The following program is written in the assembly language of 8085 processor: LXI B, 20DF LXI H, 2B44 MOV A, C SUB B MOV A, L ADD H INX B DCX H HLT (a) Complete Table 1 with the machine codes and memory addresses for the whole program by using 8085.jar simulator. (b) Determine the total number of bytes needed to execute the whole program. (c) State final value of all registers involved in the program.
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- If a microprocessor has a cycle time of 0.5 nanoseconds, what’s the processor clock rate? If the fetch cycle is 40% of the processor cycle time, what memory access speed is required to implement load operations with zero wait states and load operations with two wait states?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…Quesde: 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. (e) 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. (e) 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…Complete 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).Q: The contract between hardware and software is known as Instruction set architecture. Explain the working of registers with two main process of Von-Neuman architecture and highlight ISA in this working. Note: this question is related from computer architecture subject kindly solved this correctly and completly.
- Please using assembly language 8086 microprocessor Q5/Tow Matrixes 4X4 the first matrix has first-row storage from 220H to 223H, the second row starts from 430H to 433 H, the third row starts from 550H to 553H, and the fourth row starts from 661H to 664H at DS-OC30H. The second matrix has first-row storage from 336H to 339H, the second row starts from 546H to 549H, the third row starts from 666H to 669H, and the fourth row starts from 781H to 784H at DS-OB25H. Write a program to find their summation.Q: The contract between hardware and software is known as Instruction set architecture. Explain the working of registers with two main process of Von-Neuman architecture and highlight ISA in this working.Convert the C code below into RISC-V assembly language for the following two scenarios: a) Variable fis assigned to register x5. b) Variable fis assigned to register x20 Arguments a and b are available in x10 and x11 according to RISC-V convention. The return value is stored in x10 according to RISC-V convention. Return address is available in x1. Note: avoid unnecessary write operations to Stack memory long long int leaf_example (long long int a, long long int b) { long long int f; if (a == b) f=a + b; else f=a – b; return f; }
- Microprocessor Hw Q1 Execute the following code and show the contents of the registers: LDI R16,$03 LDI R17,$10 HERE: AND R16, R17 BREQ HERE ADD R16,178, Provide the type, assembly language instruction, and binary representation of instruction described by the following MIPS fields: op = 0, rs = 16, rt= 15, rd = 25, shamt = 0, function = 34For the MIPS assembly instructions below, what is thecorresponding C statement? Assume that the variables f, g, h, i, and j areassigned to registers $s0, $s1, $s2, $s3, and $s4, respectively. Assume thatthe base address of the arrays A and B are in registers $s6 and $s7,respectively. Note: for each line of MIPS code below, write the respective Ccode. After that, write the corresponding C code for the MIPS.sll $t0, $s0, 2add $t0, $s6sll $t1, $s1, 2 add $t1, $s7, $t1lw $s0, 0($t0)addi $t2, $t0, 4lw $t0, 0($t2)add $t0, $t0, $s0sw $t0, 0($t1)