Question

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?

Answer 1

16 bytes is minimum size of any block.

Step-by-step explanation:

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