A machine has 48-bit virtual addresses and 32-bit physical addresses. Pages are 8K. How many entries are needed for a conventional page table ?
The answer is 2^35 entries
Why not (2^32) / (2^13) = 2^19 entries...?
My doubt here is why does the number of page table entries depend on virtual address space rather than physical address space?
What does it mean to have more virtual address space than physical address space? The process can only access as many pages present in the physical memory. So it makes sense to have same have both virtual and physical memory address space same. But why have more virtual address space? Does it mean 2 different virtual address entries are mapped to the same physical location in the same page table? Guess that would lead to inconsistency right? Can someone help me get the bigger picture here..... I really need to internalize this concept.