some of your answers 9 Scratch Sheet 2 in case you need additional space for some of your answers LC 3 Instruction Set Entered by Mark D Hill on 03 14 2007 last update 03 15 2007 PC incremented PC setcc set condition codes N Z and P mem A memory contents at address A SEXT immediate sign extend immediate to 16 bits ZEXT immediate zero extend immediate to 16 bits Page 2 has an ASCII character table 15 14 13 12 11 10 9 8 7 6 5 4 3 2 1 0 ADD DR SR1 SR2 Addition 0 0 0 1 DR SR1 0 0 0 SR2 DR SR1 SR2 also setcc ADD DR SR1 imm5 Addition with Immediate 0 0 0 1 DR SR1 1 imm5 DR SR1 SEXT imm5 also setcc AND DR SR1 SR2 Bit wise AND 0 1 0 1 DR SR1 0 0 0 SR2 DR SR1 AND SR2 also setcc AND DR SR1 imm5 Bit wise AND with Immediate 0 1 0 1 DR SR1 1 imm5 DR SR1 AND SEXT imm5 also setcc BRx label where x n z p zp np nz nzp Branch 0 0 0 0 n z p PCoffset9 GO n and N OR z AND Z OR p AND P if GO is true then PC PC SEXT PCoffset9 JMP BaseR Jump 1 1 0 0 0 0 0 BaseR 0 0 0 0 0 0 PC BaseR JSR label Jump to Subroutine 0 1 0 0 1 PCoffset11 R7 PC PC PC SEXT PCoffset11 JSRR BaseR Jump to Subroutine in Register 0 1 0 0 0 0 0 BaseR 0 0 0 0 0 0 temp PC PC BaseR R7 temp LD DR label Load PC Relative 0 0 1 …