Question

Base no bit 0 0 Q-2 (25p): Assume that logical addresses are 16-bit long: 3 bits for segment no and 13 bits for offset. Segment table of the currently running process is given below: a) (5) How many segments can this Segment Present Modify Length process have at maximum? bit 0 00100 0111 0000 0100 0000 1000 0100 b) (15) Convert the following logical 10100 0100 0001 0010 0010 1100 1000 address to physical (real) address. 2 01100 0100 0000 0010 0000 1000 0011 0100 1001 1101 1110 0 1000 0000 0000 1110 0000 0010 0000 0 0100 0000 0000 0010 0110 0000 1100 1 0001 0000 0000 0000 1000 1100 1000 c) (5) Do you think that converting to physical address results in a segment 0 0001 1100 0000 0010 0000 1111 0000 fault? Explain it briefly. 0 1100 0000 1000 0011 0000 1001 0000 0 3 0 4 0 0 5 0 0 6 0 0 7 0

Answer 1

a)

There are 3 bits for segement, so the segment numbers can range from 0 (000) to 7 (111).

So, a maximum of 8 segments are possible.

b)

0100 1001 1101 1110

Here, the highest 3 bits denote segment number, equal to 010, equivalent to decimal 2. Also, the offset is binary 0 1001 1101 1110.

As seen from segment table, segment 2 is present in memory, and its base address is binary 0010 0000 1000 0011.

So, the physical address is given by

= (Base address) + (Offset)

= 0010 0000 1000 0011 + 0 1001 1101 1110

= Binary 0010 1010 0110 0001

= Decimal 10849

c)

When converting logical addresses to physical addresses, we can certainly get addresses which are present in segments that are not present/loaded for the particular process in hand.

This results in segmentation faults, and is an effective mechanism to prevent a process from unauthorized access of addresses (of the same/different processes).