blocks|key|319599|text|clock_t是整数类型。您不能使用%25f将其打印出来。有关差异为0的原因，请参阅Fred's+answer。|type|unstyled|depth|inlineStyleRanges|entityRanges|offset|length|data|319600|entityMap|0|LINK|mutability|MUTABLE|url|https://stackoverflow.com/questions/2134363/the-c-clock-function-just-returns-a-zero/2134389#2134389^0|14|D|0|0^^$0|@$1|2|3|4|5|6|7|L|8|@]|9|@$A|M|B|N|1|O]]|C|$]]|$1|D|3|-4|5|6|7|P|8|@]|9|@]|C|$]]]|E|$F|$5|G|H|I|C|$J|K]]]]

clock_t is an integer type. You can't print it out with %f. See <a href="https://stackoverflow.com/questions/2134363/the-c-clock-function-just-returns-a-zero/2134389#2134389">Fred's answer</a> for why the difference is 0.

blocks|key|318674|text|man+clock。它不会返回您认为的结果。还有man+gettimeofday+-它更有可能是你想要的。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|318675|entityMap^0|0|9|O|G|0^^$0|@$1|2|3|4|5|6|7|H|8|@$9|I|A|J|B|C]|$9|K|A|L|B|C]]|D|@]|E|$]]|$1|F|3|-4|5|6|7|M|8|@]|D|@]|E|$]]]|G|$]]

<code>man clock</code>. It's not returning what you think it is. Also <code>man gettimeofday</code> - it's more likely what you want.

blocks|key|316651|text|+printf("start+=+%25.20f\nend+++=+%25.20f\n",+start,+end);|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|316652|应该是：|unstyled|316653|+printf("start+=+%25d\nend+++=+%25d\n",+start,+end);|316654|entityMap^0|0|0|0^^$0|@$1|2|3|4|5|6|7|K|8|@]|9|@]|A|$B|C]]|$1|D|3|E|5|F|7|L|8|@]|9|@]|A|$]]|$1|G|3|H|5|6|7|M|8|@]|9|@]|A|$B|C]]|$1|I|3|-4|5|F|7|N|8|@]|9|@]|A|$]]]|J|$]]

<pre><code> printf("start = %.20f\nend = %.20f\n", start, end);
</code></pre>

should be:

<pre><code> printf("start = %d\nend = %d\n", start, end);
</code></pre>

blocks|key|318683|text|clock()报告使用的CPU时间。sleep()不占用任何CPU时间。所以你的结果可能是完全正确的，只是不是你想要的。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|318684|entityMap^0|0|7|I|7|0^^$0|@$1|2|3|4|5|6|7|H|8|@$9|I|A|J|B|C]|$9|K|A|L|B|C]]|D|@]|E|$]]|$1|F|3|-4|5|6|7|M|8|@]|D|@]|E|$]]]|G|$]]

<code>clock()</code> reports CPU time used. <code>sleep()</code> doesn't use any CPU time. So your result is probably exactly correct, just not what you want.

blocks|key|321617|text|调用sleep()不会占用任何CPU时间。不过，您应该看到一些不同之处。我更正了这一行中的printf类型不匹配错误：|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|321618|printf("start+=+%25.20f\nend+++=+%25.20f\n",+start,+end);|code-block|syntax|javascript|321619|它在我的机器上给出了合理的结果：|321620|start+=+1419
end+++=+1485
delta+=+66
cpu_time_used++=+0.000066000000000
CLOCKS_PER_SEC+=+1000000|321621|您可以尝试使用gettimeofday()来获取运行程序所用的实时时间。|321622|entityMap^0|2|7|0|0|0|0|7|E|0^^$0|@$1|2|3|4|5|6|7|S|8|@$9|T|A|U|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|V|8|@]|D|@]|E|$I|J]]|$1|K|3|L|5|6|7|W|8|@]|D|@]|E|$]]|$1|M|3|N|5|H|7|X|8|@]|D|@]|E|$I|J]]|$1|O|3|P|5|6|7|Y|8|@$9|Z|A|10|B|C]]|D|@]|E|$]]|$1|Q|3|-4|5|6|7|11|8|@]|D|@]|E|$]]]|R|$]]

Calling <code>sleep()</code> isn't going to use up any CPU time. You should see a little difference, though. I corrected your printf type mismatch bug in this line:

<pre><code>printf("start = %.20f\nend = %.20f\n", start, end);
</code></pre>

And it gave reasonable results on my machine:

<pre><code>start = 1419
end = 1485
delta = 66
cpu_time_used = 0.000066000000000
CLOCKS_PER_SEC = 1000000
</code></pre>

You might try <code>gettimeofday()</code> to get the real time spent running your program.

blocks|key|321624|text|你可能需要|type|unstyled|depth|inlineStyleRanges|entityRanges|data|321625|double+get_wall_time(){+++++++++struct+timeval+time;+++++++++if+(gettimeofday(&time,NULL)){++++++++++++++return+0;+++++++++}+++++++++return+(double)time.tv_sec+%2B+(double)time.tv_usec+*+.000001;+++++}和用法类似于|offset|length|style|CODE|321626|​|321627|double+wall0+=+get_wall_time();+++++++++double+cpu0++=+get_cpu_time();+++++++++for(long+int+i+=+0;+i<=10000000;i%2B%2B){+++++++++++++++++func1();+++++++++}+++++++++double+wall1+=+get_wall_time();+++++++++double+cpu1++=+get_cpu_time();+++++++++cout+<<+"Wall+Time+=+"+<<+wall1+-+wall0+<<+endl;+++++++++cout+<<+"CPU+Time++=+"+<<+cpu1++-+cpu0++<<+endl;|321628|代替clock()|321629|仅作为时钟，并且仅基于性能计数器计算在CPU中花费的时间。但是你可以使用上面的函数得到结果。为了验证您的应用程序，请使用time命令运行它|321630|像time+./a.out一样|321631|time命令的输出：|321632|real++++0m5.987s+user++++0m0.674s+sys+++++0m0.134s|321633|和自定义函数Wall+Time+=+5.98505+CPU+Time++=+0.8的输出|321634|entityMap^0|0|0|5J|0|0|0|9K|0|2|7|0|0|1|C|0|0|0|1E|0|6|Z|0^^$0|@$1|2|3|4|5|6|7|Z|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|10|8|@$D|11|E|12|F|G]]|9|@]|A|$]]|$1|H|3|I|5|6|7|13|8|@]|9|@]|A|$]]|$1|J|3|K|5|6|7|14|8|@$D|15|E|16|F|G]]|9|@]|A|$]]|$1|L|3|M|5|6|7|17|8|@$D|18|E|19|F|G]]|9|@]|A|$]]|$1|N|3|O|5|6|7|1A|8|@]|9|@]|A|$]]|$1|P|3|Q|5|6|7|1B|8|@$D|1C|E|1D|F|G]]|9|@]|A|$]]|$1|R|3|S|5|6|7|1E|8|@]|9|@]|A|$]]|$1|T|3|U|5|6|7|1F|8|@$D|1G|E|1H|F|G]]|9|@]|A|$]]|$1|V|3|W|5|6|7|1I|8|@$D|1J|E|1K|F|G]]|9|@]|A|$]]|$1|X|3|-4|5|6|7|1L|8|@]|9|@]|A|$]]]|Y|$]]

You probably need 

<code>double get_wall_time(){
 struct timeval time;
 if (gettimeofday(&amp;time,NULL)){
 return 0;
 }
 return (double)time.tv_sec + (double)time.tv_usec * .000001;
 }
</code>
and usage something like 

<code>double wall0 = get_wall_time();
 double cpu0 = get_cpu_time();
 for(long int i = 0; i&lt;=10000000;i++){
 func1();
 }
 double wall1 = get_wall_time();
 double cpu1 = get_cpu_time();
 cout &lt;&lt; "Wall Time = " &lt;&lt; wall1 - wall0 &lt;&lt; endl;
 cout &lt;&lt; "CPU Time = " &lt;&lt; cpu1 - cpu0 &lt;&lt; endl;
</code>

instead of 
 <code>clock()</code>

as clock only and only counts time spent in CPU only based on performance counters.
but you can get result by using above function.
just to verify your application run it with time command

like <code>time ./a.out</code>

output of time command :
<code>real 0m5.987s
user 0m0.674s
sys 0m0.134s
</code>

and output of custom function
<code>Wall Time = 5.98505
CPU Time = 0.8
</code>

The C <code>clock()</code> function just returns me a zero. I tried using different types, with no improvement... Is this a good way to measure time with good precision?

<pre><code>#include &lt;time.h&gt;
#include &lt;stdio.h&gt;

int main()
{
 clock_t start, end;
 double cpu_time_used;

 char s[32];

 start = clock();

 printf("\nSleeping 3 seconds...\n\n");
 sleep(3);

 end = clock();

 cpu_time_used = ((double)(end - start)) / ((double)CLOCKS_PER_SEC);

 printf("start = %.20f\nend = %.20f\n", start, end);
 printf("delta = %.20f\n", ((double) (end - start)));
 printf("cpu_time_used = %.15f\n", cpu_time_used);
 printf("CLOCKS_PER_SEC = %i\n\n", CLOCKS_PER_SEC);

 return 0;
}
</code></pre>

<blockquote>
<pre><code>Sleeping 3 seconds...

start = 0.00000000000000000000
end = 0.00000000000000000000
delta = 0.00000000000000000000
cpu_time_used = 0.000000000000000
CLOCKS_PER_SEC = 1000000
</code></pre>
</blockquote>

Platform: Intel 32 bit, RedHat Linux, gcc 3.4.6

The C `clock()` function just returns a zero

Linux

C clock()函数只返回一个零。我试着使用不同的类型，没有任何改进...这是一种精确测量时间的好方法吗？#include <time.h>#include <stdio.h>int main(){ clock_t start, end; double cpu_time_used; char s[3...

问函数的作用是返回一个零
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