Consider the MIPS instruction sequence shown in Figure2 (where RO, R1, R2, R3, R4, and R6 are 32-bit general-purpose registers): (a). Identify the instructions that are supported by the write-back (WB) stage of the system. (b). What is the memory address of the store instruction at line 2 (assume that R2 and R3 are initialized with Ox00000008, and Ox00000005 respectively)? Need a short quick answer. i 1. ADD R1, R2, R3 2. SW R4, 5 (R1) 3. LW R4, 8 (RO) 4. SUB R3, R4, R6 5. SW R2, 9 (R3) 6. LW R4, 8 (R2)
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- A(n) ________________ instruction always alters the instruction execution sequence. A(n) ______________ instruction alters the instruction execution sequence only if a specified Condition is true.QuedT: Choose the correct answer: [ Opcode, funct3 and funct7/6 in instruction format are used to identify the: (a) function. (b) instruction. (e) branch. (d) memory address. The register that hold the address of the current instruction being executed is called: (a) saved register. (b) global pointer. (e) stack pointer. (d) program counter. Placing the executable file into the memory for execution by the processor is the role of (a) assembler. (b) linker. (e) loader. (d) compiler. The part which responsible for transmitting the data to/from the processor is: (a) control unit. (b) Datapath. (c) data bus. (d) memory. Parallel hardware cannot be used for faster division because: (a) subtraction is conditional on sign of remainder. (b) multiplication is conditional on sign of remainder. (c) subtraction is conditional on sign of divisor. (d) multiplication is conditional on sign of divisor. we cannot slower the clock cycle to fit the floating-point adder algorithm into one clock cycle…Choose the correct answer: Opcode, funct3 and funct7/6 in instruction format are used to identify the: ● (a) function. (b) instruction. (c) branch. (d) memory address. The register that hold the address of the current instruction being executed is called: (a) saved register. (b) global pointer. (e) stack pointer. (d) program counter. Placing the executable file into the memory for execution by the processor is the role of: (a) assembler. (b) linker. (e) loader. (d) compiler. ● The part which responsible for transmitting the data to/from the processor is: (a) control unit. (b) Datapath. (c) data bus. (d) memory. ● ● Parallel hardware cannot be used for faster division because: (a) subtraction is conditional on sign of remainder. (b) multiplication is conditional on sign of remainder. (c) subtraction is conditional on sign of divisor. (d) multiplication is conditional on sign of divisor. ● we cannot slower the clock cycle t fit the floating-point adder algorithm into one clock cycle…
- Computer Science A computer uses a memory of 64K words with 16 bits in each word.It has the following registers: PC, AR, TR, AC, DR and IRA memory-reference instruction consists of two words: an 16-bit operation-code(one word) and an address field (in the next word).a-List the sequence of microoperations for fetching a memory reference instructionand then placing the operand in DR. Start from timing signal To.b-Design the logic control gates arrangement to perform the fetch instructions.Consider the MIPS instruction sequence shown in Figure2 (where RO, R1, R2, R3, R4, and R6 are 32-bit general-purpose registers): (a). Identify the instructions that are supported by the write-back (WB) stage of the system. (b). What is the memory address of the store instruction at line 2 (assume that R2 and R3 are initialized with Ox00000008, and 0x00000005 respectively)? Need a short quick answer. i 1. ADD R1, R2, R3 2. SW R4, 5 (R1) 3. LW R4, 8 (RO) 4. SUB R3, R4, R6 5. SW R2, 9 (R3) 6. LW R4, 8 (R2)A subtraction instruction takes two operands, subtracts the first from the second, and the result goes into the second. Write a 16-bit subtraction instruction, where... The first operand is stored at the memory address contained within the %edx register. (NOTE: To be clear, the %edx register contains the memory address OF the operand, not the operand itself! Recall the syntax of "indirect addressing," to use here.) The second operand -- also where the result shall be stored -- is stored directly in the 16-bit %bp register. Type the appropriate assembly language instruction here:
- Suppose that the following instructions are found at the given locations in memory: 20 LDA 50 21 BRP 22 22 STO 51 50 100 51 100 Choose the contents of the registers: PC, MAR, MDR, IR, A at the end of fetch-execute cycle for instruction 22: PC MAR MDR IR AComplete the following table: MIPS Instruction op code rs rt rd shamt funct imm. /address Hexadecimal Representation add $t4, $s2, $s1 addi $s0, $t0, 123 lw $s6, -88($t7) Note: In MIPS register file, temporary registers $t0-$t7 have indices 8-15 (respec- tively). Also, the saved registers $s0-$s7 have indices 16-23 (respectively).Determine the specific type of addressing mode (SMALL LETTERS only) and compute for the address/es. If applicable, determine the content of the destination after the execution of the instruction. Otherwise, NA. For physical address and content of the destination, use CAPITAL LETTERS,NO SPACE/S in between and NO need to include "H" or the unit. If the given register are fewer than required, just put zeroes in the most significant place. A -effective address B - physical address/es e.g. 19000-19001 C- is for addressing mode type, e.g. direct D- content of the destination after execution e.g. if register: AX=1234 if memory (lower address first): 12000=34;12001=12 5(2).JPG AX =6FFAH DX =BFA8H SS = 6FFEH SI = 2BCAH IP = FADEH ВХ %-9CF5H CS =5FFEH ES = 8DBEH BP = 6EEFH CX =3FFEH DS =AF9EH DI = 5EC9H SP = 9AABEH MOV [ESI*2 + CBECH], EDX EFFECTIVE ADDRES: Blank 1 PHYSICAL ADDRESS/ES: Blank 2 ADDRESSING MODE : Blank 3 CONTENT OF THE DESTINATION: Blank 4
- Determine the specific type of addressing mode (SMALL LETTERS only) and compute for the address/es. If applicable, determine the content of the destination after the execution of the instruction. Otherwise, NA. For physical address and content of the destination, use CAPITAL LETTERS,NO SPACE/S in between and NO need to include "H" or the unit. If the given register are fewer than required, just put zeroes in the most significant place. A -effective address B - physical address/es e.g. 19000-19001 C- is for addressing mode type, e.g. direct D- content of the destination after execution e.g. if register: AX=1234 if memory (lower address first): 12000=34;12001=12 3(2).JPG ... AX =6FFAH DX =BFA8H SS = 6FFEH SI %3D 2BCАН IP = FADEH ВХ -9CF5H CS =5FFEH ES = 8DBEH BP = 6EEFH CX =3FFEH DS =AF9EH DI = 5EC9H SP = 9AABEH MOV [BX+SI+ADBCH], ESP EFFECTIVE ADDRES: Blank 1 PHYSICAL ADDRESS/ES: Blank 2 ADDRESSING MODE : Blank 3 CONTENT OF THE DESTINATION: Blank 4Determine the specific type of addressing mode (SMALL LETTERS only) and compute for the address/es. If applicable, determine the content of the destination after the execution of the instruction. Otherwise, NA. For physical address and content of the destination, use CAPITAL LETTERS,NO SPACE/S in between and NO need to include "H" or the unit. If the given register are fewer than required, just put zeroes in the most significant place. A -effective address B - physical address/es e.g. 19000-19001 C - is for addressing mode type, e.g. direct D - content of the destination after execution e.g. if register: AX=1234 EFFECTIVE ADDRES: PHYSICAL ADDRESS/ES: ADDRESSING MODE : CONTENT OF THE DESTINATION: if memory (lower address first): 12000=34;12001=12Determine the specific type of addressing mode (SMALL LETTERS only) and compute for the address/es. If applicable, determine the content of the destination after the execution of the instruction. Otherwise, NA. For physical address and content of the destination, use CAPITAL LETTERS,NO SPACE/S in between and NO need to include "H" or the unit. If the given register are fewer than required, just put zeroes in the most significant place. A -effective address B - physical address/es e.g. 19000-19001 C - is for addressing mode type, e.g. direct D - content of the destination after execution e.g. if register: AX=1234 if memory (lower address first): 12000=34;12001=12 EFFECTIVE ADDRES: PHYSICAL ADDRESS/ES: ADDRESSING MODE : CONTENT OF THE DESTINATION: