Explain specifically why the following instruction set may not work correctly. mov ах, -10 mov сх, 7 imul CX mov qword [qAns], rax You may assume the variable is declared appropriately.
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![Question 14
Explain specifically why the following instruction set may not work correctly.
mov
ах, -10
mov
сх, 7
imul
CX
mov
qword [qAns], rax
You may assume the variable is declared appropriately.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F32a1012c-241a-4e79-abfd-64c33aa6f411%2Fcf3fa3a7-6ad8-4032-950c-88d1c5f4b0b8%2F47o7yz2_processed.jpeg&w=3840&q=75)
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- MIPS Simulator QtSpim: You are to have a complete program in MIPS assembly language that behaves exactly as the included C program. This program contains four functions in addition to the main() one. Your solution must contain all five C routines as they have been coded in the example. Make sure to run the program in MIPS and show the same output on MIPS as well to make sure there are no errors. Below is the five C routines and attached is the image of what the output must print out on QtSpim. #include <stdio.h> int getMax(int arr[], int n){int mx = arr[0];for (int i = 1; i < n; i++)if (arr[i] > mx)mx = arr[i];return mx;}void countSort(int arr[], int n, int exp){int output[n];int i, count[10] = { 0 };for (i = 0; i < n; i++)count[(arr[i] / exp) % 10]++;for (i = 1; i < 10; i++)count[i] += count[i - 1];for (i = n - 1; i >= 0; i--) {output[count[(arr[i] / exp) % 10] - 1] = arr[i];count[(arr[i] / exp) % 10]--;}for (i = 0; i < n; i++)arr[i] = output[i];}void…Mutual Exclusion: Select all of the following statements that are true. The Peterson Algorithm is a purely software-based solution to the mutual exclusion problem. The 'test_and_set' instruction is atomic. Mutex Locks are an API offered by operating systems to protect critical sections. The Peterson Algorithm does not rely on the mechanism of busy waiting. In order to enter a critical section protected by Mutex Locks, the release() operation must be executed. F3 F4 -!!! F5 F6 8 F8 (DELL) F9 W prt sc F10 home F11 > end F1210.- Consider the following code and pictureLoop1MOVLW 0x32MOVWF REG2DECFSZ REG2,FGOTO LOOP1 If the system clock frequency is aMHz and each machine cycle consumes 4 cycles of it. Calculate the delay time of the LOOP1 loop. Include the execution difference time of the DECFSZ instruction in the last cycle. Select an answerA) 0.6.sB) 6msC)6usD) 60us
- PLEASE DON'T USE STRUCTS Using C programming language write a program that simulates a variant of the Tiny Harvard Architecture. In this implementation memory (RAM) is split into Instruction Memory (IM) and Data Memory (DM). Your code must implement the basic instruction set architecture (ISA) of the TinyMachine Architecture: 1 -> LOAD 2->ADD 3->STORE 4 -> SUB 5 ->IN 6-> OUT 7-> END 8 -> JMP 9 -> SKIPZ Each piece of the architecture must be accurately represented in your code (Instruction Register, Program Counter, Instruction Memory (IM), MAR1, MDR-1(MAR-1 and MDR-1 are connected to the IM). Data Memory, MAR-2, MDR2 (MAR-2 and MDR-2 are connected to the DM), and Accumulator. Instruction Memory will be represented by an integer array and each instruction will use 2 elements of the array(one for OP and the other one for address) Data Memory will be represented by an integer array and each data value uses an element of the DM array. Your Program Counter will begin pointing to the first…Memory Address Memory Content Øx20000013 Øx79 Øx20000012 Өхcd Øx20000011 Øxa3 Øx20000010 Oxfd ØX2000000F Øx12 According to the memory view given below, if RO = 0x20000004, after executing the LDMIA r0, {r7, r3, r9} instruction, which of the following are the contents of the r3 and r0 registers, respectively (data layout big endian)? Øx2000000E Øxod ØX2000000D Øxe0 Øx2000000C Өxeb ØX2000000B Øx2c ØX2000000A Øx1a Øx20000009 Øxdc Øx20000008 Ox11 Øx20000007 Өx34 Øx20000006 Øxce Øx20000005 Øxfd Øx20000004 Øxe4 Øx20000003 Øxa2 Øx20000002 Øx3f Øx20000001 Øx06 Øx20000000 Øxf5 Lütfen birini seçin: O A. R3= 0X2C1ADC11 ro =0x20000010 O B. R3= 0X34CEFDE4 r0 =0x20000010 O C. R3= 0X11DC1A2C ro =0x2000000C O D. R3= 0XE4FDCE34 ro =0x20000010 O E. R3= 0XE4FDCE34 ro =0x20000004The codes below are disassembled from a program. Write an invalid target for the “jmp” instruction such that the code can be considered as anti-disassembly. Jmp ??? loc_2: call 15FF2A1h OR [ecx], d1 inc eax jmp short loc_3 db 0E8h Please just give me a simple answer
- It is the compiler's job to associate program variables with registers. Take, for instance, the assignment statement from our earlier example: f (g + h) (i + j): The variables f, g, h, i, and j are assigned to the registers X19, X20, X21, X22, and X23, respectively. What is the compiled LEGv8 code?What is the function of this instruction "WDTCTL a) it configures and sets the watchdog timer b) it stops the watchdog timer c) it sets the watchdog timer d) it configures and clears the watchdog timer = WDTPW | WDTCONFIG", where **#define WDTCONFIG (WDTCNTCL WDTSSEL)**creating an assembly program that performs addition and subtraction write assembly code that does the following: Create a DWORD variable called "sum". Its initial value doesn't matter. Set the registers eax, ebx, ecx, and edx to whatever values you like Perform the following arithmetic: (eax + ebx) - (ecx + edx) Move the result of the above arithmetic into the sum variable. I am attaching an image of starting point of the code
- It is the compiler's job to associate program variables with registers. Take, for instance, the assignment statement from our earlier example: f = (g + h) (i+j): The variables f, g, h, i, and j are assigned to the registers X 19, X20, X21, X22, and X 23, respectively. What is the compiled LEGV8 code?Assume a CPU with a fixed 32-bit instruction length has the following instruction format:opcode mode [operand1] [operand2] [operand3]The mode encodes the number of operands and each operand’s mode. For instance, one mode indicates three registers, another indicates two registers and an immediate datum, another indicates a main memory reference, etc. Assume there are 94 instructions and 22 modes. Answer the following.a. One mode indicates three registers. How many registers can be referenced in this mode?b. One mode indicates two registers and an immediate datum in two’s complement. Assuming there are 32 registers, what is the largest immediate datum that can be referenced?c. One mode has a destination register and a source memory address (an unsignednumber). Assuming 16 registers, what is the largest memory reference available?d. One mode has two memory addresses, both using base displacement. In both, the basesare stored in index registers and the displacements are specified in the…37. Describe the operation that is performed by the following instruction sequence. MOV BL, [CONTROL_FLAGS] AND BL, 08H XOR BL, 08H MOV [CONTROL_FLAGS], BL
