
Create a
The memory array can be filled with any values of your choice. The program should work by taking user input of a memory address (index). This input represents the memory data that should be cached. Your program will check the cache to see if the item is already cached. If it is not, your program should count a cache miss, and then replace the item currently in the cache with the data from the inputted address. Allow the user to input addresses (in a loop), until they so choose to end the program. The program should output the number of cache misses upon ending.

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- Assume variables have logical addresses with 16-bit page numbers and 16-bit offset using the memory configuration below. (Note that each hexidecimal is 4 bits long and Ox means hexadecimal radix) Logical Address Format Physical Memory Physical Address (starting) Page Table Size (dec) 65536 65536 Frame Size (hex) Oxppppdddd Page Frame 2 Ox10000 Ox10000 Охс000 pppp: page number dddd: page offset 1 1 Oxd000 2 Охе000 Ox10000 65536 3 Oxf000 Ox10000 65536 Translate the following addresses: What is the physical address for 0x0002ffff ? What is the physical address for 0x0000abcd ?arrow_forwardGiven the following: Logical Memory size of 1000 Physical Memory size of 2000 Page (and frame) size of 100 Block A contains data for a program Select Block A’s size and its starting point in both memories. Then write the page table for Block A based on your selections. See below for the layout of both memories and an example of Block A of size 200. Logical Memory Physical Memory location/ page location/frame 0 to 99/ 0 0 to 99/ 0 100 to 199 /1 Block A 100 to 199/ 1 200 to 299/ 2 Block A 200 to 299/ 2 300 to 399/ 3 300 to 399/ 3 400 to 499/ 4 400 to 499/ 4 500 to 599/ 5 500 to 599/ 5 600 to 699/ 6 600 to 699/ 6 700 to 799/ 7 700 to 799/ 7 800 to 899/ 8 800 to 899/ 8 900 to 999/ 9 900 to 999/ 9 1000 to 1099/ 10 1100 to 1199/ 11 1200 to 1299/ 12 1300 to 1399/ 13 Block A 1400 to 1499/ 14 Block…arrow_forwardFind C Language Program Fill in the Gaps by CodeChum Admin The foundation for our building is slowly collapsing. We need to reinforce it by filling in the gaps! Our foundation is made up of a bunch of different integers, with values that could still be filled in between. Your job is to figure out which values can be filled in between the existing ones. Input 1. Size of the integer array Constraints The value is > 0. 2. Elements of the integer array Constraints The elements are in ascending order and each element is within the range: 0 <= n <= 2,147,483,647 Output The first line will contain a message prompt to input the size of the array. The second line will contain a message prompt to input the elements of the array. The last line contains all the "gaps" in the array. Enter the size: 5 Enter the elements: 1 3 4 7 9 Gaps = 2 5 6 8arrow_forward
- Operating sys :In the following producer and consumer problem using semaphores, explain with an example (semaphore values), how a producer sends a signal to a consumer, so that the consumer does not have to check the buffer all the time. /*program, Producer/Consumer */ semaphore n = 0; /* no of items in the buffer */ semaphore s = 1; /* ME semaphore */ void Producer () { while(TRUE) { produce(); wait(s); append(); signal(s); signal(n); } } void Consumer () { while(TRUE) { wait(n); wait(s); take(); signal(s); consume(); } }arrow_forwardThe Tower of Hanoi Problem Tower of Hanoi is a mathematical game consisting of three pegs (P1, P2 and P3) and a stack of disks of different diameters. Disks can slide onto any peg. The game starts with all disks stacked on P1 and ends at the point where all disks stacked on P3. The game player is required to move all disks from P1 to P3 using P2 as a buffer. Three rules must be followed when playing the game (1) Only one disk may be moved at a time. (2) Each move involves taking a disk on the top of a peg and place it on the top of another peg. (3) A disk of a larger diameter should never be placed on top of a disk of a smaller diameter. The diagrams below demonstrate the starting state and goal state of the game with 5 disks. starting sate p1 to goal state p3 Requirements In this assignment, students are required to solve the Tower of Hanoi (with five disks) using state space search algorithms implemented in Python. Two state space search algorithms: (1) a blind search (depth-first…arrow_forwardProblem D. Find URLsHTML is a language for representing documents designed to be displayed by a web browser. In many browsers, you can see the HTML source code by right-clicking somewhere on the page and clicking “View Page Source”. If you try this on a web page with links to other websites, you’ll notice that the URL of the link is usually formatted in the following way: href="https://some.website.com/subfolder/more_stuff.txt" Write a function find_url(html) that takes in a string of html text that contains exactly one external link URL formatted as above, and returns just the URL string (in the above example, that would be https://some.website.com/subfolder/more_stuff.txt). You can assume that the only place in the string where the substring href=" occurs is right before the URL, and that the next quotation mark after that point denotes the end of the URL. Hints: ● The .find method and string slicing will likely make this easier. ● Remember that in order to use a double quote…arrow_forward
- Part A For each byte sequence listed, determine the Y86 instruction sequence it encodes. If there is some invalid byte in the sequence, show the instruction sequence up to that point and indicate where the invalid value occurs. For each sequence, the starting address, then a colon, and then the byte sequence are shown. 0x100: 30f3fcfffff40630008000000000000 0x100: 30f3fcfffffff irmovq $-4,%rbx Ox10a: 40630008000000000000 | rrmovq %rsi,0×80A(%rcx) O0x115: 00 halt Ox100: 30f3fcffffffff irmovq $-4,%rbx Ox10a: 40630008000000000000 | rmmovq %rsi,0x800(%rbx) Ox114: 00 halt 0x100: 30f3fcfffffffff rrmovq $-8,%rbx Ox109: 40630008000000000000 | rmmovq %rsi,0x800(%rbx) 0x200: a06f800c020000000000000030f30a00000000000000 0x113: 00 halt 0x100: 30f3fcffffffffff irmovq $-4,%rbx 0x200: a06f pushq %rsi 0x10a: 40630008000000000000 | rmmovq %rsi,0x800(%rbx) Ox202: 800c02000000000000 call proc Ox116: 00 halt 0x20b: 00 halt 0x20c: proc: Submit Request Answer 0x20c: 30f30a00000000000000 | irmovq $10,%rbx…arrow_forwardTrue or false: Temporal locality is the tendency for a program to access a memory address shortly after accessing a 'nearby' memory addressarrow_forwardAssembly Language x86: (Micro Macro): templet: .386.model flat, stdcall.stack 4096ExitProcess PROTO, dwExitCode: DWORD.data day BYTE 0month BYTE 0year WORD 0 .codemain PROC main ENDPINVOKE ExitProcess, 0END main Question: What bit string repersents April 1, 2024? Examples: Date: 00111 (year) 1100 (month) 10010 (day)00111 = 18, 1100 = 12, 10010 = 30, 1980 + 30 = 2010 = 2010 Dec 18mov ax, 00111110010010b; 2010 Dec 10; 30 12 18 DAY:mov dx, ax and dx, 0000000000011111bmov day, dl (18) ; 00111 Month: mov dx, ax and dx, 00000001111000000b shr dx, 5 ; 000000000000 1100 (12) mov month, dl Year:mov dx, axshr dx, 9 ; 011110and dx, 0000000001111111bmov year, dl (30) 10010arrow_forward
- The memory location at address 00002001 contains the memory variable in binary form. What is the data memory variable in hexadecimal form? MEMORY 1110 1011 00002001 1110 1010 00002000 1110 1001 00001999 1110 1000 00001998 1110 0111 00001997 1110 0110 00001996 DATA ADDRESS The data memory variable in hexadecimal form is E7. a. b. The data memory variable in hexadecimal form is EA. The data memory variable in hexadecimal form is EB. C. The data memory variable in hexadecimal form is E9. Od.arrow_forwardCan someone please help and explain the C code that I would need for this. I am new to C and would like some help. I also do not know how to check the storage for the cache on my microsoft surface 2 computer. Define Data Structures Large Arrays: Define a large array in memory. The size should be chosen based on the cache size of the target processor. Choose a size that is larger than the L3 cache of the processor. You will need to look up the sizes of the L1, L2, and L3 for the processor you are using. Access Area: Access areas within the L1, L2, L3, and larger than the L3 cache. You must read memory the same number of times for each, but simply access more memory for each. Sequential Access: Access the elements of the first array sequentially. This pattern benefits from spatial locality, as once a cache line (containing a part of the array) is loaded into the cache, subsequent accesses to nearby elements will be faster. Random: Access the elements of the second array in a…arrow_forwardProblem DescriptionThe Tower of Hanoi ProblemTower of Hanoi is a mathematical game consisting of three pegs (P1, P2 and P3) and a stack of disks of different diameters. Disks can slide onto any peg. The game starts with all disks stacked on P1 and ends at the point where all disks stacked on P3. The game player is required to move all disks from P1 to P3 using P2 as a buffer. Three rules must be followed when playing the game(1) Only one disk may be moved at a time.(2) Each move involves taking a disk on the top of a peg and place it on the top of another peg. (3) A disk of a larger diameter should never be placed on top of a disk of a smaller diameter. The diagrams below demonstrate the starting state and goal state of the game with 5 disks.Starting state Goal stateP1 P2 P3 P1 P2 P32RequirementsIn this assignment, students are required to solve the Tower of Hanoi (with five disks) using state space search algorithms implemented in Python.Two state space search algorithms: (1) a blind…arrow_forward
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