2. (14 points) The following pseudocode swaps two values in a data structure if both entries...

   

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2. (14 points) The following pseudocode swaps two values in a data structure if both entries (holding the values) are not null. If either entry is NULL, the data structure is left as before the swap was attempted. The routine must appear to occur atomically and must be highly concurrent - it must allow multiple swaps between unrelated entries in parallel. You may assume that index1 and index2 never refer to the same entry. SwapNonNull(index1, data[index1]->lock. datal = getData(index1); data[index2] ->lock. Acquire(); data2 getData(index2); if (data1 == NULL || data2 else { temp data1->val data2->val temp; } index2) { Acquire(); = return; data1->val; /* May involve disk I/0 */ /* May involve disk I/0 */ NULL) return; == data2->val; storeData (data1); /* Involves disk I/0 */ storeData (data2); /* Involves disk I/0 */ Release(); data[index1]->lock. data[index2]->lock. Release (); State whether the aforementioned routine either (1) works, (ii) doesn't work, or (iii) is dangerous - that is, sometimes works and sometimes doesn't. If the implementation does not work or is dangerous, explain why (there may be several errors) and show how to fix it so it does work. 2. (14 points) The following pseudocode swaps two values in a data structure if both entries (holding the values) are not null. If either entry is NULL, the data structure is left as before the swap was attempted. The routine must appear to occur atomically and must be highly concurrent - it must allow multiple swaps between unrelated entries in parallel. You may assume that index1 and index2 never refer to the same entry. SwapNonNull(index1, data[index1]->lock. datal = getData(index1); data[index2] ->lock. Acquire(); data2 getData(index2); if (data1 == NULL || data2 else { temp data1->val data2->val temp; } index2) { Acquire(); = return; data1->val; /* May involve disk I/0 */ /* May involve disk I/0 */ NULL) return; == data2->val; storeData (data1); /* Involves disk I/0 */ storeData (data2); /* Involves disk I/0 */ Release(); data[index1]->lock. data[index2]->lock. Release (); State whether the aforementioned routine either (1) works, (ii) doesn't work, or (iii) is dangerous - that is, sometimes works and sometimes doesn't. If the implementation does not work or is dangerous, explain why (there may be several errors) and show how to fix it so it does work.

Answer rating: 100% (QA)

The correct option is iii Is dangerous because if any one ... View the full answer