Below shows the following hexidecimal values: Address 1000: 13 Address 1001: 03 Address 1002: C5 Address 1003: 00 Provide the format and assembly language instruction for the following hexadecimal values
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Below shows the following hexidecimal values:
Address 1000: 13
Address 1001: 03
Address 1002: C5
Address 1003: 00
Provide the format and assembly language instruction for the following hexadecimal values.
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- A(n) __________ is a storage location implemented in the CPU.What memory address (0 through 256) is represented by each of the following hexadecimal numbers: (a) 0A16 (b) 3F16 (c) CD16the available space list of a computer memory is specified as follows: 9 start address block address in words 100 50 200 150 450 600 1200 400 determine the available space list after allocating the space for the stream of requests consisting of the following block sizes: 25,100,250,200,100,150 use i) first fit ii) best fit and iii) worst fit algorithms
- 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.Assembly Language x86: (Micro Macro): templet: .386.model flat, stdcall.stack 4096ExitProcess PROTO, dwExitCode: DWORD.data .codemain PROC main ENDPINVOKE ExitProcess, 0END main Question: What are the reults (in decimal) of the following operations? A: 1111001000101111 multiplied by 0000001011010010 (these are signed numbers) B: 01011110110101101010010100101001 devided by 1111111111011010 (these are signed number)(hint: use idiv (signed), div (unsigned) ) Examples: Multiply: mul bx, 10000 mov ax, 355 mul bx, shl edx, 16 mov dx, ax Division:mov eax, 2000000 mov bx, 500mov edx, eax shr edx, 16 div bx,Show how the following values would be stored by byteaddressable machines with 32-bit words, using little endian and then big endian format. Assume that each value starts at address 10 . Draw a diagram of memory for each, placing the appropriate values in the correct (and labeled) memory locations.Q.) 0x14148888
- Assume the following values are stored at the indicated memory addresses and registers Address Value 0x100 0xaaa 0x104 0x123 0x108 0x12 0x10c 0x10 Register Value %eax 0x100 %ecx 0x1 %edx 0x3 Fill up the following table: %eax 0x104 $0x108 (%eax) 4(%eax) 9(%eax,%edx) 260(%ecx,%edx) 0xFC(,%ecx,4) (%eax,%edx,4)Code a descriptor that describes a memory segment that begins at location 0005CF00h and ends at location 00060EFFh. The memory segment is a data segment that grows upward in the memory system and can be written. The segment has a user level privilege (lowest) and has not been accessed. The descriptor is for an 80386 microprocessor.A user program before being executed goes through several steps. Binding of symbolic addresses to relocatable addresses happens along the way. Discuss the different ways in which these binding/ mapping of instructions and data to memory addresses can happen.
- instruction is in the first picture please give me only implementation of int L1lookup(u_int32_t address) and int L2lookup(u_int32_t address) cacheSim.h #include<stdlib.h>#include<stdio.h>#define DRAM_SIZE 1048576typedef struct cb_struct {unsigned char data[16]; // One cache block is 16 bytes.u_int32_t tag;u_int32_t timeStamp; /// This is used to determine what to evict. You can update the timestamp using cycles.}cacheBlock;typedef struct access {int readWrite; // 0 for read, 1 for writeu_int32_t address;u_int32_t data; // If this is a read access, value here is 0}cacheAccess;// This is our dummy DRAM. You can initialize this in anyway you want to test.unsigned char * DRAM;cacheBlock L1_cache[2][2]; // Our 2-way, 64 byte cachecacheBlock L2_cache[4][4]; // Our 4-way, 256 byte cache// Trace points to a series of cache accesses.FILE *trace;long cycles;void init_DRAM();// This function print the content of the cache in the following format for an N-way cache with M Sets// Set 0…The 32-bit number 52AB43FC (in hexadecimal) is stored in abyte-addressable memory starting at physical address FE08 (in hex) using Little-Endian notation. The byte(value ni hexadecimal) stored at theaddress FE0B will beAssembly 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) 10010