A system with 2-level page table, the frame size is 4KB; the address register is 32 bits, the row size of the page table is 4 bytes; the address register is divided into (p1, p2, offset) for fast address translation (no need to use multiplication and addition for calculating the physical address). Given a reference =4200746, the corresponding (p1, p2, offset) is?
Question
A system with 2-level page table, the frame size is 4KB; the address register is 32 bits, the row size of the page table is 4 bytes; the address register is divided into (p1, p2, offset) for fast address translation (no need to use multiplication and addition for calculating the physical address). Given a reference =4200746, the corresponding (p1, p2, offset) is?
Solution
To solve this problem, we first need to understand the structure of the address in a two-level paging system. The 32-bit address is divided into three parts: p1, p2, and offset.
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Determine the size of the offset: The frame size is 4KB, so the offset is 12 bits (since 2^12 = 4096, which is 4KB).
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Determine the size of p1 and p2: The row size of the page table is 4 bytes, which means each entry can point to 2^12 = 4096 different pages. So, both p1 and p2 are 10 bits each (since 2^10 = 1024, which is close to 4096).
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Convert the reference to binary: The reference is 4200746. Convert this to binary, you get 100000000110001101010.
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Split the binary into (p1, p2, offset): From the right, take the first 12 bits as the offset, the next 10 bits as p2, and the remaining bits as p1. So, you get p1 = 100, p2 = 0000011000, offset = 0110101010.
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Convert back to decimal: Convert these binary numbers back to decimal to get the final answer. So, p1 = 4, p2 = 8, offset = 682.
So, the corresponding (p1, p2, offset) for the reference 4200746 is (4, 8, 682).
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