Assume we are dealing with multilevel page table here Page Size is 4KB, Page table entry is 4B and the outer page table size is 256B then if the no of levels is 1 so Virtual Address is= 256 KB. but
when there are 2 levels of paging then please check my procedure to solve it. Virtual address space=no. of pages x page size.
6 |
10 |
but how to get offset?? |
here 6 is no of bit in outer table of second split and 10 is the no of bit in first split.