If there are 32 segments, each of size 1 K byte, then the logical address should have
| A. | 13 bits |
| B. | 14 bits |
| C. | 15 bits |
| D. | 16 bits |
Answer : C Explanation : Th specify a particular segment, 5 bits are required (since 25 = 32). . Having selected a page, to select a particular byte one needs 10 bits (since 210 = 1 k byte). So, totally 5 + 10 =15 bits are needed. |
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Option: A Explanation : Explanation will come here. Explanation will come here. Explanation will come here. Explanation will come here. Explanation will come here. |
