Q1) If BX=1000, DS=0200, SS=0100, CS=0300 and AL=EDH, for the following instruction: MOV [BX] + 1234H, AL Find the physical address in the memory. 10:1
Q: (a) Set a base address for a generic PIO at 0x10008000 called pioBase. Assume this is just an 8-bit…
A: Change in memory location an interrupt is set to send signals, when we change in memory.
Q: 3. Translate following program into compiled MIPS code. Assume that g, c, i are in $s0, $s1, $s2…
A: Given: We have to translate the following program into Compiled MIPS code . Assume that g, c , i…
Q: if BX=1000, DS=0400, and AL=EDH, for the following instruction: MOV [BX] + 1234H, AL. the physical…
A: The answer will be:- 6234H
Q: e instruction, Add #45,R1 does Adds 45 to the value of Rl and stores it in R1 Adds the value of 45…
A: Add #45, R1 is instruction for addition.
Q: 2.12 Assume that registers $s0 and $s1 hold the values 0x80000000 and OXD0000000, respectively.
A:
Q: CPU Datapath The following figure shows the overall datapath of the simple 5-stage CPU we have…
A: ANSWER: a) MUX 1 Input 1: Register A input 2:Register B b) MUX 2 Input 1: Register C input…
Q: 7. Suppose that. DS = 0200H, BX = 0300H, and DI = 400H Determine the memory address accessed by each…
A: Given data, DS= 0200H BX= 0300H DI= 400H To find :- Memory address for following instructions at…
Q: 3) The physical address is the actual location within the RAM. It is pu bus by the CPU to be decoded…
A: 1. Logical address =CS:IP =426:A436 2.offset address = IP…
Q: 2. a) Explain the following code and indicate in each case the type of addressing mode used. i) СМА…
A: Given second step different types of instructions are explained, addressing mode of 8086are…
Q: 7-lf we assume we place the following MIPS code starting at location 8000 in memory, what is the…
A: 1) SLT $t2 $zero $t0 Binary: 00000000000010000101000000101010 Hex: 0x0008502a 2) BNE $t2 $zero…
Q: onsider the code sequence below lw $t1, 4($t0) add $s2, $t1, $t2 lw $t3, 16($t0) add $s3,…
A: Memory operands - Data transfer command: A command to move data to and from memory Registered…
Q: Let R15=0x0000 00FF. The contents of the memory location at an address equal to the last 5 digits of…
A: R15 =0x0000 00FF. The contents of the memory location at an adress equal to the last digits of your…
Q: (B) - Identify the addressing mode for the following instructions then compute the physical address…
A: The physical address of any instruction can be calculated by the formula: PA =Segment reg…
Q: Assume the following contents of registers and quadwords in memory: Location Value %rax Ox8 %rbx Ox1…
A: leaq s,d Here, “s” denotes source, “d” denotes destination. This instruction loads the address of…
Q: Q1: what is the addressing mode for the following instructions: 1. SUB A,B,C 2. OR A,B 3. ADD R1 4.…
A: Instructions Addressing Mode Explanation 1. SUB A,B,C Implied addressing mode The SUB is a…
Q: Discussion: 1. Two bytes - sizcd BCD integers are stored at the symbolic addresses NUMI and NUM2…
A: Instruction sequence to generate the difference (NUM2-NUM1) and storing it at NUM3
Q: 1) Write simple instructions (simple program) that will add two consecutive bytes of data storied…
A: Ans 1) CMP and SUB, both the instructions subtract one from the other. But the difference is, CMP…
Q: Assume the following values are stored at the indicated memory addresses and registers: Address…
A: Firstly movl moves a long (32-bits) from source to destination. Here we are given :
Q: 1. The hypothetical machine of Figure 3.4 also has two I/O instructions: In these cases, the 12-bit…
A:
Q: 3.i) Assume that the following registers contain these hex contents: RO = 0XF123, RI - 0x3456, and…
A: Answer: I have given answer in the handwritten format.
Q: 14- Change the content of memory location [300h] to FFh without using MOV instruction. Use just one…
A: Algorithm : Move 300h into CX register Move CX into DS segment (now we are in 300h data segment)…
Q: (i) Identify the addressing mode used in each instruction in the following code segment, and give…
A: The way of specifying data to be operated by an instruction is known as addressing modes. This…
Q: 1) into the data memory at address stored in ($s0). Hint: In this problem, the third byte value in…
A: Note: We are given the data in bytes so de defined the variable size by bytes "db"
Q: Q:find the actual address for the ..il following instruction assume X=38 and R index=DDCE8 hex LOAD…
A: Given: X = 38 Ri = DCE8
Q: Q:find the actual address for the following instruction assume X= (27)hex and R index=DBC9 LOAD…
A: Question:
Q: 8086Microprocessor 1. Write a piece of code that exchanges a block of 236 bytes stored at locations…
A: Write 8085 code exchanges a block of bytes to another block Algorithm: Take a count equal to 256…
Q: To get the physical address from the logical address generated by CPU we use ____ . a. MAR b. MMU c.…
A: Task :- Choose the correct option for given question.
Q: 22. Suppose that DS = 0200H, BX = 0300H, and DI = accessed by each of the following instructions,…
A: The Answer is
Q: 1) Write simple instructions (simple program) that will add two consecutive bytes of data storied…
A: In Direct addressing mode, we directly give the address of the location, where we want to retrieve…
Q: QI) I BX=1000, DS-0200, SS=0100, CS-0300 and AL=EDH, for the following instruction: MOV [BX] +…
A: Here is Solution for Above Problem :: Q1). Given Data : BX = 1000 DS = 0200 SS = 0100 CS = 0300 AL…
Q: QUESTION ONE (1) 1. The hypothetical machine has two I/O instructions: 0011 = Load AC from I/O 0111…
A: 34 Opcode Address (a) Instruction format Magnitude (b) Integer format Program counter (PC) = Address…
Q: Write simple instructions (simple program) that will add two consecutive bytes of data storied with…
A: In a direct addressing mode, the data to be worked upon is in a memory location and as an operand,…
Q: The states of the instruction cycle involve operand address calculation which means that the…
A: Algorithms and algorithmic problem resolving that can concern as a central place in computer science…
Q: Q:find the actual address for the following instruction assume X=38 and R index=DCE8 hex LOAD X(Ri),…
A: Solution:-
Q: (B)- Identify the addressing mode for the following instructions then compute the physical address…
A: 1)LDS [FFH],SI Register indirect addressing mode. In this addressing mode, The address field of…
Q: Microprocessor Hw Q1 Execute the following code and show the contents of the registers: LDI R16,$03…
A: A CPU or processor register is one of a small set of data holding places that are part of the…
Q: Question 4 Endianness Assume that a snippet of memory is shown below (in hex), starting with the…
A: 4. Given, Address starts with 0x10. Data is : | 77 | AB | 69 | CA | 0D | F0 | 12 | BE | The system…
Q: 2. a) Explain the following code and indicate in each case the type of addressing mode used. i) CMA…
A: This insertions are related to microprocessor. Above question explained in step send.
Q: Explain utiat haprns PUSH Bx instruction executes. Make sure to show where BH and BL are stored…
A: Hey there, I am authorised to answer any one question at a time when there are multiple questions…
Q: Instruction Description LD RI, 45(R2) Read data from memory and store in R1. Memory address is…
A: Answer:-
Q: Example: The content of PC in the simple computer is 3AFU. The content of AC is 6ECSH. The content…
A: List of the initial conditions that are given below: AR IR 3AF 932E 32E 09AC…
Q: Q3/ Design a machine with a byte addressable main memory of 216 bytes and block size of 8 byte.…
A: Given that, Size of main memory= 216 bytes Block size= 8 bytes Number of lines= 32
Q: 8 Find the physical address of the memory locations referred in the following instructions if…
A: 8086, via its 20 bit address bus, can address 220 = 1,048,576 or 1 MB of different memory locations.…
Q: Write at most two instructions to move ONLY the fourth byte value in the register ($s1) into the…
A: This is how you can do this easily. Note: We are given the data in bytes so de defined the variable…
Q: 52-Two word wide unsigned integers are stored at the physical memory addresses 00A00 and 00A02,…
A: Given:- Two word wide unsigned integers are stored at the physical memory addresses 00A00 and 00A02…
Q: 1- The instruction : MOV [Dx+SI], Ax is allowed T 2- The instruction : MOV ES:[SI], Ax is not…
A: 1. True The instruction is valid 2. False The instruction is invalid, since in based index…
Q: For the MIPS assembly instructions below, what is the corresponding C statement? Assume that the…
A: Actually, registers are used to stores the data/information.
Q: iv) Using the following instruction format, a total of registers can be addressed 8 7 10 OP code…
A: 4) 4.10
Q: 2. MIPS C a. (.. -) Implement the following code in MIPS assembly. Assume variables 'm' and 'n' are…
A:
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- Suppose a computer system uses 16-bit addresses for both its virtual and physical addresses. In addition, assume each page (and frame) has size 256 bytes. a) Suppose that the OS uses a two-level page table. Draw the page table. (Assume that frames 7 through 221 are free, so you can allocate space for the page table there.) In addition, suppose that the page-table directory storage comprises a whole number of consecutive full frames. (For examples: if the directory entry is 2 bytes, the entry’s storage comprises 1 frame; if the directory entry is 260 bytes, the entry’s storage comprises 2 consecutive frames.) b) What is the size of the two-level page table c) Now, translate the virtual addresses shown in question I(e) to physical addresses for the two-level page table. Show how you obtain your answers. 0x0389 0xDF78 0x0245 0x8012CA_6 We study the properties of cache memory, and for reasons of easier design and efficient circuits, we assume that the cache capacity is 2i Bytes, and cache line size is 2j Bytes, with i and j being natural numbers: (a) How many bits should the tag field have? And can the tag field contain 0 bit (i.e., be empty)? Elaborate (b) Repeat the above for the index field. (c) Repeat the above for the byte-offset field. (d) Finally, depict a figure showing a cache line, indicate what fields it possibly has, state the possible sizes of these fields, and explain the uses of these fields.Suppose a computer system uses 16-bit addresses for both its virtual and physical addresses. In addition, assume each page (and frame) has size 256 bytes. Could you hand draw the page table, if possible a) Suppose that the OS uses a two-level page table. Draw the page table. (Assume that frames 7 through 221 are free, so you can allocate space for the page table there.) In addition, suppose that the page-table directory storage comprises a whole number of consecutive full frames. (For examples: if the directory entry is 2 bytes, the entry’s storage comprises 1 frame; if the directory entry is 260 bytes, the entry’s storage comprises 2 consecutive frames.) b) What is the size of the two-level page table
- 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…6. Assume a computer has 32-bit integers. Show how the value 0x0001122 would be stored sequentially in memory, starting at address 0x000, on both a big endian machine and a little endian machine, assuming that each address holds one byte. Address Big Endian Little Endian0x000 0x001 0x002 0x003Consider the following image that represents part of the memory of a 16-bit address space that has an addressability of 2 bytes (like LC-3): A memory location can store an address. We call that memory location's contents a "pointer" since it's an address that "points" to another memory location. G.) Interpret the contents at address 0x0C0B as a pointer. (Enter hex like the following example: 0x2A3F) H.) What are the contents of the memory location that the pointer above is pointing to? (Enter hex like the following example: 0x2A3F) Another reference : LC-3 Opcodes in Hex ADD 0x1 JMP 0xC LDR 0x6
- Consider the block of three-address code Identify leaders and generate basic blocks To = a * b T1 = To + j T2 = T0 * T1 T3 = b[T2] j = T3 T4 = j + 2 T5 = c[T4] T6 = a * b * 10 i = T6 sub=i If i <= 15 GOTO (1) note: subject: compailer concepts deptt: cs/ItFill in blank Suppose that linear page table is used where the memory addresses are 12-bit binary numbers and the page size is 256 bytes. If a virtual address in binary format is 101000011100, then the VPN (virtual page number) in binary format will be ---------Input file sample.txt contains a hex dump of some data in the following format:<address> <byte1> <byte2> ... <byte16>Example Input File:00000000 54 68 69 73 20 69 73 20 61 6e 20 65 78 61 6d00000010 70 6c 65 20 6f 66 20 68 65 78 20 64 75 6d 70...Construct a pipeline using Ubuntu bash to discard the first column (address) and to reformat bytesinto a single column:
- CA_10 Let the virtual address be V bits and the virtual addtess space be byte-addressable, the page size be P KB (and P is a power of 2), and the the main memory size be MM MB(where [MM MB]) is divide into [P KB]). (d)How many of the virtual memory bits need to be translated? (e) How many bits will be produced if the virtual-to-pyysical address translation is "successful" (f) How many bits does a physical address have, and how are each of these bits obtained?Assume you now have 1kB of memory, i.e. the memory address space runs from 0 to 1023. The starting address of the first word is 0, the second word is 4, the third word is 8, and so on. The last word comprising 4 bytes resides in addresses 1020, 1021, 1022, 1023. Thus, the last word starts at 1020, which is a multiple of 4. Now assume the same 1kB of memory but now, word size is 64 bits. The starting address of the first word is 0, the second, third, and last word starts at? my answer: (correct) the second is equal to = 8 the third is equal to = 16 help me find the last wordAssume you now have 1kB of memory, i.e. the memory address space runs from 0 to 1023. The starting address of the first word is 0, the second word is 4, the third word is 8, and so on. The last word comprising 4 bytes resides in addresses 1020, 1021, 1022, 1023. Thus, the last word starts at 1020, which is a multiple of 4. Now assume the same 1kB of memory but now, word size is 64 bits. The starting address of the first word is 0, the second, third, and last word starts at?