The following tables contains memory dump and contents of few registers, as follows: k**** add cont reg cont Ox100 Oxff eax Ox100 Ox104 OXCD есx 0x1 Ox108 Ox22 edx 0x3 Ох10C Ох33 **** *** What is the value of the following operand 4[eax] ? Select one: a. OXFF b. Ox104 с. ОхCD d. Ov102
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- Convert given code to LEGv8 code:int f, g, y //global 64-bit variablesint sum (int a, int b) { // at memory address X0+1000.return (a +b)} int main (void) // at memory address X0 + 800{f=2;g=3;y= sum (f, g);return y;}Convert this code, making valid assumptions about registers and register use. Notethat brackets and global variable declarations are not affecting the addresses of the instructionsin memory.Below is a list of 32-bit memory address references, given as memory addresses. 12, 720, 172, 8, 764, 352, 760, 56, 724, 176, 744You would like to access a cache with the given memory addresses. The size of cache is 23 = 8-blocks. Your task is to: (1) find out the binary address, (2) fill out the tag and index for each memory address and (3) indicate whether the access is hit or miss in the following table:Consider the following C++ code snippet: int cards[3]; for (int i = 0; i < 3; i++) cards[i] = 0; Choose the best statement: A) This code will compile and run but causes an out-of-range memory write. B) This code will compile and has no apparent issues. C) This code will not compile as written. D) This code shows an example of dynamic memory allocation.
- (a) Find the value of AX and EAX registers.dataarrayW WORD 7000h,8000h,9000harrayD DWORD 1,2,3,4.codemov ax,[arrayW+2] ; AX =mov ax,[arrayW+4] ; AX =mov eax,[arrayD+4] ; EAX =(b) write a program that adds the following three bytes:.datamyBytes BYTE 80h,66h,0A5hWrite programs that will accomplish the desired tasks listed below, using as few lines of code as possible e) Divide the data in RAM location 3EH by the number 12H ; put the quotient in R4 and the remainderin R5.f) Divide the data in RAM location 15H by the data in RAM location 16H ; put the quotient inexternal RAM location 7CH and the remainder in in external RAM location 7DH.g) Double the number in register R2, and put the result in register R3 (high byte) and R4 (low byte).Use logic operations.h) OR the contents of ports 1 and 2 ; put the result in external RAM location 0100Hi) Set Port 0, bits 1, 3, 5 and 7 to one, set the rest to zero.j) Clear bit 3 of RAM location 22H without affecting any other bit.k) Invert the data on the port 0 pins and write the data to port 1.Write programs that will accomplish the desired tasks listed below, using as few lines of code as possible. a) Add the bytes in RAM locations 34H and 35H ; put the result in register R5 (LSB) and R6 (MSB)b) Subtract the content of R2 from the number F3H ; put the result in external RAM location 028BH.c) Subtract the content of R1 from R0 ; put the result in R7.d) Multiply the data in RAM location 22H by the data in RAM location 15H ; put the result in RAMlocation 19H (low byte) and 1AH (high byte).e) Divide the data in RAM location 3EH by the number 12H ; put the quotient in R4 and the remainderin R5.f) Divide the data in RAM location 15H by the data in RAM location 16H ; put the quotient inexternal RAM location 7CH and the remainder in in external RAM location 7DH.
- Write programs that will accomplish the desired tasks listed below, using as few lines of code as possible. a) Add the bytes in RAM locations 34H and 35H ; put the result in register R5 (LSB) and R6 (MSB)b) Subtract the content of R2 from the number F3H ; put the result in external RAM location 028BH.c) Subtract the content of R1 from R0 ; put the result in R7.d) Multiply the data in RAM location 22H by the data in RAM location 15H ; put the result in RAMlocation 19H (low byte) and 1AH (high byte).e) Divide the data in RAM location 3EH by the number 12H ; put the quotient in R4 and the remainderin R5.f) Divide the data in RAM location 15H by the data in RAM location 16H ; put the quotient inexternal RAM location 7CH and the remainder in in external RAM location 7DH.g) Double the number in register R2, and put the result in register R3 (high byte) and R4 (low byte).Use logic operations.h) OR the contents of ports 1 and 2 ; put the result in external RAM location 0100Hi) Set Port 0, bits…Given the following data definitions, the address of the first variable var1 is given at 0x1001 1000 (hexadecimal). Using little endian, show the memory dump below. Fill each box below with a byte of the allocated memory in hexadecimal. You may use the ASCII table provided. .data var1: .byte 3, -2, ‘A’ var2: .half 1, 256, 0xFFFF var3: .word 0x3DE1C74, 0xFF .align 3 str1: .asciiz "CPS2390" DATA SEGMENT: ADDRESS5 Consider the following code snippet: .datamyVar byte 4, 8, 5, 7.codemov esi, offset myVaradd esi, 2 What is the value at the memory location esi is pointing to? In other words, what value is referenced by [esi]? 6 Consider the following code snippet: .datamyArray word 01F2h, 1111h, 0ABDh, AAAAh.codemov ebx, offset myArrayadd ebx, 2 What is the value at the memory location ebx is pointing to? In other words, what value is referenced by [ebx]? 7 Consider the following code: .dataarray word 1, 2, 3, 4, 5.codemov eax, offset arraymov ecx, lengthof arrayloopstart:add byte ptr [eax], 1add eax, 2loop loopstart What are the values in the array after this code is finished executing?
- You are required to make changes in the below programs and introduce the use of compaction where required. #include<stdio.h> #include<conio.h> main() { int ms, bs, nob, ef,n, mp[10],tif=0; int i,p=0; clrscr(); printf("Enter the total memory available (in Bytes) -- "); scanf("%d",&ms); printf("Enter the block size (in Bytes) -- "); scanf("%d", &bs); nob=ms/bs; ef=ms - nob*bs; printf("\nEnter the number of processes -- "); scanf("%d",&n); for(i=0;i<n;i++) { printf("Enter memory required for process %d (in Bytes)-- ",i+1); scanf("%d",&mp[i]); } printf("\nNo. of Blocks available in memory -- %d",nob); printf("\n\nPROCESS\tMEMORY REQUIRED\t ALLOCATED\tINTERNAL FRAGMENTATION"); for(i=0;i<n && p<nob;i++) { printf("\n %d\t\t%d",i+1,mp[i]); if(mp[i] > bs) printf("\t\tNO\t\t---"); else { printf("\t\tYES\t%d",bs-mp[i]);tif = tif + bs-mp[i]; p++; } } if(i<n) printf("\nMemory is Full, Remaining Processes cannot be accomodated"); printf("\n\nTotal…If the value 8 is stored in the memory location designated by address 5, what is the functional difference between writing the value 5 into cell number 6 and copying the contents of cell 5 into cell 6?Using MIPS assembly language PLEASE HELP ME FIX THE FOLLOWING CODE RUNNING WITH FOLLOWING ERRORS: Error in C:\Users\supre\Documents\Assignment9.asm line 56 column 15: "$t2": operand is of incorrect typeError in C:\Users\supre\Documents\Assignment9.asm line 73 column 15: "$s2": operand is of incorrect typeError in C:\Users\supre\Documents\Assignment9.asm line 77 column 15: "$s2": operand is of incorrect typeAssemble: operation completed with errors. Heres the code: # Program to generate Hamming code for an 8-bit byte .dataprompt: .asciiz "Enter an 8-bit positive integer (0-255): "hamming_label: .asciiz "The Hamming code is: " .text.globl mainmain: # Prompt the user to input an 8-bit positive integerli $v0, 4 # system call for printing stringla $a0, prompt # load string addresssyscall # Get the user inputli $v0, 5 # system call for reading integersyscallmove $s0, $v0 # store the input in s0 # Create the data word and initialize the parity bits to 0li $s1, 0 # s1 stores the Hamming codeli…