Recent questions tagged memory-management

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For each of the following decimal virtual addresses, compute the virtual page number and offset for a $4-KB$ page and for an $8 KB$ page$:20000, 32768, 60000.$
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What is the difference between a physical address and a virtual address?
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Consider a swapping system in which memory consists of the following hole sizes in memory order: ... for first fit? Now repeat the question for best fit, worst fit, and next fit.
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A swapping system eliminates holes by compaction. Assuming a random distribution of many holes and many data segments and a time to read or write a $32-bit$ ... is part of a hole and that the highest word in memory contains valid data.
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In Fig. $3-3$ the base and limit registers contain the same value, $16,384$. Is this just an accident, or are they always the same? It is just an accident, why are they the same in this example?
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The $IBM\:360$ had a scheme of locking $2-KB$ blocks by assigning each one a $4-bit$ key and having the CPU compare the key on every memory reference to ... $PSW$. Name two drawbacks of this scheme not mentioned in the text.
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Modern operating systems decouple a process address space from the machine’s physical memory. List two advantages of this design.
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I have read that paging does not suffer from external fragmentation as the frames and the pages are all of the equal sizes, but when we store a last ... , right? So why is it said that paging does not have external fragmentation?
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What is the mode(User/Kernel) of the Operating System when a Page Table or TLB are being accessed?
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i write this program, during initialization of array i given the size as 11 means the number of elements stored in an array is 11.as we know array is not assigned a value of index 12 and ... ; } for(i=0;i<20;i++) { printf("%d\n",a[i]); } }
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how to know how much size is allocated while dynamically allocating memory to the pointer variable with malloc.i just write the below code to know the size of ptr2 after dynamically ... of *ptr2 is == %d\n",(sizeof(ptr2)/sizeof(int))); } }
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Consider a system that allocates pages of different sizes to its processes. What are the advantages of such a paging scheme ? What modifications to the virtual memory system provide this functionality ?
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Consider the Intel address-translation scheme shown in Figure 8.22.a. Describe all the steps taken by the Intel Pentium in translating a logical address into a ... are they? If not, why is this scheme not used by every manufacturer ?
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Compare the segmented paging scheme with the hashed page table scheme for handling large address spaces. Under what circumstances is one scheme preferable to the other ?
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Consider the hierarchical paging scheme used by the $VAX$ architecture. How many memory operations are performed when a user program executes a memory-load operation ?
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Consider the following segment table: $Segment$ $Base$ $Length$ ... addresses for the following logical addresses ?$a. 0,430$b. 1,10$c. 2,500$d. 3,400$e. 4,112$
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Explain why sharing a reentrant module is easier when segmentation is used than when pure paging is used.
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Why are segmentation and paging sometimes combined into one scheme ?
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Consider a paging system with the page table stored in memory.a. If a memory reference takes 50 nanoseconds, how long does a paged memory reference take ?b. If we ... -table entry in the $TLBs$ takes 2 nanoseconds, if the entry is present.)
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Consider a logical address space of $256$ pages with a $4-KB$ page size, mapped onto a physical memory of $64$ frames.$a$. How many bits are required in the logical address ?$b$. How many bits are required in the physical address ?
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The BTV operating system has a $21-bit$ virtual address, yet on certain embedded devices, it has only a $16-bit$ physical address. It also has a $2-KB$ ... following ?$a$. A conventional, single-level page table$b$. An inverted page table
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Assuming a $1$ $KB$ page size, what are the page numbers and offsets for the following address references (provided as decimal numbers):$a. 3085$b. 42095$c. 215201$d. 650000$e. 2000001$
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Program binaries in many systems are typically structured as follows. Code is stored starting with a small, fixed virtual address, such as $0$. The code segment ... ?$a$. Contiguous memory allocation$b$. Pure segmentation$c$. Pure paging
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Explain why address space identifiers ($ASIDs$) are used.
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Compare paging with segmentation with respect to how much memory the address translation structures require to convert virtual addresses to physical addresses.
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Although Android does not support swapping on its boot disk, it is possible to set up a swap space using a separate SD nonvolatile memory card. Why would Android disallow swapping on its boot disk yet allow it on a secondary disk ?
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Explain why mobile operating systems such as iOS and Android do not support swapping ?
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On a system with paging, a process cannot access memory that it does not own. Why ? How could the operating system allow access to other memory ? Why should it or should it not ?