Write instructions starting at address 200H, to store the data 78H into 50 consecutive memory location starting from the address B0H. Note : In this task indirect adressing mode should be used
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Write instructions starting at address 200H, to store the data 78H into 50 consecutive memory location starting from the address B0H.
Note : In this task indirect adressing mode should be used
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- Which statement is correct for the memory segments in 8086 microprocessor? a. For Stack segment, the offset address can be a direct value. b. The offset address for program instructions is taken from SP register. c. Variables created in the Assembly program are stored in Data segment. d. Segment address for executing sub-routines is taken from ES register.Write instructions to store Registers R0, R1, and R2 to 3 consecutive bytes starting at address 1000H in data memory using the equate directive to name each byte location in data memoryA 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.
- 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.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…Assume the processor makes a request to the cache, and the request reaches the cache when a block from the write buffer is being sent back to main memory. What should come next?
- 6. Assume that two numbers: dividend and divisor are saved in memory address M1 and M2 respectively. Quotient and remainder should be saved in R1 and R2 respectively. Write assembly language instructions and then list microoperations for each instruction and list the control signals required to be activated for each microoperation. MBR is used as buffer for any register to register transfer operation. Signal Description: Control signals operation Comments C0 MAR to RAM (through address bus) C1 PC to MBR C2 PC to MAR C3 MBR to PC C4 MBR to IR C5 RAM to MBR C6 MBR to ALU C7 Accumulator to ALU C8 IR to MAR C9 ALU to Accumulator C10 MBR to Accumulator C11 Accumulator to MBR C12 MBR to RAM (through data bus) C13 IR to Control Unit C14 MBR to R1 C15 MBR to R2 C16 MBR to R3 C17 MBR to R4…A 5-KB request on a system using the Buddy system for kernel memory allocation will be allocated ________KB segment.Let's assume that CPU want to read a hexadecimal value stored at a specific location in the main memory. The 32-bi address of the location is 1AC34045. Write all the steps that the CPU take to perform the read operation.
- In sim8085. c) Execute the given 8085 instructions and attach the memory view before and after execution. Take value of your choice at the required memory locations. Code: LXI H, 2000H MOV A,M ADI 02H INX H MOV M,A HLTAssume the processor sends a request to the cache, and the request reaches the cache while a block is being written back to main memory from the write buffer. What should happen next?Why is it preferable to use fragmented memory address translation rather than just translating the addresses in question?