2) How should the pointer with segment base address equal to A00016and offset address 55FF16 be stored at an even-address boundary starting at 0000816? Is the double word aligned or misaligned? Address Memory (hexadecimal) 00008 AO 0000A 00 00009 55 00008 FF
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- translate the following MIPS code to C. Assume that the variables i, j, and k areassigned to the registers $s0, $s1, and $s2, respectively. Assume that the baseaddress of the array A is in registers $s6.Loop: blt $s0, $s1, Exitbge $s1, $s2, Exitaddi $s1, $s1, 5j LoopExit:addi $t0, $zero, 4ble $s0, $t0, Donesll $t1, $s0, 2add $t2, $s6, $t1sw $zero, 0($t2)Done:Show the content of the individual bytes allocated in memory in hexadecimal for the following declarations. Assuming that the address of I is 404000h, what are the addresses of J, K, and L? What is the total number of allocated bytes?Need crt Answer else downvoted Translate the following into three address code while (A <B) do If (C<D) then X = Y + Z.
- A set of eight data bytes is stored in memory locations starting from XX70H. Write a program to add two bytes at a time and store the sum in the same memory locations, low order sum replacing the first byte and a carry replacing the second byte. If any pair does not generate a carry, the memory location of the second byte should be cleared. Data(H) F9, 38, A7, 56,98,52, 8F, F2Convert 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.49. Which of the following is/are restriction(s) in classless addressing ? a. The number of addresses needs to be a power of 2 b. The mask needs to be included in the address to define the block c. The starting address must be divisible by the number of addresses in the block d. All of the above
- Please help! Please take a look at the code I have provided as a attachment to this question. Please comment each line of code of the differences between program 1 and 2. What happens when you subtract one pointer from another? Is it subtracting the two addresses, or something else? How do you know? What happens when you increment a pointer? Compare and contrast how C-strings are laid out in memory compared to arrays of integers. What's one similarity and one difference? Show how the individual bytes of this integer array are laid out in memory. Thanks!Modify vaddr.c and call it vaddr2.c . The new program should output the page number and offset for the given address using 1MB page size.#include #include int main(int argc, char *argv[]) { unsigned long page, offset, address; if(argc != 2) exit(1); address= atoll(argv[1]); page = address >> 12; /*calculating pages number*/ offset = address & 0xfff; /*calculating remaining offset*/ printf("The address %lu contains: \n", address); printf("page number = %lu\n",page); printf("offset = %lu\n", offset); return 0; }Write a function in MIPS assembly that copy the value of the 8th element of a word array to the 5th element. The starting address of the array is in $s0. Note thatthe index of the array starts with 0; i.e., the index of the 1st element in the array is 0. Please only modify $t0 in the code you write.
- The array sum function below is called on an array of length four starting at address B. List, in order, the data addresses referenced by this function during execution. add1: add $v0, $0, $0beq1: beq $a1, $0, jr1addi1: addiu $a1, $a1, -1lw1: lw $t0, 0($a0)add2: add $v0, $v0, $t0addi2: addiu $a0, $a0, 4beq2: beq $0, $0, beq1jr1: jr $raModify the code below: Add code to display both sum and diff in a hexadecimal format on the screen, all bytes must be stored in a reversed sequence order. code: .386 ; Tells MASM to use Intel 80386 instruction set..MODEL FLAT ; Flat memory modeloption casemap:none ; Treat labels as case-sensitive INCLUDE IO.H ; header file for input/output .STACK 100h ; (default is 1-kilobyte stack) .const ; Constant data segment .DATA ; Begin initialized data segment op1 QWORD 0A2B2A40675981234h ; first 64-bit operand for additionop2 QWORD 08010870001234502h ; second 64-bit operand for addition sum DWORD 3 dup(?) ; 96-bit sum = ????????????????????????h op3 DWORD 2h, 0h, 0h ; 96-bit operand to subtract: 20000000200000002h .CODE ; Begin code segment_main PROC ; Beginning of code;-----------------------------------------------------------------------------; add two 64 bit numbers and store the result as 96 bit sum;-----------------------------------------------------------------------------mov EAX,…Assume that arrX array is already defined and allocated in memory, its base address 0xabcd3000 and already stored in $a2 and its number of elements in $a3. Answer each of the next questions as required. arrX: .word 0x99,0x20,0x73,0x40,0x50,0x69, ... Please write question numbers and answer parts in this question in order. Q4) Write NO more than 3 MIPS instructions to decrement the fourth element of arrX: ______________________________ Q5) Write no more than 6 instructions to calculate $v0= first element - last element in arrX.