操作系统导论-课后作业-ch6

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补充：测量作业

1. 测试函数gettimeofday

代码如下：

#include <stdio.h>
#include <sys/time.h>
#include <stdlib.h>//return us
int time_diff(struct timeval* t1, struct timeval* t2) {return (t2->tv_sec - t1->tv_sec) * 1000000 + (t2->tv_usec - t1->tv_usec);
}int main() {int i;struct timeval curr, now;if (gettimeofday(&curr, NULL) == -1) {fprintf(stderr, "gettimeofday error\n");exit(1);}for (i = 0; i < 10; ++i) {if (gettimeofday(&now, NULL) != 0) {fprintf(stderr, "gettimeofday error\n");exit(1);}printf("time elaps %d us\n", time_diff(&curr, &now));}return 0;
}

执行结果如下：

最多精确到us，并且大致精确（并不完全精确），需要多迭代几次减少误差。

2. 测试系统调用时间

2.1 测试读取0字节耗时

本处以执行0字节的读取为例来测试此系统调用的时间：

#include <stdio.h>
#include <sys/types.h>
#include <sys/stat.h>
#include <fcntl.h>
#include <sys/time.h>
#include <stdlib.h>
#include <unistd.h>int time_diff(struct timeval* t1, struct timeval* t2) {return (t2->tv_sec - t1->tv_sec) * 1000000 + (t2->tv_usec - t1->tv_usec);
}int main() {int i;char buf[2];int fd = open("./tmp.txt", O_CREAT | O_APPEND | O_RDWR, S_IRWXU);struct timeval curr, now;if (gettimeofday(&curr, NULL) == -1) {fprintf(stderr, "gettimeofday error\n");exit(1);}for (i = 0; i < 100; ++i) {if (read(fd, buf, 0) != 0) {fprintf(stderr, "read error\n");exit(1);}}if (gettimeofday(&now, NULL) != 0) {fprintf(stderr, "gettimeofday error\n");exit(1);}printf("time elaps %d us\n", time_diff(&curr, &now));return 0;
}

执行结果如下：

执行了100次后的结果为19us，也就是每次执行调用耗时约0.2us。

2.2 测试write和read总时间

此处测试write和read管道的总时间，代码更改如下（这里是为实验三做准备）：

#include <stdio.h>
#include <sys/types.h>
#include <sys/stat.h>
#include <fcntl.h>
#include <sys/time.h>
#include <stdlib.h>
#include <unistd.h>int time_diff(struct timeval* t1, struct timeval* t2) {return (t2->tv_sec - t1->tv_sec) * 1000000 + (t2->tv_usec - t1->tv_usec);
}int main() {int i;char buf[] = "a";int fds[2];struct timeval curr, now;if (gettimeofday(&curr, NULL) == -1) {fprintf(stderr, "gettimeofday error\n");exit(1);}pipe(fds);for (i = 0; i < 20000; ++i) {if (write(fds[1], buf, 1) != 1) {fprintf(stderr, "read error\n");exit(1);}if (read(fds[0], buf, 1) != 1) {fprintf(stderr, "read error\n");exit(1);}}if (gettimeofday(&now, NULL) != 0) {fprintf(stderr, "gettimeofday error\n");exit(1);}printf("time elaps %d us\n", time_diff(&curr, &now));return 0;
}

测试结果如下：

可见在没有上下文切换的情况下，20000次的读取和写入需要11326 us。

3. 测量上下文切换成本

其原理是通过设置两个管道，第一个进程写入数据后读取数据，此时陷入阻塞，进程被挂起，切换到第二个进程，而第二个进程读取数据后，再次写入管道后自身进行读取操作进而陷入阻塞状态，系统再度切换到第一个进程执行，如此往复循环可以测试得到具体的切换成本。（多核系统注意调用sched_setaffinity函数）

#include <stdio.h>
#include <sys/types.h>
#include <sys/stat.h>
#include <fcntl.h>
#include <sys/time.h>
#include <stdlib.h>
#include <unistd.h>
#include <sys/wait.h>int main() {int i;char buf[] = "a";int fds1[2], fds2[2];struct timeval begin, end;pipe(fds1);pipe(fds2);int rc = fork();if (rc == -1){fprintf(stderr, "fork error\n");exit(1);} else if (rc == 0) {close(fds1[0]);close(fds2[1]);if (gettimeofday(&begin, NULL) == -1) {fprintf(stderr, "gettimeofday error\n");exit(1);}printf("test begin at %lu s and %lu us\n", begin.tv_sec, begin.tv_usec);for (i = 0; i < 10000; ++i) {if (write(fds1[1], buf, 1) != 1) {fprintf(stderr, "read error\n");exit(1);}if (read(fds2[0], buf, 1) != 1) {fprintf(stderr, "read error\n");exit(1);}}close(fds1[1]);close(fds2[0]);exit(0);}close(fds1[1]);close(fds2[0]);for (i = 0; i < 10000; ++i) {if (read(fds1[0], buf, 1) != 1) {fprintf(stderr, "read error\n");exit(1);}if (write(fds2[1], buf, 1) != 1) {fprintf(stderr, "read error\n");exit(1);}}if (gettimeofday(&end, NULL) != 0) {fprintf(stderr, "gettimeofday error\n");exit(1);}printf("test end at %lu s and %lu us\n", end.tv_sec, end.tv_usec);close(fds1[1]);close(fds2[0]);wait(NULL);return 0;
}

测试结果如下所示：

系统总用时的计算公式如下：
$$t_{all}=t_{switch\_all}+t_{read\_and\_write\_all}$$
这其中包含20000次对管道的读取和写入以及20000次的进程切换，通过计算可得
$$t_{switch}=\frac{t_{switch\_all}}{count}=\frac{t_{all}-t_{read\_and\_write\_all}}{count}$$
最终得到切换所需要的总切换开销为26114us而单次切换的开销大约为1.3us。

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