Operating Systems

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SegmentationCPU


Example


Sharing of Segments


Protection Associate valid/invalid bit with each segment table entry to indicate if the referenced segment is part of the process address space or not Read, write, and execute bits to define legal operations on a segment


Paged Segmentation Divide every segment in a process into fixed size pagesNeed for a page table per segmentCPUs memory management unit must support both segmentation and paging


Paged Segmentation 1324logical memory 5physical memory


Paged Segmentation 1324logical memory 5physical memory12030123...

10

126 1273112610page table


Paged Segmentation Logical address is still , with s used to index the segment tableEach segment table entry consist of the tuple The logical address is legal if d < segment-length


Paged Segmentation Segment offset, d, is partitioned into two parts: p and d, where p is used to index the page table associated with segment, s, and d is used as offset within a page


Paged Segmentation p indexes the page table to retrieve frame, f, and physical address (f,d) is formedsdpdindex segment tableindex page tableoffset within the page p


Paged Segmentation


GE 345 processorLogical address = 34 bits Page size = 1 KBs is 18 bits and d is 16 bitsSize of p and d, largest segment size, and max. number of segments per process?


Largest segment= 2d bytes= 216 bytesMaximum number of pages per segment= 216 / 1 K = 64|p| = log2 64 bits= 6 bits|d| = log2 1 K= 10 bitsMaximum number of segments per process= 2s = 218


sdpd18 bits6 bits10 bits


Consider a process with its segment 15 having 5096 bytes. The process generates a logical address (15,3921). Is it a legal address? How many pages does the segment have? What page does the logical address refer to?


Is it a legal address? YesHow many pages does the segment have? ceiling[5096/1024]= 5What page does the logical address refers to? ceiling[3921/1024]= 4(i.e., page number 3)


What are the value of d and the physical address if page number 3 (i.e., the fourth page) is in frame 12?d = 3921 3*1K= 849Physical address= 12*1K + 849 = 13137


1539213849spddpage table for segment 150123412


15 39213 849392112 12 84913137


IBM OS/2, Microsoft Windows, and LinuxPaged segmentation with two-level pagingLogical address = 48 bits 16-bit selector and 32-bit offsetPage size = 4 KB


4-byte page table entry32-entry TLB, covering 32*4K (128 KB) memory TLB Reach


13-bit Segment #16-bit Selectorgp32-bit Offsets2-bit field for specifying the privilege level1-bit field to specify GDT or LDT


Real Mode20-bit physical address is obtained by shifting left the Selector value by four bits and adding to it the 16-bit effective address


Operating Systems

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