Suppose we have already the page table shown below, what happens when CPU generates address 3000? Does it generate a page fault? Valid Addresses Frame Bit 0 - 1023 Page 0 : 1: 1024 - 2047 2: 2048 - 3071 3 : 3072 - 4095 4: 4096 - 5119 5 : 5120 - 6143 6 : 6144 - 7167 7: 7168 - 8191 Page 0 1 1 1 3 Page 2 Table 3 4 1 1 2

Suppose we have already the page table shown below, what happens when CPU generates address 3000? Does it generate a page fault? Valid Addresses Frame Bit 0 - 1023 Page 0 : 1: 1024 - 2047 2: 2048 - 3071 3 : 3072 - 4095 4: 4096 - 5119 5 : 5120 - 6143 6 : 6144 - 7167 7: 7168 - 8191 Page 0 1 1 1 3 Page 2 Table 3 4 1 1 2

Systems Architecture

7th Edition

ISBN:9781305080195

Author:Stephen D. Burd

Publisher:Stephen D. Burd

Chapter3: Data Representation

Section: Chapter Questions

Problem 13RQ: How is an array stored in main memory? How is a linked list stored in main memory? What are their...

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Suppose a malloc implementation returns 8-byte aligned addresses and uses an explicit free list where the next and previous pointers are each 32-bits. Blocks have a 32-bit header and 32-bit footer, where the low-order bit of the header and footer are used to indicate whether the block is allocated (1) or free (0). Furthermore, the block size (which includes the header, payload, footer, and any necessary padding) is rounded up to the nearest multiple of 8, and this size (in bytes) is stored in the header and footer. Assume any padding must be between the payload and the footer. a)If we call malloc(1), what block size will be allocated, in bytes? b)Using the same condition, assuming we've already called malloc(1), if the heap used by malloc starts at address 0x4000 (16384 in decimal), what address would be returned if we then called malloc(32)?
Consider the following runtime stack: BEFORE 00001000 00000006 00000FFC 000000A5 00000FF8 00000001 00000FF4 00000002 ESP 00000FF0 What would be the value of ESP after popping the 32-bit value shown below below off of the stack? 00000002 00000FFC 00000FF0 00000FF8 00000002
The STACK is a dynamic data structure. The 80x86 computer controls its stack via stack pointer ESP. Whenever you PUSH data onto the stack segment memory using PUSH EBX, the 80x86 will transfer data by: Oa. Da. Decreasing the stack pointer ESP by 4. Ob. Increasing the stack pointer ESP by 2. Oc. Increasing the stack pointer ESP by 4. Od. Decreasing the stack pointer ESP by 2.
Memory address Data According to the memory view given below, if RO = Ox20008002 then LDRSB r1, [r0, #-4] is executed as a result of r1 = ?(data overlay big endian)? Øx20008002 ØXA1 Øx20008001 ØXB2 Øx20008000 Øx73 ØX20007FFE ØXD4 ØX20007FFE Lütfen birini seçin: O A. R1 = 0X7F O B. R1 = Oxffffffd4 O C. R1 = Oxffffff7F O D. R1=0XD4000000 O E. R1 = 0XD4
I have a 128-bit computer where addresses are 128 bits wide. I want to implement virtual memory with paging for this computer, and I want to use a page size of 4KB. I will design it so that a page table consumes exactly one page. How many page table entries (PTEs) fit within a page?
Given that SS=2400, SP=8631H, AX=4FA6H, and DX=8C3FH. What the contents are of registers AH, DL and SP after the execution of the following lines of instructions?PUSH DXPUSH AXPOP DXPOP AXAlso find the physical address of the memory location pointed by the stack pointer, SP, after the execution of the above instructions?
Suppose your stack was allocated to be between addresses 0x20002000 and 0x200022FF. Imagine the stack pointer was currently pointing at 0x20002138. How many more words can you push to the stack and not have a problem for the following 4 stack policies: a)Empty ascending b)Empty descending c)Full ascending d)Full descending
Quiz 5: In this problem we want to set the control signals of the datapath shown below (also in in slide # 1 of "chapter3_single_cycle_datapaths.pptx") so that it supports execution of a new instruction called swi. Single Cycle Datapath: PC Read Instru- address ction [31-0] Instruction memory Sns Add Ins 1 [25-21] 1 [20-16] [15-11]. 1[10-0] RegWrite Read register 1 Read register 2 Write register Write data Read data 1 Read data 2 Read Ins Write 3ns Sign extend 2ns MemWrite Read Read address data Write address Read Gns Write data Write 10ns ins ALUSTO1 MemRead ALU Result 2ns ALUOP1 -XEWO) ins ALUSrc2 ALUSrc3 x=3 ins ALU Result 2ns ALUOP2 swi rd, rs, rt, imm # Memory [R[rs]]= R[rt], R[rd] =R [rs]+R [rt]+Imm #this instruction copies contents of "rt" register into the main memory addressed by the "rs" register. In the same cycle it add "rs" and "rt" register contents along with the "imm" field of the instruction and writes the final result into the "rd" register. You are NOT allowed to…
The STACK is a dynamic data structure. The 80x86 computer controls its stack via stack pointer ESP. Whenever you PUSH data onto the stack segment memory using PUSH EBX, the 80x86 will transfer data by: Decreasing the stack pointer ESP by 4. O b. Incrcasing the stack pointer ESP by 2. Decreasing the stack pointer ESP by 2. Increasing the stack pointer ESP by 4.
The UNIX kernel will dynamically grow a process's stack in virtual memory as needed, but it will never try to shrink it. Consider the case in which a program calls a C subroutine that allocates a local array on the stack that consumes 10K. The kernel will expand the stack segment to accommodate it. When the subroutine returns, the stack pointer is adjusted and this space could be released by the kernel, but it is not released. Explain why it would be possible to shrink the stack at this point and why the UNIX kernel does not shrink it.
IN ASSEMBLY 80X86 Write a program to copy one array of size 24 to another array of size 24 using string instructions. Write 3 versions of this code. One code must copy byte at a time. One code must copy word at a time and one code must copy double word at a time. Cut and paste the array in memory to show your code is working. IN ASSEMBLY 80X86
Consider the following runtime stack: BEFORE 0000100 0 00000006 ESP 00000FFC 00000FF 8 00000FF 4 00000FF 0 What would be the value of ESP after pushing the 32-bit value shown below onto the stack? 000000A5 000000A5 00000FF8 00001020 00000FFC