Question: In________memory binding are changed in such manner that all free memory area can be merged to form a single free memory area.
Answer:
It is achieved by packing all allocated areas towards once end of the memory.
MCQ: In________memory binding are changed in such manner that all free memory area can be merged to form a single free memory area.
Correct Answer:
A. Memory Paging
Explanation:
It is achieved by packing all allocated areas towards once end of the memory.
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Related Questions
- 1. Which of the following statements stands true for memory compaction method. <ol> <li>it involves movement of code and data in the memory.</li> <li>it is feasible only if computer system provides relocation register;the relocation can be achieved by simply changing the address in the relocation register</li> <li>it does not involves movement of code and data in the memory</li> <li>it does not involves use of relocation register</li> </ol>
- 2. Buddy system and power of 2 allocators leads to ___________
- 3. In the Binary buddy system the sizes of memory blocks are ___________ and separate free lists are maintained for blocks of ___________sizes.
- 4. Which of the following statement is true for buddy system allocators? <ol> <li>Buddy System splits and recombines memory blocks in a predetermined manner during allocation and deallocation.</li> <li>No splitting of blocks takes place, also no effort is made to coalesce adjoining blocks to form larger blocks; when released, a block is simply returned to its free list.</li> <li> When a request is made for m bytes, the allocator first check the free list containing blocks whose size is 2<sup>i</sup> for the smallest value of i such that 2<sup>i</sup> ≥ m.if the free list is empty ,it checks the list containing blocks that are higher the next higher power of 2 in size on so on. an entire block is allocated to a request.</li> <li> When a request is made for m bytes. the system finds the smallest power of 2 that is ≥ m. Let this be 2<sup>i</sup>.if the list is empty, it checks the lists for block of size 2<sup>i+1</sup>.it takes one block off this list and splits it into two halves of size 2<sup>i</sup>.it put one of these blocks into the free list of size 2<sup>i</sup>,and uses the other block to satisfy the request.</li> </ol>
- 5. Which of the following statement is true for Power-of-2 allocators? <ol> <li>Buddy System splits and recombines memory blocks in a predetermined manner during allocation and deallocation.</li> <li>No splitting of blocks takes place, also no effort is made to coalesce adjoining blocks to form larger blocks; when released, a block is simply returned to its free list.</li> <li> When a request is made for m bytes, the allocator first check the free list containing blocks whose size is 2<sup>i</sup> for the smallest value of i such that 2<sup>i</sup> ≥ m.if the free list is empty ,it checks the list containing blocks that are higher the next higher power of 2 in size on so on. an entire block is allocated to a request.</li> <li> When a request is made for m bytes. the system finds the smallest power of 2 that is ≥ m. Let this be 2<sup>i</sup>.if the list is empty, it checks the lists for block of size 2<sup>i+1</sup>.it takes one block off this list and splits it into two halves of size 2<sup>i</sup>.it put one of these blocks into the free list of size 2<sup>i</sup>,and uses the other block to satisfy the request.</li> </ol>
- 6. The buddy and power of 2 allocators are faster than the first-fit ,best-fit, next-fit allocators
- 7. The power of 2 allocators is faster than the Buddy System allocators.
- 8. Contiguous memory allocation is the classical memory allocation model in which
- 9. Contiguous memory allocation faces the problem of
- 10. In Contiguous memory allocation _________ has no cure
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