Using a system call to print a string of text, Change this MIPS code to RISC-V Assembly code.
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Using a system call to print a string of text, Change this MIPS code to RISC-V Assembly code.
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- C++ LANGUAGE Dynamic Memory Allocation Practice I Write a program that swaps the values of X and Y with malloc. Output Before swap X:412 Before swap Y: 623 After swap X: 623 After swap Y:·412MIPS 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…*C PROGRAM FOR MICROPROCESSOR* Write a C program that will use 7-segement LEDS to display the number of key on the keypad when it is pressed. The display needs to stay on the 7-segment LEDS until the next key is pressed. Interrupt method is not required.
- In simple words, describe the final data pointer register.Select common examples of when an assembly programmer would want to use the stack: to pass arguments to save return address for CALL local variables temporary save area for registers applications which have FIFO nature, such as customers waiting in a bank queueint total; int i; total = 0; for (i = 10; i > 0; i--) { total + i; } The following translates this into ARM assembler: /* -- sum-to-ten.s */ .text .global start _start: mov rº, #0 mov r1, #10 again: end: add r0, r0, r1 subs r1, r1, #1 bne again mov r7, #1 swi 0 @r0 := 0 @r1 = 10 @r0 := r0 + r1 @ r1 := r1 - 1 @loop again if we haven't hit zero @setup exit @exit In the above assembly language program, we first use the MOV instruction to initialize RO at O and R1 at 10. The ADD instruction computes the sum of RO and R1 (the second and third arguments) and places the result into RO (the first argument); this corresponds to the total += i; line of the equivalent C program. Note the use of the label at the beginning of the ADD instruction. The label is simply the relative memory address of where that instruction is when the assembler loads it into memory. The subsequent SUBS instruction decreases R1 by 1. To understand the next instruction, we need to understand that in addition to the…
- Topic: Assembly Language using 8086 emulator, exe template Please write the following code in Assembly Language and use comments to explain.Microprocessor lab tasks - Solve the problems in assembly language using emu8086 solution format .MODEL SMALL .STACK 100H .DATA ; DEFINE YOUR VARIABLES HERE .CODE MAIN PROC MOV AX, @DATA MOV DS, AX ; YOUR CODE STARTS HERE ; YOUR CODE ENDS HERE MOV AX, 4C00H INT 21H MAIN ENDP END MAIN Collapse :white_tick: 1 Problems : Task 01 a program that takes in 3 digits as input from the user and finds the maximum Sample input: 1st input: 1 2nd input: 2 3rd input: 3 Sample Output: 3 Task 02 Take two digits as input from the user and multiply them. If the result is divisible by 2 and 3 both, print "Divisible". Otherwise, print "Not divisible" Sample input: 1st input: 5 2nd input: 6 Result is 30 Sample Output: Divisible Sample input: 1st input: 5 2nd input: 2 Result is 10 Sample Output: Not divisibleMicroprocessor Quiz Write an assembly language program to find the minimum value of a byte from a string of bytes. >
- 8- The memory unit of a computer has 2.00E+20 words. The computer has instruction format with four fields: 3-An operation code field, a mode field to specify one of 4 addressing modes, a register address field to specify one of 65 processor registers, and a memory address. - Specify the number of bits in each field if the instruction occupies one memory word of 32 bits. Opcode Mode Register AR ii-Specify the size of the memory word and the number of bits in each field if the available number of opcodes is increased to 32. word Opcode Mode Register AR Find the size of the new memory in K Bytes (1K-1024 Bytes) Memory size in KBytes Memory size in KBytes8- The memory unit of a computer has 2.00E+20 words. The computer has instruction format with four fields; 3- An operation code field, a mode field to specify one of 4 addressing modes, a register address field to specify one of 65 processor registers, and a memory address. i- ii- Specify the number of bits in each field if the instruction occupies one memory word of 32 bits. Opcode Mode Register AR Specify the size of the memory word and the number of bits in each field if the available number of opcodes is increased to 32. word Opcode Mode Register AR iii- Find the size of the new memory in K Bytes (1K=1024 Bytes) Memory size in K Bytes Memory size in K BytesPatti the Programmer loads the EBX register with a pointer value as shown below: static: iValue: int16 := 0; begin progExample; mov(&iValue, EBX); // Patti uses [EBX] end progExample; After she has finished using EBX, must Patti the Programmer FREE EBX? Non of the choices listed are correct No Yes The answer will depend on the value of iValue
