With a 2^32 address space and 4K ( 2^12 ) page sizes, this leave 2^20 entries in the page table. At 4 bytes per entry, this amounts to a 4 MB page table, which is too large to reasonably keep in contiguous memory. ( And to swap in and out of memory with each process switch. ) Note that with 4K pages, this would take 1024 pages just to hold the page table!
Please explain me how they are calculating 1024 pages at the end? what is 4bytes per entry in this? Thanks in advance.