the hexadecimal address of dueDate ?
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The following data segment starts at memory address 1000h (hexadecimal)
.data
printString BYTE "ASSEMBLY IS FUN",0
moreBytes BYTE 25(DUP)0
dateIssued DWORD ?
dueDate DWORD ?
elapsedTime Word ?
What is the hexadecimal address of dueDate ?
a. 1045h
b. 1029h
c.1010h
d. 102Dh
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- Suppose a computer using fully associative cache has 220 words of main memory and a cache of 128 blocks, where each cache block contains 16 words. (a) How many blocks of main memory are there? (b) What is the format of a memory address as seen by the cache, that is, what are the sizes of the tag and word fields? (c) To which cache block will the memory reference 01D872_{16} map?Suppose a computer using direct mapped cache has 4M byte of byte-addressable main memory, and a cache of 512 blocks, where each cache block contains 64 bytes. a) How many blocks of main memory are there? b) What is the format of a memory address as seen by the cache, i.e., what are the sizes of the tag, block, and offset fields? c) To which cache block will the memory address 0x00007266 map?Suppose a computer using fully associative cache has 224 bytes of byte-addressable main memory and a cache of 128 blocks, where each cache block contains 64 bytes. a) How many blocks of main memory are there? b) What is the format of a memory address as seen by the cache, i.e., what are the sizes of the tag and offset fields? c) To which cache block will the memory address 0x01D872 map?
- What are the contents of the 25 memory bytes starting at address A, in hex, on a machine that uses Little Endian? (Hint: “a”=061h) Consider the following .data segment: A dw 0AAFFh B db 051h, 0CCh, 0EEh C times 3 dw -23 D dd -177 E db "e", -5, "c", 0 F times 2 dw -17 G dw 0EEhSuppose a computer using fully associative cache has 4 Gbytes of byte-addressable main memory and a cache of 256 blocks, where each cache block contains 32 bytes. a) How many blocks of main memory are there? b) What is the format of a memory address? Provide the names and the sizes of the fields. c) To which cache block will the memory address 0x01752 map?Write a service routine which resets all elements of an array that resides in memory location from A000 H to A0FF H with DS equal to 0000 H. The service routine address is CS:IP where CS is 2000 H and IP is 0100H. Assume the interrupt type that is called is 50 (x8086- nano)
- Suppose a computer using direct-mapped cache has 232 (that's 232)232) bytes of byte-addressable main memory, and a cache size of 512 bytes, and each cache block contains 64 bytes. How many blocks of main memory are there? What is the format of a memory address as seen by cache, i.e. what are the sizes of the tag, block, and offset fields? To which cache block will the memory address 0x13A4498A map?Suppose a computer using direct mapped cache has 4G Bytes of main memory and a cache of 256 Blocks, where each cache Block has 16 Words, and Word Size is 4 Bytes. a)How many blocks of main memory? b)What is the format of a memory address as seen by the cache (Tag, Block and Word fields)? c)To which cache block will the memory reference 0000146A in Hex?Suppose a computer using direct mapped cache has 2^32 bytes of byte-addressable main memory, and a cache size of 512 bytes, and each cache block contains 64 bytes. a) how many blocks of main memory are there? b) what is the format of a memory address as seen by cache, i.e., what are the sizes of the tag, block, and offset fields? c) to which cache block will the memory address 13A4498A map?
- 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…Suppose a computer using direct mapped cache has 232232 words of main memory and a cache of 1024 blocks, where each cache block contains 32 words. How many blocks of main memory are there? What is the format of a memory address as seen by the cache, that is, what are the sizes of the tag, block, and word fields? To which cache block will the memory reference 000063FA16 map?
