Q6 ) (Memory Organization) A computer provides each process with 65,536 bytes (2^16) of address space divided
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Q6 ) (Memory Organization) A computer provides each process with 65,536 bytes (2^16) of address space divided into pages of 4096 (2^12) bytes each.
- A particular program has a text size of 32,768 (2^15)bytes
- a data size of 16,386 (2^14+2) bytes
- and a stack size of 15,870 (2^13+ 7678) bytes.
Will this program fit in the machine’s address space? Suppose that instead of 4096 (2^12) bytes, the page size were 512 (2^9) bytes, would it then fit? Each page must contain either text, data, or stack, not a mixture of two or three of them.
Expert Answer:
Given that Allocated address space for each process is 65536 which is 2 16 bytes Page size is 4096 b... View the full answer
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