计算cpu空闲时间 [英] calculation of cpu idle time
问题描述
嗨
我有计算cpu空闲时间的问题，如果空闲时间是
，那么我必须关闭系统。请有人告诉我这个想法
请
hi
i have problem to calculate idle time of cpu and if idle time is
more then i have to shut down the system. can anyone tell me the idea
to so that please
推荐答案
ra ****** @ gmail.com 写道：
嗨
i计算cpu的空闲时间有问题，如果空闲时间是多少b $ b，那么我必须关闭系统。任何人都可以告诉我这个想法
以便请b
hi
i have problem to calculate idle time of cpu and if idle time is
more then i have to shut down the system. can anyone tell me the idea
to so that please
你的意思是如果CPU似乎处于空闲状态，你会把它关闭吗？ br />
关闭它不是C本身可以做的事情，但是如果你的C有'b
'系统'
函数，或者如果你可以把运行这个程序的命令放在一个
批处理文件中，你没问题。
只做一个系统（关闭 ;）或者说关闭在批处理文件中，
无论什么命令关闭你的计算机。
现在要弄清楚它是否空闲，这有点棘手。但是可行，如果
你的C有一个函数可以返回当前时间，最好是
小增量。我们的想法是在时间（）返回值之间形成间隔
的直方图，例如：
#include< time.h>
#define MaxMsec 1000
unsigned long int Histo [MaxMsec];
int i，j; time_t现在，Prev，Elap，Base;
for（i = 0; i< MaxHisto; i ++）Histo [i] = 0;
上一页= 0; Base = Prev = time（）;
for（j = 1; j< 100000; j ++）{
Now = time（） - Base; Elap = Now - Prev; Prev = Now;
If（Elap< 0）fprintf（stderr，Time to backwardwards !! \ n）
else
If（Elap> = MaxMsec）Elap = MaxMsec;
Histo [Elap] ++;
}
....然后扫描直方图。如果CPU非常空闲，您应该在非常低或零毫秒处看到一个
大峰值。如果有其他重要的任务在运行，你会看到其他峰值，通常在5到50美元/ b $ b毫秒范围内。 （许多OS的调度算法将时间分配为5到
50毫秒）。
可以肯定的是，绕过该代码循环它运行了几秒钟。
是的，不保证可以在任何地方工作，但似乎在
Windows和Linux上工作。
You mean if the CPU seems to be idle, you shut it down?
Shutting it down isnt something C can do by itself, but if your C has
the "system"
function, or if you can put the command that runs this program in a
batch file, you''re okay.
Just do a system( "shutdown") or put "shutdown" in the batch file,
whatever command shuts down your computer.
Now to figure out if it''s idle, that''s a bit tricky. But doable, if
your C has a function that returns you the current time, preferably in
small increments. The idea is to form a histogram of the intervals
between the values time() returns, something like:
#include <time.h>
#define MaxMsec 1000
unsigned long int Histo[ MaxMsec ];
int i, j; time_t Now, Prev, Elap, Base;
for( i = 0; i < MaxHisto; i++ ) Histo[i] = 0;
Prev = 0; Base = Prev = time();
for( j = 1; j < 100000; j++ ) {
Now = time() - Base; Elap = Now - Prev; Prev = Now;
If( Elap < 0 ) fprintf( stderr, "Time going backwards!!\n" )
else
If( Elap >= MaxMsec ) Elap = MaxMsec;
Histo[ Elap ] ++;
}
.... then scan the histogram. If the CPU is very idle, you should see a
big peak at very low or zero msec. If there are significant other
tasks running, you''ll see other peaks, usually in the range of 5 to 50
msec. (Many OS''s scheduling algorithms allocate time in slices of 5 to
50 msec).
To be sure, put a loop around that code so it runs for several seconds.
Yes, not guaranteed to work anywhere, but seems to do the job on
Windows and Linux.
Ancient_Hacker在08/23/06 11:45写道：
Ancient_Hacker wrote On 08/23/06 11:45,:
[...]
现在要弄清楚它是否闲置，这有点棘手。但是可行，如果
你的C有一个函数可以返回当前时间，最好是
小增量。我们的想法是在时间（）返回值之间形成间隔
的直方图，例如：
[Millikan油滴实验]
[...]
Now to figure out if it''s idle, that''s a bit tricky. But doable, if
your C has a function that returns you the current time, preferably in
small increments. The idea is to form a histogram of the intervals
between the values time() returns, something like:
[the Millikan oil-drop experiment]
这是一个糟糕的建议，编码严重。
-
Er ********* @sun.com
This is terrible advice, and badly coded.
--
Er*********@sun.com
Eric Sosman写道：
Eric Sosman wrote:
Ancient_Hacker写于08/23/06 11:45，：
Ancient_Hacker wrote On 08/23/06 11:45,:
[...]
现在要弄清楚它是否闲置，这有点棘手。但是可行，如果
你的C有一个函数可以返回当前时间，最好是
小增量。我们的想法是在时间（）返回值之间形成间隔
的直方图，例如：
[Millikan油滴实验]
[...]
Now to figure out if it''s idle, that''s a bit tricky. But doable, if
your C has a function that returns you the current time, preferably in
small increments. The idea is to form a histogram of the intervals
between the values time() returns, something like:
[the Millikan oil-drop experiment]
这是一个糟糕的建议，而且编码很糟糕。
-
Er ********* @ sun.com
非常感谢您的意见。
如果符合以下条件，您的意见会更加重要：
（1）您有一个可行的选择。
（2你解释了为什么这是严重的原因。编码。我看到一个蠢 - 我们没有检测到正确倒退的时间。否则有一些方法可以做一个简单的直方图。请随时教育我。
Thank you so much for your opinion.
Your opinion would have more weight if:
(1) You had a viable alternative.
(2) You explained why this is "badly" coded. I see one goof-- we''re
not detecting time going backwards correctly. Otherwise there are darn
few ways to do a simple histogram. Please feel free to educate me.
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