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Memory Management Questions and Answers

Answer: C

Correct answer: either low or high memory (depending on the location of interrupt vector)

The operating system is placed at whichever end of memory the interrupt vector occupies, since the vector must be reachable and is fixed by the hardware. On most machines that is low memory, but the principle rather than the address is what matters.

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183.

In repeat-until loop 'UNTIL Option>0 and Option<8 AND INT(Option) = Option' , INT function is classified as

Answer: B

Correct answer: function

INT takes a value and returns its integer part, so it is a function - here used inside the loop's condition to test whether Option is a whole number.

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184.

In internal fragmentation, memory is internal to a partition and :

Answer: B

Correct answer: is not being used

Internal fragmentation is the space inside an allocated block that the process does not use. It is allocated but wasted, which is what distinguishes it from external fragmentation, where the waste lies between blocks.

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Answer: D

Correct answer: A program may fail if the required library version is changed or missing

With dynamic linking the library is not bound into the executable but located at run time, so a program depends on finding a compatible version present on the system. If the library is missing, moved or replaced by an incompatible version, the program fails to run - the difficulty known informally as dependency hell. Its advantages are substantial, which is why it is universal. Executables are far smaller, since library code is not duplicated into each. One copy of a library in memory serves every process using it, mapped into each address space. And a library can be patched - for a security fix, say - and every program benefits without relinking. Version management is handled by numbering, so several versions can coexist and each program binds to the one it needs. Static linking avoids the dependency entirely by copying the library in, at the cost of size and of losing the shared updates. That is why it is still chosen for self-contained or embedded software.

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Answer: A

Correct answer: each process is contained in a single contiguous section of memory

Contiguous allocation means each process occupies one unbroken run of addresses. That is what makes it simple - a base and a limit describe the whole process - and also what causes external fragmentation.

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