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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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Check the exlanation

Explanation:

1) add $s0,$s0,$s1 \rightarrow R-format

   

opcode                 rs        rt      rd          shamt      function

000000            $s0       $s1    $s0        00000    100000

000000            10000  10001  10000    00000  100000

  = 0X42118020

2) lw $s0,0X20($t7)\rightarrow I-format

opcode                     rs      rt            offset

100000                  $s0     $t7          0X20

100000               10000    01111       0000000000010000

  = 0X820F0010

3) addi $s0,$s0,-10 \rightarrow I-format

opcode             rs           rt            offset

001000            $s0         $s0       -10

001000             10000   10000    1111111111110110

= 0X2210FFF6

4) ori $s1,$a0,100 \rightarrow I-format

opcode             rs        rt            offset

001101             $s0     $a0         100

001101            10000  00100    0000000001100100

= 0X36040064

5) j 0X400C \rightarrow J-format

opcode  rs

000010  0X400C

000010  0000000000100000000001100

= 0X0800400C

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