Suppose we have the main memory following Big Endian ordering and it is shown in the following table. (All the data contents are in hexadecimal and main memory is byte addressable) address 8 9 10 11 12 13 14 15 data 24 FD 03 FF 32 20 7D 6B If we run the following commands: li $t2, 10 //assume same syntax as in MARS Ih $t3, 0($t2)
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- Answer the given question with a proper explanation and step-by-step solution. Consider the following source code, where k,l and m are constants declared with #define. The struct lnode is as defined as: struct lnode { char *str; // points to 32-character string struct lnode *next; // points to next node address struct lnode *prev; // points to prev node address }; You will need to determine the values for k, l and m, which are the dimensions of a 3-D array named A. struct lnode A[k][l][m]; int store_ele(int h, int i, int j, struct lnode dest) { A[h][i][j] = dest; return sizeof(A); } On compiling this program (with gcc -O2 -S -fno-asynchronous-unwind-tables) ,gcc generates the following assembly code for the store_ele function: store_ele: endbr64 movslq %edi, %rdi movslq %edx, %rdx movslq %esi, %rsi movdqu 8(%rsp), %xmm0 movq %rdi, %rax leaq (%rdx,%rdx,2), %rdx salq $6, %rax subq %rdi, %rax salq $4, %rax leaq (%rax,%rdx,8), %rax leaq (%rsi,%rsi,4), %rdx leaq (%rsi,%rdx,4), %rdx leaq…) Consider the following sequence of virtual memory references (in decimal) generatedby a single program in a pure paging system:100, 110, 1400, 1700, 703, 3090, 1850, 2405, 2460, 4304, 4580, 3640a. Derive the corresponding reference string of pages (i.e. the pages the virtual addressesare located on), assuming a page size of 1024 bytes. (Assume that page numberingstarts at 0)b. For the page sequence derived above, determine the number of page faults for each ofthe following page replacement strategies, assuming that two (2) page frames areavailable to the program.i. LRUii. FIFOiii. OPT (Optimal)Assume a program of size 460 Bytes, and its virtual address sequence is as following 10 11 104 170 73 309 185 245 246 434 458 364 1.Suppose the page size is 100 Bytes, please write out the reference string. 2.Use the above reference string to calculate the number of page missing using the following algorithms: LRU, FIFO, OPT. We suppose the page frame is initially empty, and there is a total of 200 Bytes of physical memory.
- Let's say that p is a pointer to memory and the next four bytes in memory (in hex) beginning at p's address are: 89 AB CD EF. After the following code is run on a little-endian computer what are the values of x, y, and the four bytes in memory beginning at p's address. Answer for x and y in four digits of upper-case hexadecimal (eg, 0AC6). Answer for memory as eight digits of upper-case hexadecimal with a single space between each byte (eg, 89 AB CD EF). x: y: mem at p:Can u explain this code in detail. I will Upvote : #include<stdio.h> #include<stdlib.h> #include<time.h> struct MainMem { int start_add; int pn; }; int main() { int s,i,totPages,pno,offset,n,r,arr[200]={0},phy_add,fno[100]; struct MainMem mm[100]; srand((unsigned)time(NULL)); printf("Logical Address To Physical Address\n"); printf("Enter the Size of File : "); scanf("%d",&n); printf("Enter the Page Size : "); scanf("%d",&s); totPages=n/s; for(i=0;i<totPages;i++) { r=rand()%totPages; if(arr[r] == 1) { i--; continue; } arr[r]=1; mm[i].pn=r; mm[i].start_add=i*s; fno[r]=i; } printf("\n*************\n"); printf("The Structure of Main Memory\n"); printf("***********\n"); printf("Frame\tPage\nNumber\tNumber\n------\t------\n"); for(i=0; i < totPages; i++) { printf("%d\t%d\n",i,mm[i].pn); } printf("***********\n"); printf("Enter The Logical Address\nPage Number : "); scanf("%d",&pno); printf("Offset : "); scanf("%d",&offset); if(pno >= totPages…5.How many bytes in the memory will be occupied by short int array[10] 10 Bytes 20 Bytes 40 Bytes 60 Bytes 6.Given the definition void myFunc(int a, int* b , int* c)which statement is true The values of the aliases a, band c are available to the function Only the addresses of the variables b, care available to the function myFunccannot change the variable whose value is copied to a myFunc can change the value of the variable whose address is copied into b all of the above
- A.) What are the contents in hex of the memory location at the following address in binary: 0000 1100 0000 1111? (Enter hex like the following example: 0x2A3F) In all computer models, the contents of a memory location can be interpreted as data. B.) Interpret the contents at address 0xFFFF as a two's complement integer in base 10. C.) Interpret the contents at address 0x0C0C as two ASCII characters. D.) Interpret the contents at the same address as C.) above as an unsigned integer in base 10.Use C, C++, python or matlab to develop a program whose main routine accepts two parameters n and k, i.e. when you invoke your program from the shell, you pass it two parameters, n and k, where n >=16 and k >=8and is in powers of 2 (e.g. 8, 16, 32, etc.). Your main routine shall generate a random page trace of length n, where the page numbers have values ranging from 0 to ? − 1. Develop a subroutine within your program that implements the FIFO page replacement algorithm (as a separate function within your program). The function shall accept a page trace and a parameter f for the number of frames allocated. Your main routine shall then apply the random page trace to the subroutine implementing the page replacement algorithm, multiple times (using only one trace, randomly generated), passing a parameter f (number of page frames used) that ranges from 4 to k. Your main routine shall then record the number of page faults for each run (i.e. for each f).Run your program using a page…Draw a picture illustrating the contents of memory, given the following data declarations: You need to mark all the memory addresses. Assume that your data segment starts at 0x1000 in memory. Name: .asciiz "Jim Bond!"Age: .byte 24Numbers: .word 11, 22, 33Letter1: .asciiz 'M' In this format HexadecimalAddress, Hex Value, Character/Number/Symbol, Binary Value, Decimal Value ALREADY HAVE! marking all the memory addresses. Assuming that the data segment starts at 0x1000 in memory. The Memory Layout looks like Byte by Byte Address Data 0x1000 4a 0x1001 61 0x1002 6d 0x1003 65 0x1004 73 0x1005 00 0x1006 18 0x1007 00 0x1008 0b 0x1009 00 0x100a 00 0x100b 00 0x100c 21 0x100d 00 0x100e 00 0x100f 00 0x1010 14 0x1011 00 0x1012 00 0x1013 00 0x1014 4d arrow_forward Step 2 In mips 1 word is equal to 4 bytes. Address Data…
- Assume you now have 1kB of memory, i.e. the memory address space runs from 0 to 1023. The starting address of the first word is 0, the second word is 4, the third word is 8, and so on. The last word comprising 4 bytes resides in addresses 1020, 1021, 1022, 1023. Thus, the last word starts at 1020, which is a multiple of 4. Now assume the same 1kB of memory but now, word size is 64 bits. The starting address of the first word is 0, the second, third, and last word starts at? my answer: (correct) the second is equal to = 8 the third is equal to = 16 help me find the last wordAssume you now have 1kB of memory, i.e. the memory address space runs from 0 to 1023. The starting address of the first word is 0, the second word is 4, the third word is 8, and so on. The last word comprising 4 bytes resides in addresses 1020, 1021, 1022, 1023. Thus, the last word starts at 1020, which is a multiple of 4. Now assume the same 1kB of memory but now, word size is 64 bits. The starting address of the first word is 0, the second, third, and last word starts at?Giventhe following assignment of some program’s virtual pages to physical pages in a system with 4 KiB byte pages, what physical memory address corresponds to virtual address 20000? (All values are given in decimal.)