Sample Solutions to Project 1

Transcription

1 CS Introduction to Operating System Spring 2016 Dr. Zhizhang Shen 1 An example Sample Solutions to Project 1 Below is a simple example involving the fork() call. /home/plymouth/zshen > more testp1.c #include <stdlib.h> #include <sys/types.h> #include <unistd.h> int globalvar = 6; /* external variable in the initialized data */ char buf[] = "Write to the standard output\n"; int localvar; /* local variable in the initialized data */ pid_t pid; localvar = 88; if (write(stdout_fileno, buf, sizeof(buf)-1)!= sizeof(buf)-1){ printf("write error"); exit(1); printf("now, the fork starts\n"); if ( (pid = fork()) < 0){ printf("fork error\n"); exit(1); else if (pid == 0) { globalvar++; localvar++; //1. //2. else sleep(2); printf("pid = %d, globalvar = %d, localvar = %d\n", getpid(), globalvar, localvar); 1

2 /home/plymouth/zshen > cc testp1.c /home/plymouth/zshen >./a.out Write to the standard output Now, the fork starts pid = 24661, globalvar = 6, localvar = 88 pid = 24662, globalvar = 7, localvar = 89 /home/plymouth/zshen > 1.1 A few questions Read through the code and understand what each and every line is doing. Play with the above program, write down the output, and then answer the following questions. 1. What is the logically next stuff after this fork() call in the above example? A solution: It is the assignment to pid, followed by the comparison <0, i.e., pid = fork() 2. What are the child and parent pids when you execute the above program? A solution: Mines are 3824, and 3825, respectively. 3. Which parts are done by both child and parent processes, which by child only, and which by the parent only? A solution: Both the parent and the child processes will do the above assignment and several comparisons, and the last printf and exit(0). Only the child process will do the else if part, i.e., the variable increment part, and only the parent process will sleep(2). 4. Why does only the child process increment the two variables? Why are both the global and the local variables affected? A solution: Because its pid equals 0, thus doing the whole segment, where both the global and the local variables are changed. 5. What happens if you take out the comment made in front of as marked with 1., i.e., at the end of the pid==0 segment, and why? A solution: The child process will finish right there, thus it will not do the last printf. 2

3 6. What happens if you take out the comment made in front ofelse sleep(2), as marked with 2, and why? A solution: Then, because of the uncertainty of process scheduling, the parent process might do the printf first. 7. When we take off the comment in front of the else sleep(2) part, which process (child or parent) executes the sleep command? Why is that? A solution: As we explained earlier, it is the parent process that will sleep, since its pid is greater than 0, thus fitting else. 2 Another example More realistically, a parent wants to create a process to get something done, and will actually wait until the newly created process to complete. The following code, consisting of two files, demonstrates this situation. /home/plymouth/zshen > more parent.c /* The parent.c file */ #include <sys/wait.h> #include <stdlib.h> if(fork()==0){ //Notice the following name of "child" is not an accident execve("child", NULL, NULL); //1. printf("i am done\n"); printf("process[%d]: Parent in execution...\n", getpid()); //2. sleep(1); //3. if(wait(null)>0) printf("process[%d]: Parent detects terminating child \n", getpid()); printf("process[%d]: parent terminating...\n", getpid()); /home/plymouth/zshen > cc parent.c /home/plymouth/zshen > more child.c /* The child.c file */ printf("process[%d]: child in execution...\n", getpid()); sleep(2); 3

4 printf("process[%d]: child terminating...\n", getpid()); /home/plymouth/zshen > cc child.c -o child /home/zshen >./a.out Process[4234]: Parent in execution... Process[4235]: child in execution... Process[4235]: child terminating... Process[4234]: Parent detects terminating child Process[4234]: parent terminating... /home/plymouth/zshen > 2.1 What does it do? The parent program will create a process, which executes the child program using the system function execve(3), as contained in child.c. The parent then makes a system call of wait to block itself until the OS kernel signals the process to continue again, because one of its child processes has just terminated. 2.2 A few more questions Read through the code and understand what each and every line is doing. Remove the three comments in front of the lines as marked with 1, 2 and 3, and play with the above program, write down the output, and then answer the following questions. 1. Find out the general format, and discuss the various usage, of those strange functions, particularly, execve(), exit(), fork(), vfork(), getpid(), and wait(). Notice that I ignored the input parameters for all these functions. 2. When you take out the comment made in front ofprintf("i am done\n");as marked with 1, i.e., at the end of the pid==0 segment, what happens? Why the string I am done does not print? A solution: Nothing happens. Particularly, I am done won t be printed, since execve() will not return when successfully executed, as you should have found in answering the above question When you add back the comment made in front of sleep(1); as marked with 2, what happens? Why is that? A solution: The output pattern might change, as child process might print out its stuff first. 4. When you add back the comment made in front of if(wait(null)>0) as marked with 3, what happens? Why is that? 4

5 A solution: The parent process will hang, since it has to wait for the child process to complete first, as you should have found out in answering 1. 3 Now, it is your turn... Write a program to calculate the sum of the areas of a triangle (base = 3, height = 4), a circle (radiu = 5), and a square (side = 4) 1, such that each of them shall be passed to a child process to calculate, then the result will be sent back for the parent process to sum up and report. A solution (Matt w/ B): This one uses vfork(), which is the same as fork(), except that, with vfork(), child processes (threads) inherit data space as the parent process. So this is actually similar to the thread concept that we have been talking about in the lectures. (Notice that, as it claims, Unix only creates one thread for each process. Thus, officially, it is not a thread but a process.) Check out what happens if we use fork() here. /home/zshen > more shapesum.c #include <stdlib.h> #include <sys/types.h> #include <math.h> int sum = 0; int i; for(i = 0; i <= 3; i++){ if(i==1 && vfork()==0){ int tribase = 3; int triheight = 4; printf("calculating area of the triangle...\n"); sleep(1); sum += (0.5 * (tribase*triheight)); else if (i==2 && vfork()==0) { int radius = 5; 1 You should have learned, somewhere between preschool and high school, how to calculate those areas, given, for a triangle, its base b, and height, h; for a circle, its radius, r; and for a square, its side length, s. Just in case you have forgotten, it is b h for a triangle; π r 2 for a circle; and you-know-what for a square, 2 respectively. 5

6 printf("calculating area of the circle...\n"); sleep(1); sum += (M_PI * (radius*radius)); else if (i==3 && vfork()==0) { int sqside = 4; printf("calculating area of the sqaure...\n"); sleep(1); sum += (sqside*sqside); printf("the sum of these shapes is: %d\n", sum); /home/zshen > cc shapesum.c /home/zshen >./a.out Calculating area of the triangle... Calculating area of the circle... Calculating area of the sqaure... The sum of these shapes is: 100 /home/zshen > 6