In the request paging storage management solution, if a user space has 16 pages, the page length is 1KB, and the existing page table is as follows, the physical address corresponding to the logical address 0A1F (H) is ().

In the request paging storage management solution, if a user space has 16 pages, the page length is 1KB, and the existing page table is as follows, the physical address corresponding to the logical address 0A1F (H) is ().

Two ways:

1. Logical address% 1024 (that is, the page size of 1k) = address offset
2. Logical address / 1024 (ie 1k page size) = page number
3. According to the page number, check the page table to get the block number
4. Physical address = block number X page size + address offset

The page length is 1KB 2^10=1k. The page length is 10 bits,
so the logical address 0A1F (H) is converted into binary bits 0000 1010 0001 1111 (underlined as a page)
. The 2 (binary 10) in front of it is the page number. Look up the table to find a block The number is 3, so the physical address is 0000 1110 0001 1111 (0E1F)
because it is paging storage management, so the condition that the page size is equal to the page frame size is hidden.