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 Consider the following segment table in segmentation scheme :

$\begin{array}{|l|l|l|l|l|}\hline  \textbf{Segment ID} & \textbf{Base} & \textbf{Limit} \\\hline \text{0} & \text{200} & \text{200} \\\hline \text{1} & \text{5000} & \text{1210} \\\hline \text{2} & \text{1527} & \text{498} \\\hline \text{3} & \text{2500} & \text{50} \\\hline  \end{array}$

 What happens if the logical address requested is – Segment Id $2$ and offset $1000$?

  1. Fetches the entry at the physical address $2527$ for segment Id $2$
  2. A trap is generated
  3. Deadlock
  4. Fetches the entry at offset $27$ in Segment Id $3$
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Best answer
6 votes
6 votes

Answer: (b)

Explanation: Offset 1000 in Segment Id 2 is crossing the Segment Id 2 memory area. Here Segment Id 2 memory area is 1527 to 1527 + 498 = 2025.

selected by
4 votes
4 votes
an os trap signal will be generated. as offset tell how much deeper we have to go in the segment. as we have given 1000 as offset while the segment size is 498 . it will try to read another process memory which should not be allowed in any case
0 votes
0 votes
It should be a TRAP  as the limit   > offset  here for id 2
Answer:

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