Question

We decide to include metadata and padding for each page table entry such that each entry in our page table is exactly 32 bits long. For the purposes of this class, we assume that our page table is structured like an array of entries, with a slot for each virtual address. How many page table entries are required for the full page table?

Answer 1

To determine the number of page table entries required for a full page table, you must know the virtual address space size. With a 32-bit address and 4KB pages, you would need 2^20 page table entries, assuming each entry is 32 bits long.

Step-by-step explanation:

The question is asking how many page table entries are required for a full page table, given that each entry is 32 bits long. To answer this, we need to know the size of the virtual address space, as that will dictate the number of entries in the page table.

Typically, the number of entries in a page table is equal to the number of pages in the virtual memory. For example, if we have a 32-bit address space and a page size of 4KB (2^12 bytes), we can calculate the number of pages by dividing the address space by the page size: 2^32 / 2^12 = 2^20 pages.

Thus, there would be 2^20 page table entries required for the full page table, with each entry being 32 bits long.