Memory address Data According to the memory view given below, if RO = Ox20008002 then LDRSB r1, [ro, #-4] is executed as a result Øx20008002 ØXA1 Øx20008001 ØXB2 Øx20008000 Øx73 of r1 = ?(data overlay big endian)? ØX20007FFE ØXD4 ØX20007FFE Lütfen birini seçin: O A. R1 = OX7F O B. R1 = Oxfffffd4 O C. R1 = OxffffffZE. O D. R1=0XD4000000 O F. R1 = OXD4
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- Create a program in C++ which simulates a direct cache. The memory array that contains the data to becached is byte addressable and can contain 256 single byte entries or lines. The cache has only 8 entriesor lines. The Data field in each line of the cache is 8 bits. Since the data stored in each line of the cache isonly 8 bits, there is no need for a line field. Only a tag field is needed which is log2(256) = 8 bits.The memory array can be filled with any values of your choice. The program should work by taking userinput of a memory address (index). This input represents the memory data that should be cached.Check the cache to see if the item is already cached. If it is not, your program should counta 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.Buffer overlow :computer security In a C program, we print the address of relevant variables and arrays and get the following: 0xbfffe7b8 i 0xbfffe7bc length 0xbfffe7c0 hash_ptr 0xbfffe7ce targetuid 0xbfffe7d5 userid 0xbfffe817 pw 0xbfffe858 t 0xbfffe899 hashhex 0xbfffe8da target 0xbfffe91b hash 0xbfffe95c buffer The program executes the following instruction: strcpy(pw, buffer); You want to exploit the buffer overflow vulnerability by putting in buffer a string aaaaaaaaaaaaa ... 0c0beacef8877bbf2416eb00f2b5dc96354e26dd1df5517320459b1236860f8c with the goal of putting the hash 0c0beacef8877bbf2416eb00f2b5dc96354e26dd1df5517320459b1236860f8c variable target. If you count offset as 0 for the first a, 1 for the second a, and so on, what should be the offset of 0, the first character of the hash? In other words, how many a's should you put before the first zero character? (You can give the answer in decimal or in hexadecimal, but please specify.)variable memory address S arr[0] arr[1] arr[2] ptrl ptr2 3000 1000 1004 1008 4000 4004 int S= 7; int arr[] = {1,2,3}; int * ptr1 = &S; int * ptr2 = arr; *ptr1 = 4; ptr1 = ptr2+2; *(ptr1+1) = *ptr2; cout<Let's say that p is a pointer to memory and the next six bytes in memory (in hex) beginning at p's address are: aa bb cc dd ee ff. What value would be in x if the following code is run on a little- endian computer? uint16_t *q uint16_t x = (uint16_t *)p; q[0]; aa aabb bbaa aabbccdd ddccbbaaQ) What will be stored in %esp (stack pointer) after the disassembled code below? %eip Ox080483dc %esp Oxff9bc960 • call 8048394 (a) 0x080483dc (b) 0x080483dd (c) 0x08048394 (d) 0xff9bc960 (e) 0xff9bc95c 60 Beip: program counter (address for the next instruction to be executed)Create a program in C++ which simulates a direct cache. The memory array that contains the data to be cached is byte addressable and can contain 256 single byte entries or lines. The cache has only 8 entries or lines. The Data field in each line of the cache is 8 bits. Since the data stored in each line of the cache is only 8 bits, there is no need for a line field. Only a tag field is needed which is log2(256) = 8 bits. 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.Write a subroutine that finds the total number of elements in an array that are divisible by 8. The starting address of the array is passed in Z pointer and the array count is passed in r16. The count of elements that are divisible by 8 is returned in r22. Also write a test program to test this subroutine. The test program should define an array of 40 elements in program memory (immediately after the test program) to test this subroutine. use atmel studio pleaseProgram this in java the state and variable display must be:Unsorted Partition Reference Index: 0, Traversing Index: 7, Current Traversing Index Value: mango, Next Index Value: plum, Swapping Condition: FalseCurrent Array: ['apple', 'avocado', 'orange', 'banana', 'strawberry', 'pineapple', 'plum', 'mango'] The input:["apple", "avocado", "orange", "banana", "strawberry", "pineapple", "plum", "mango"]Note:The sorted partition is on the left side. Then, the order is increasing based on the number of vowels of a string.the output must be like this:['plum', 'apple', 'strawberry', 'mango', 'orange', 'banana', 'avocado', 'pineapple']Please include steps and comments. Starter Code: title Title of Program (skel.asm) .model small ; one data segment, one code segment.stack 100h ; reserves 256 bytes for the stack.386 ; for 32 bits.data ; start definition of variables ; your variables go here .code ; start code portion of programmain proc ; start of the main procedure mov eax,@data ; load the address of the data segment into eax mov ds,eax ; load the address of the data segment into ds ; the two previous instructions initalize the data segment ; your code goes here ; the following two instructions exit cleanly from the program mov eax,4C00h ; 4C in ah means exit with code 0 (al) (similar to return 0; in C++) int 21h ; exitmain endp ; end procedure end main ; end programGiven the following code which could execute at the beginning of a procedure, drag each component into its proper location on the stack. $p--- > can be used to indication the address of a pointer p on the stack. addi $sp, $sp, -8 $fp, 4($sp) addi $fp, $sp, 8 $ra, e($fp) SW $s1, -8($fp) Sra > Sip --> Ssp $sp $s1 $s0 Sra Sfp 14Computer organization and assembly language Please help me with this. I have to write line by line what each line of codes does. CODE IS BELOW: .model small .386 .stack 100h .data msg1 db 13, 10, "Enter any number --> ", "$" msg2 db "Enter an operation +,- * or / --> ",13, 10, "$" msg3 db "The Operation is --> ", "$" msg4 db "The result is --> ", "$" By_base dd 21 by_10 dd 10 ; 32 bits variable with initial value = 10 sp_counter db 0 ; 8 bits variable with initial value of zero disp_number dd 0 ; 32 bits variable with initial value = 0 disp_number2 dd 0 disp_number3 dd 0 op_type db 0 last_key dd 0 ; 32 bits variable with initial value of zero remainder db 0 .code main proc mov ax,@data;set up datasegment movds,ax mov dx,offset msg1 call display_message callm_keyin calloperation mov dx,offset msg1 calldisplay_message callm_keyin cmpop_type, "+" jnz short skip_plus callop_plus skiP_plus: cmp op_type, "-" jnz short skip_minus callop_minus…3. In the StackGuard approach to solving the buffer overflow problem, the compiler inserts a canary value on the memory location before the retum address in the stack. The canary value is randomly generated. When there is a return from the function call, the compiler checks if the canary value has been overwritten or not. Do you think this approach would work? Why or why not?SEE MORE QUESTIONS