假设磁盘块与缓冲区大小相同，每个盘块读入缓冲区的时间为15μs，由缓冲区送至用户区的时间是5μs，在用户区内系统对每块数据的处理时间为1μs。若用户需要将大小为10个磁盘块的Doc1文件逐块从磁盘读入缓冲区，并送至用户区进行处理，那么采用单缓冲区需要花费的时间为（请作答此空）μs；采用双缓冲区需要花费的时间为______μs。
- A.150
- B.151
- C.156
- D.201
正确答案及解析
正确答案
解析
这要理解流水线分段的原理，例如，流水线之所以能分成取址、执行两段，是因为第1个线程执行的时候，第2个线程可以去取址，这种叠加的关系。 题目中，单缓冲区的时候，前两段描述，即读入缓冲区、由缓冲区送到用户区，都需要用到单缓冲区，此时，必须等待线程1完成这两段， 即线程1到第三段：处理磁盘块的时候，线程2才能进入第1段，因此前两段必须合并成流水线中的一段，因为前两段无法让多个线程并行叠加执行。因此单缓冲区是两段，第一段20us，第二段1us，用流水线公式计算：21+（10-1）*20=201 双缓冲区就不存在这种情况，当线程1执行完第1段：读入缓冲区后，进入第2段：缓冲区送到用户区时，线程2就可以立即开始第1段，因此双缓冲区这三段正好是可以叠加执行，不用合并。因此双缓冲时直接按流水线时间计算：15+5+1+（10-1）*15=156
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