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Consider the following table that represents part of the memory of a 16-bit address space that has an addressability of 2 bytes: ADDRESS CONTENTS … … 0x0C10 0101 0010 0111 1011 0x0C0F 0000 0000 1010 0101 0x0C0E 1000 0000 0110 0111 0x0C0D 0110 1010 0001 1111 0x0C0C 0100 0011 0111 0011 0x0C0B 0000 1100 0001 0000 … … The table above shows the addresses in hex (base 16) and the contents at the corresponding address in binary (base 2). A.) Interpret the contents at address 0x0C0C as two ASCII characters. [ascii] B.) Interpret the contents at the same address as A.) above as an unsigned integer in base 10. [unsigned] C.) Assume address 0x0C0F holds the more significant digits of a value (the high 16 bits) and address 0x0C0E holds the least significant digits of the same value (the low 16 bits). Interpret the resultant value an RV32I instruction to determine the operation. (Enter either ADD, JR, LW, or OTHER if it is not one of the first 3) [instr1] Recall that a memory location can store an address. We call that memory location’s contents a “pointer” since it’s an address that “points” to another memory location. D.) Interpret the contents at address 0x0C0B as a pointer.(Enter hex like the following example: 0x2A3F) [ptrvalue] E.) What are the contents of the memory location that the pointer above is pointing to?(Enter hex like the following example: 0x2A3F) [ptevalue]   REFERENCE: Partial Table of Hex to ASCII Characters: 20 sp 30 0 40 @ 50 P 60 ` 70 p 21 ! 31 1 41 A 51 Q 61 a 71 q 22 ” 32 2 42 B 52 R 62 b 72 r 23 # 33 3 43 C 53 S 63 c 73 s 24 $ 34 4 44 D 54 T 64 d 74 t Table of RV32I Opcodes in Hex: ADD 0x33 JR 0x67 LW 0x03

Assume the operating system for our simulator has had new service routines added to it. Consider the following code fragment that requests one of those new service routines: LW a2, %lo(B)(t0) ADDI a0, zero, 34 LOOP: ECALL ADDI a2, a2, -1 Also consider the following partial trap vector table (aka jump table):    ADDRESS      CONTENTS   0x0000_007C 0x0000_0708 0x0000_0080 0x0000_0528 0x0000_0084 0x0000_0914 0x0000_0088 0x0000_0958 A. What address is used by the code above to access the trap vector table?  [trpaddr]B. What is the starting address of the service routine called by the code above?  [srvaddr]C. If the symbol table shows LOOP is address 0x0003_5850, what is the return address from the service routine?  [retaddr]D. What is the machine code for the instruction the service routine uses to return to the code above?  [mc] Enter the addresses and the machine code as 8-digit hex values in RV32I format like the following: 0x0000_01F3

Consider the following table that represents part of the memory of a 16-bit address space that has an addressability of 2 bytes: ADDRESS CONTENTS … … 0x0C10 0101 0010 0111 1011 0x0C0F 1111 1111 1111 0101 0x0C0E 0111 0101 0001 0011 0x0C0D 0110 1010 0001 1111 0x0C0C 0111 0000 0101 0100 0x0C0B 0000 1100 0000 1101 … … The table above shows the addresses in hex (base 16) and the contents at the corresponding address in binary (base 2). A.) Interpret the contents at address 0x0C0C as two ASCII characters. [ascii] B.) Interpret the contents at the same address as A.) above as an unsigned integer in base 10. [unsigned] C.) Assume address 0x0C0F holds the more significant digits of a value (the high 16 bits) and address 0x0C0E holds the least significant digits of the same value (the low 16 bits). Interpret the resultant value an RV32I instruction to determine the operation.(Enter either ADD, JR, LW, or OTHER if it is not one of the first 3) [instr1] Recall that a memory location can store an address. We call that memory location’s contents a “pointer” since it’s an address that “points” to another memory location. D.) Interpret the contents at address 0x0C0B as a pointer.(Enter hex like the following example: 0x2A3F) [ptrvalue] E.) What are the contents of the memory location that the pointer above is pointing to?(Enter hex like the following example: 0x2A3F) [ptevalue]   REFERENCE: Partial Table of Hex to ASCII Characters: 20 sp 30 0 40 @ 50 P 60 ` 70 p 21 ! 31 1 41 A 51 Q 61 a 71 q 22 ” 32 2 42 B 52 R 62 b 72 r 23 # 33 3 43 C 53 S 63 c 73 s 24 $ 34 4 44 D 54 T 64 d 74 t Table of RV32I Opcodes in Hex: ADD 0x33 JR 0x67 LW 0x03

Consider the program below that displays 4 characters. .origin 0x0001_0000 START: lui s0, %hi(START) addi s1, s0, %lo(CHARS) lb a1, 3(s1) addi a0, zero, 10 ecall lw t0, %lo(N)(s0) add t1, s1, t0 lb a1, 0(t1) addi a0, zero, 10 ecall addi t4, s0, %lo(N) lw t0, 4(t4) add t1, s1, t0 lb a1, 0(t1) addi a0, zero, 10 ecall addi t5, s0, %lo(DONE) lb a1, 15(t5) addi a0, zero, 10 ecall DONE: halt N: .word 5 .word 7 CHARS: .string “oadewlts” Trace the execution of the program above and answer the following questions. A. What is the first character that this program displays? [ch1]B. What is the second character that this program displays? [ch2]C. What is the third character that this program displays? [ch3]D. What is the fourth character that this program displays? [ch4]

The following assumes our implementation of the RV32I architecture. Consider following table that represents several of the 32-bit registers in the register file: REGISTER CONTENTS (binary) R10 0000 0000 0000 0000 0000 1010 0001 1000 R11 0000 0000 0000 0000 0000 1010 0010 0000 R12 0000 0000 0000 0000 0000 1010 0010 0100 R13 0000 0000 0000 0000 0000 1100 0010 1000 R14 0000 0000 0000 0000 0000 1100 0011 0100  R15 0000 0000 0000 0000 0000 1100 0011 1000 Also consider the following table that represents part of the memory of a 32-bit address space that has an addressability of 1 byte: ADDRESS (hex) CONTENTS (binary) 0x00000C38 0000 0000 1101 0101 0000 0110 0011 0011 0x00000C34 0000 0001 1111 0101 0010 0110 0000 0011 0x00000C30 0000 0000 0000 0111 0000 0000 0110 0111 0x00000C2C 0000 0000 0000 0111 1000 0000 0110 0111 0x00000C28 0000 0001 1000 0101 0010 0110 1000 0011 … … 0x00000A24 0000 0000 1010 1000 0000 0101 1011 0011 0x00000A20 0000 0001 1001 0101 1010 0110 1000 0011 0x00000A1C 0000 0000 1011 0101 0000 1000 0011 0011 0x00000A18 0000 0000 0000 1000 0000 0000 0110 0111 … … 0x0000033C 0000 0000 0000 0101 1000 0000 0110 0111 0x00000338 0000 0000 0000 0101 0000 0000 0110 0111 0x00000334 0000 0000 0000 0111 0000 0000 0110 0111 0x00000330 0000 0000 0000 0110 0000 0000 0110 0111 … … Assume the PC has the address 0x00000334 when FETCH phase begins. After that first instruction executes, which one describes the second instruction to execute?   REFERENCE: Table of RV32I Instructions: 31 25 24 20 19 15 14 12 11 7 6 0 ADD 0000000 rs2 rs1 000 rd 0110011 JR imm[11:0] rs1 000 00000 1100111 LW imm[11:0] rs1 010 rd 0000011

Complete the following program that only adds integers that are odd positives. The program considers the NON-ZERO integers at labels N1, N2, and N3 and stores the sum of the appropriate integers in the memory at label SUM. It uses a subroutine at label CHKN to check each integer. .origin 0x0001_0000 START: LUI t0, %hi(START) LW a0, %lo(N1)(t0) [missing_line1] L1: ADDI a1, a0, 0 LW a0, %lo(N2)(t0) ADDI a6, t0, %lo(CHKN) [missing_line2] JR a6 L2: ADD a1, a1, a0 LW a0, %lo(N3)(t0) [missing_line3] L3: ADD a1, a1, a0 SW a1, %lo(SUM)(t0) HALT CHKN: ADDI a4, a0, 0 # move in to a4 for testing [missing_line4] ANDI a4, a4, 0x0001 # right most bit is always 1 if odd [missing_line5] WRONG: ANDI a0, a0, 0 # zero out wrong values RETURN: RET N1: .word # value not shown N2: .word # value not shown N3: .word # value not shown SUM: .word 0

Interpret the following 7 bit binary sequence  1101001  as specified below: base 10 value for unsigned integer [unsigned] base 10 value for sign magnitude integer [signmag] base 10 value for  fixed-point real number with 2 digits to the right of the binary point. [fixpnt] character value for ASCII [ascii] Partial Hex to ASCII Table: 20 sp 30 0 40 @ 50 P 60 ` 70 p 21 ! 31 1 41 A 51 Q 61 a 71 q 22 ” 32 2 42 B 52 R 62 b 72 r 23 # 33 3 43 C 53 S 63 c 73 s 24 $ 34 4 44 D 54 T 64 d 74 t 25 % 35 5 45 E 55 U 65 e 75 u 26 & 36 6 46 F 56 V 66 f 76 v 27 ‘ 37 7 47 G 57 W 67 g 77 w 28 ( 38 8 48 H 58 X 68 h 78 x 29 ) 39 9 49 I 59 Y 69 i 79 y 2A * 3A : 4A J 5A Z 6A j 7A z 2B + 3B ; 4B K 5B [ 6B k 7B { 2C , 3C < 4C L 5C \ 6C l 7C | 2D - 3D = 4D M 5D ] 6D m 7D } 2E . 3E > 4E N 5E ^ 6E n 7E ~ 2F / 3F ? 4F O 5F _ 6F o 7F del

The skin is deep to the bones of the body.

Match the plane of motion with the associated axis of rotation 

Please select whether this statement is true or false for your experience today. I was able to access and use the calculator function on this Respondus Browser Quiz.