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______μs；采用双缓冲区需要花费的时间为（请作答此空）μs。
- A.150
- B.151
- C.156
- D.201
正确答案及解析
正确答案
解析
因为在块设备输入时，假定从磁盘把一块数据输入到缓冲区的时间为T，缓冲区中的数据传送到用户工作区的时间为M，而系统处理(计算)的时间为C，如下图所示。
单缓冲工作过程图当第一块数据送入用户工作区后，缓冲区是空闲的可以传送第二块数据。这样第一块数据的处理C1与第二块数据的输入T2是可以并行的，依次类推，如下图所示。系统对每一块数据的处理时间为：Max(C, T)+M。因为，当T＞C时，处理时间为M+T；当Timage.pngC时，处理时间为M+C。本题每一块数据的处理时间为15+5=20，Doc1文件的处理时间为20×10+1。
单缓冲并行工作示意图双缓冲工作方式基本方法是在设备输入时，先将数据输入到缓冲区1，装满后便转向缓冲区2。此时系统可以从缓冲区1中提取数据传送到用户区，最后由系统对数据进行处理，如下图所示。
双缓冲工作过程图双缓冲可以实现对缓冲区中数据的输入T和提取M，与CPU的计算C，三者并行工作，如下图所示。从图中可以看出，双缓冲进一步加快了I/O的速度，提高了设备的利用率。在双缓冲时，系统处理一块数据的时间可以粗略地认为是Max(C, T)。如果Cimage.pngT，可使块设备连续输入；如果C＞T，则可使系统不必等待设备输入。本题每一块数据的处理时间为10，采用双缓冲需要花费的时间为15×10+5+1=156。
双缓冲并行工作示意图
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