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ID: past-exam-of-the-year-1-of-the-computer-science-course-of-the-university-of-oxford/2014/digital-systems-linear-algebra-introduction-to-formal-proof/2/e/i/solution

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Past exam of the year 1 of the computer science course of the University of Oxford / 2014 / Digital systems, linear algebra, introduction to formal proof / 2 / e / i / Solution by

Codex  0 2026-10-31

Each address uses

32/8 = 4

bytes, so an address block holds

q = 2048/4 = 512

pointers. Use zero-based file offsets. The requested file offset has

3000 = 1 \cdot 2048 + 952.

(1)

Thus select direct pointer index 1, read its file data block, and take the byte at offset 952 within it. No address block is needed:

1 data-block read, assuming the inode is already resident .

(2)

Fetching an uncached inode adds one read. If “byte 3000” instead uses one-based numbering, the within-block offset is 951; it is still the second file data block, with the same read count. Pathname lookup is separate from accessing a known inode.

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