At Westonci.ca, we make it easy for you to get the answers you need from a community of knowledgeable individuals. Get precise and detailed answers to your questions from a knowledgeable community of experts on our Q&A platform. Explore comprehensive solutions to your questions from knowledgeable professionals across various fields on our platform.
Sagot :
Answer:
The page field is 8-bit wide, then the page size is 256 bytes.
Using the subdivision above, the first level page table points to 1024 2nd level page tables, each pointing to 256 3rd page tables, each containing 64 pages. The program's address space consists of 1024 pages, thus we need we need 16 third-level page tables. Therefore we need 16 entries in a 2nd level page table, and one entry in the first level page table. Therefore the size is: 1024 entries for the first table, 256 entries for the 2nd level page table, and 16 3rd level page table containing 64 entries each. Assuming 2 bytes per entry, the space required is 1024 * 2 + 256 * 2 (one second-level paget table) + 16 * 64 * 2 (16 third-level page tables) = 4608 bytes.
First, the stack, data and code segments are at addresses that require having 3 page tables entries active in the first level page table. For 64K, you need 256 pages, or 4 third-level page tables. For 600K, you need 2400 pages, or 38 third-level page tables and for 48K you need 192 pages or 3 third-level page tables. Assuming 2 bytes per entry, the space required is 1024 * 2 + 256 * 3 * 2 (3 second-level page tables) + 64 * (38+4+3)* 2 (38 third-level page tables for data segment, 4 for stack and 3 for code segment) = 9344 bytes.
Explanation:
16 E the answer
Thank you for trusting us with your questions. We're here to help you find accurate answers quickly and efficiently. Thank you for visiting. Our goal is to provide the most accurate answers for all your informational needs. Come back soon. We're here to help at Westonci.ca. Keep visiting for the best answers to your questions.
