eksamensprepp Flashcards

(24 cards)

1
Q

what is inside code - memory

A

OS instantiates code, and other static data

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2
Q

what are in the stack

A

used for function calls, local variables

Fast, structured (LIFO)

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3
Q

what are in the heap

A

dynamic memory (e.g., malloc, new)

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4
Q

page and page frame size

A

page - fixed size block in virtual memory
page frame - fixed size block physical memory

same size (4KB)

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5
Q

what does condition variable do

A

producer to wait until there is space
consumer to wait until there is an item in the buffer

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6
Q

producer consumer problem - what is the goal

A

We need to ensure that when a producer is placing an item in the buffer, then at the same time consumer should not consume any item

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7
Q

what is the critical section i producer-consumer problem

A

the buffer

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8
Q

solve producer consumer

A

needs 2 semaphores - 1 full 1 empty

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9
Q

semaphore value = 0 -> then calls wait -> what happens

A

nothing, cant call wait, bc the thread sleeps

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10
Q

difference between SSD and HDD

A

SSD shorter life span - bc limited write cycles

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10
Q

must all queues in MLFQ use the scheduling algorithm

A

no

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11
Q

method to fix external fragmentation

A

compact - small blocks into one big

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12
Q

internal vs external fragmentation

A

Internal fragmentation = unused space inside an allocated block(pages)

External fragmentation = unused space between allocated blocks (heap, stack)

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13
Q

what is not shared among threads

A

stack registers of threads

each thread has its own stack

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14
Q

fork - hva returnerer den

A

returnere to forskjellige prossesser

parent - 0 >
child - returner 0

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15
Q

solution for critical section

A

mutual exlusion, bounded waiting, progress

16
Q

what does progress mean in ensuring critical section is safe

A

thread want to enter the critical section, and no one is inside
→ then some process must be allowed to proceed

17
Q

what is bounded waiting

A

here must be a limit (bound) on how many times other processes can enter their critical sections before a waiting process gets its turn

18
Q

hvor mange bytes er 1GB

19
Q

waiting time

A

waiting = turnaroud - execution time