Convert the +virtual address 0x12345678 into a physical address solution?

Please show your work

Suppose that a page contains 4Kbytes. A virtual address and a physical address are both 32 bits wide. Here is the TLB: Page Table Index Virtual Page Number Physical Page Number
0 0x00678 0x00027
1 0x01234 0x00543
2 0x12345 0x00200
3 0x45678 0x00005

There are some missing parts to your question. Specifically, what sort of MMU is in use. We will assume a single level of page tables, and a 4KiB page size.

Given that the virtual address V is 0x12345678, and that the page size P is 4KiB.

A memory access is made to location V. The processor splits the address V into two sections. The upper bits represent the page table, and the lower bits the page offset.

Since there are 4KiB bytes per page, 12 bits are needed for the page offset. The page table T (upper 20 bits) is then 0x12345 and the offset O (lower 12 bits) within that page is 0x678.

Now, since we have a single level page table, that page table would need 1 million entries to cover a 4GiB address space. We index into the page table, and either find that there is no entry for T, in which case we issue a "Segmentation fault" type error message and terminate the program, or that we have a valid entry, or a paged entry.

If we have a paged entry, the data for the page is resident in the swap file. We choose another page to swap out (using LRU - least recently used - approximation), and swap in the page, mark the entry valid, and continue. Note that we receive an exception from the processor on access, and we will have to restart the instruction in this case.

If we have a valid entry, we retrieve the physical base address A (say 0x54321) from the descriptor in the virtual address translation table, then add O (the offset) to it, resulting in the final physical address of 0x54321678.

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