Write MIPS assembly code implementing the following C/C++ statement: a[k*2-4] = a[k]*2 + 7;
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You may use only the instructions we covered in class: add, sub, addi, lw, sw, and the same rules as mentioned earlier still apply
Write MIPS assembly code implementing the following C/C++ statement:
a[k*2-4] = a[k]*2 + 7;
You may assume that no bounds checking is needed.
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- Q2/Write a program with a specific programming language to insert a user program into memory and allocate locations for it using Reg Fence once and use relocatio again e 1:41 Q3/Apply that if the program is 5kb, 22kb, or 40kb, what will happen in the implementation of C ++ language 2 1:41Note: assembly language(8086 microprocessor) Write a program that takes a number N (2 digit decimal) as input. Then the user will enter that many numeric values ranging from 0 to 9 as input. Here the user may enter a negative number too. The program then finds a pair of values whose summation is closest to zero.Explanation: Here at first the user enters 10, which is a 2 digit decimal number. That means the user now has to enter 10 negative or positive digits ranging from 0 to 9. Then the program will calculate the sum of all possible pairs and find the pair which gives the minimum sum. Here from the given example, we see that the sum of -2 and -4 which is -6 is the least sum.Input:10 -2, 3, 7, 4, -4, 7, -8, 0, 9, 9 Output: Pair of values which has the smallest sum = -4, -2Lab Task: You are required to make changes in the above programs and introduce the use of compaction where required. Write code to simulate Worst-Fit Algorithm Write a C program to simulate the MVT and MFT memory management techniques. Write a C program to simulate the following contiguous memory allocation techniques Worst-fit Best-fit First-fit TASK1: DESCRIPTION MFT (Multiprogramming with a Fixed number of Tasks) is one of the old memory management techniques in which the memory is partitioned into fixed size partitions and each job is assigned to a partition. The memory assigned to a partition does not change. MVT (Multiprogramming with a Variable number of Tasks) is the memory management technique in which each job gets just the amount of memory it needs. That is, the partitioning of memory is dynamic and changes as jobs enter and leave the system. MVT is a more ``efficient'' user of resources. MFT suffers with the problem of internal fragmentation and MVT suffers with…
- *Written in MASM Assembly 80x86 no c++ no python etc. allowed even if it supports** There will be a function called getdouble. This function will simply double any number which is currently in eax and store the result in eax. There will be a function called gettriple. This function will simply triple any number which is currently in eax and store the result in eax. There will be a function called getoddeven. This function will check if the value in eax is even. IF it is even, it will call the getdouble function. IF it is odd, it will get the gettriple function. (Note: edx stores remainder after you divide) Your main program should ask the first user for a name as well as for a number. You should then call the getoddeven function. That function will either double or triple the initial value entered by the user. Display the name and the final result for this first user. Your program will then do the same for a second user for a name as well as for a number. You will again call the…The following instruction set is supported by a simple processor, which is similar to what we discussed in the class, with a few new instructions added. The format of most instructions is defined as follows. bits 15:14 13:10 9 8:6 5:3 2:0 field unused opcode w srcl src2 dst where the fields are defined as follows. opcode : operation to be performed by the processor write back ALU output to register file (1= yes, 0 = no) address of the first ALU operand in the register file address of the second ALU operand in the register file address in the register file where the output is written w: srcl: src2: dst: For opcodes BEQ, BLEZ and JUMP, the 6 least significant bits (5:0) give an address in the instruction memory, which is byte-addressed. The opcode HALT has all operand bits (9:0) being 0. When an instruction has only two operands, the field for the unused operand is filled with 0-bits. For example, bits (5:3) for SLL are all zero because src2 is not used. The opcode and meaning of these…Example: The Problem Input File Using C programming language write a program that simulates a variant of the Tiny Machine 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 Tiny Machine Architecture: //IN 5 //OUT 7 //STORE O //IN 5 //OUT 7 //STORE 1 //LOAD O //SUB 1 55 67 30 55 67 1 LOAD 2- ADD 3> STORE 4> SUB 5> IN 6> OUT 7> END 8> JMP 9> SKIPZ 31 10 41 30 //STORE O 67 //OUT 7 11 /LOAD 1 //OUT 7 //END 67 70 Output Specifications Each piece of the architecture must be accurately represented in your code (Instruction Register, Program Counter, Memory Address Registers, Instruction Memory, Data Memory, Memory Data Registers, and Accumulator). Data Memory will be represented by an integer array. Your Program Counter will begin pointing to the first instruction of the program. Your simulator should provide output according to the input file. Along with…
- Answer the given question with a proper explanation and step-by-step solution. Prime Counter Implement an OpenMP program that counts the number of primes between 1 and N, including N. The following is a set of examples to test your code: Please provide a C++ and C code formatI need a unique solution for the following question: Note: please use stack in C language. In this project you will implement a program that simulates a simple text editor with the undo/redo functionalities allowed. The program will accept statements, one at a time with a maximum of 100 characters per line.There will be some special commands as follows:1. undo: this will undo (i.e., remove) that last entered statement;2. redo: this will redo (i.e., re-add) that last removed statement;3. print: this command will print the entire stored input text;4. save: will save the text to a file called (output.txt);5. quit: will exit the program and save all results to output.txtExecution example:MyCommand > This is a test inputMyCommand > COMP2421 – Data structures & AlgorithmsMyCommand > test1MyCommand > printresult > This is a test inputCOMP2421 – Data structures & Algorithmstest1MyCommand > undoresult > This is a test inputCOMP2421 – Data structures &…Instruction: Explain the function of the program line by line thoroughly.Program: #include <iostream> using namespace std; //Swap functionvoid swap(int *xp, int *yp){ int temp = *xp; *xp = *yp; *yp = temp;} void selectionSort(int arr[], int n){ int i, j, min_idx; // One by one move boundary of // unsorted subarray for (i = 0; i < n-1; i++) { // Find the minimum element in // unsorted array min_idx = i; for (j = i+1; j < n; j++) if (arr[j] < arr[min_idx]) min_idx = j; // Swap the found minimum element // with the first element if(min_idx!=i) swap(&arr[min_idx], &arr[i]); }} //Function to print an arrayvoid printArray(int arr[], int size){ int i; for (i=0; i < size; i++) cout << arr[i] << " "; cout << endl;} int main(){ int x, n; cout << "Enter number of integers: "; cin >>n; int *arr = new int(n); for (int i = 0; i < n; i++) { cout << " Enter Integer No. " << (i + 1) << ": ";…
- B1: Implementing Arithmetic Expression ( i ) Armed with the ADD, SUB, and NEG instructions, we have the means to implement arithmetic expressions involving addition, subtraction, and negation in assembly language. In other words, we can simulate what a C++ compiler might do when reading an expression such as Rval = -Xval + (Yval - Zval); Based on the above arithmetic expression, we can construct the assembly program by the following steps: The following signed 32-bit variables will be used: Rval SDWORD ? Xval SDWORD 26 Yval SDWORD 30 Zval SDWORD 40 When translating an expression, evaluate each term separately and combine the terms at the end. First, we negate a copy of Xval: ; first term: -Xval mov eax, Xval neg eax ; EAX = -26 Then Yval is copied to a register and Zval is subtracted: ; second term: (Yval - Zval) mov ebx, Yval sub ebx, Zval ; EBX = -10 Finally, the two terms (in EAX and EBX) are added: ; add the terms and store: mov Rval, eax add Rval, ebx ; Rval = -36…Exercise 4: Write the MIPS compiler of the following C Code: int sum (int a, int b, int c) { int tot; tot=a+b+c return tot; }Project Summary: Write an Intel 8086 Assembly program that reads N numbers as Strings, convert them into variable sized Integer numbers, and then print the summation and average of the numbers. The program should allow the user to decide the size of the input number itself (assume integers in format but with variable size). Detailed Description: - Have the program prompt the user to input N and the size of the number then request inputting the first number, then the second and so on until N numbers are input. - Your code should allow users to select the size of the numbers, for example you can have integers with size of 1 Byte each, or you can make them 10 Bytes large. - Validation: Your code should make sure user inputs Decimal numbers only, and with predetermined size only. When a user inputs a wrong value, your code should print an error message that explains it, and then gives the user another chance to input it correctly. - When the user presses Enter, your code should read the…