题目内容

在进程运行的过程中,把选择换出页面的算法叫做页面置换算法。其中,选择的被淘汰的页面,将是暂时不使用的或者在最长时间内不再被防问的页面的算法是 (6) ,总是选择淘汰最先进入内存的页面的算法是 (7) ,而选择淘汰在最近一段时间内最久未用的页面的算法是 (8) 。在一个请求分页系统中,假如系统分配给一个作业的物理块数为3,并且此作业的页向走向为2,3,2,1,5,2,4,5,3,2,5,2,则采用FIFO和LRU算法的缺页次数分别是 (9) , (10) 。 (8)处填()。

A. 最近最久未使用置换算法
B. 最佳置换算法
C. 最少使用置换算法
D. 先进先出置换算法
E. 页面缓冲算法
F. Clock置换算法

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在UNIX操作系统中,当用户执行如下命令Link("/user/include/myfile.sh","/usr/userwang/youfile.sh")则文件名"/usr/userwang/youfile.sh"存放在()。

A. user目录文件中
B. include目录文件中
C. userwang目录文件中
D. youfile.sh的文件内容中

You really do have to wonder whether a few years from now we’ll look back at the first decade of the 21st century—when food prices spiked, energy prices soared, world population surged, tornados plowed through cities, floods and droughts set records, populations were displaced and governments were threatened by the confluence of it all—and ask ourselves. What were we thinking How did we not panic when the evidence was so obvious that we’d crossed some growth, climate, natural resource and population redlines all at once "The only answer can be denial," argues Paul Gilding, an Australian environmentalist, in a new book called The Great Disruption. "When you are surrounded by something so big that requires you to change everything about the way you think and see the world, then denial is the natural response. But the longer we wait, the bigger the response required." Gilding cites the work of the Global Footprint Network, an alliance of scientists, which calculates how many "planet Earths" we need to sustain our current growth rates. G. F. N. measures how much land and water area we need to produce the resources we consume and absorb our waste, using prevailing technology. On the whole, says G. F. N. , we are currently growing at a rate that is using up the Earth’s resources far faster than they can be sustainably replenished, so we are eating into the future. This is not science fiction. This is what happens when our system of growth and the system of nature hit the wall at once. We are now using so many resources and putting out so much waste into the Earth that we have reached some kind of limit, given current technologies. The economy is going to have to get smaller in terms of physical impact. We will not change systems, though, without a crisis. But don’t worry, we’re getting there. We’re currently caught in two loops: One is that more population growth and more global warming together are pushing up food prices, causing political instability in the Middle East, which leads to higher oil prices, thus to higher food prices and more instability. At the same time, improved productivity means fewer people are needed in every factory to produce more stuff. So if we want to have more jobs, we need more factories. More factories making more stuff make more global warming, and that is where the two loops meet. But Gilding is actually an eco-optimist. As the impact o the imminent Great Disruption hits us, he says, "our response will be proportionally dramatic, mobilizing as we do in war. We will change at a scale and speed we can barely imagine today, completely transforming our economy, including our energy and transport industries, in just a few short decades. " We will realize, he predicts, that the consumer-driven growth model is broken and we have to move to a more happiness-driven growth model, based on people working less and owning less. To correct the situation, Gilding advocates

A. stabilizing the political and economic situation.
B. learning useful lessons from wartime mobilization.
C. keeping economic growth at a sustainable rate.
D. making better use of current technologies.

若有一个仓库,可以存放P1,P2两种产品,但是每次只能存放一种产品,要求:①w=P1的数量-P2的数量②-i<w<k(i,k为正整数)若用PV操作实现P1和P2产品的入库过程,至少需要 (9) 个同步信号量及 (10) 个互斥信号量,其中,同步信号量的初值分别为 (11) ,互斥信号量的初值分别为 (12) 。 (9)处填()。

A. 0
B. 1
C. 2
D. 3

有如下程序: #include<iostream> using namespace std; Class Con char ID; public: Con():ID(’A’)cout<<1; Con(char ID) :ID(ID) cout<<2; Con(Con& c):ID(c.getID())cout<<3; char get1D()constreturn ID; ; void show(Con c)cout<<c.get1D(); int main() Con c1; show(c1); Con c2(’B’); show(c2); return 0; 执行上面程序的输出是( )。

A. 13A23A
B. 23A13B
C. 13A23B
D. 13B23A

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