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spayn [35]
3 years ago
12

Suppose that we have an implicit-free list allocator for which header and footer blocks are eight bytes each. To reduce overhead

, only free blocks have footers. The alignment requirement of this allocator is only eight bytes (instead of the typically 16 for a 64-bit system). What is the minimum size (in bytes) of any block, allocated or free?
Computers and Technology
1 answer:
Sergio039 [100]3 years ago
3 0

Answer:

16 bytes is minimum size of any block.

Explanation:

The minimum block size is 16 bytes. 4 additional bytes added for a header and blocks are allocated in multiples of 8 bytes.

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