操作系统试题

Operating Systems Fall 2009 Midterm ExaminationName Student No. ScoreProblem 1: Booting (4Points)Turn on Bochs, and running JOS, JOS will first enter the 16-bit real mode. After execute several instructions, JOS will turn into 32-bit protected mode and fall into i386_init() function .1. How does processor translate virtual address to physical address in16-bits real mode, please give an example? (2’)Real Mode: physical addr = 16 * segment + offset Example:0xf000:0xe05b16*0xf000+0x e05b = 0xf e05b2. How does processor translate virtual address to physical address in32-bits protected mode, please give an example? (2’)Protected Mode: each segment is assigned an entry in a descriptor table. The index of the entry of the segment is the selector value that is stored in segment registers.Example: (no paging) CS + base = 0x22 descriptor table[0x22] = 0x1000 Virtual address = 0x89014 Physical address = 0x1000 + 0x89014 = 0x 90014CS + base Physical addr = Segment start addr + offsetThe xv6’s system call stubs (sheet 27) are called by user-space C code as if they were ordinary C functions, so all callers are prepared for the stubs to not preserve caller-saved registers. The stubs invoke the INT instruction which vectors into kernel code that shortly jumps to alltraps (line 2755). alltraps and trapret save and restore all registers to and from a struct trapframe.1.As we know, the return value is saved in %eax, how does the systemcall return a value to user process through %eax? (2’)cp−>tf−>eax = syscalls[num](); (sheet 27)2.If the user process wants the kernel return a string of “Good Luck”, whatshould user process provide and how does the kernel return certain string to user? (3’)User process provide a buf[10] and transfer it to the kernel as a pointer. The kernel copies the “Good Luck” string into buf and returns a success. Then the user process can read the string in buf.In the backtrace function, we can print out the back trace of the program. Suppose we have following call trace:foo(int b) -> bar(int a) -> backtrace() The functions’ code address range list below.A partial code and stack are as follows:Fill in the blanks ([1-3], 2’ * 3) and answer the question. [1][2][3] . 1. What is the input argument of bar function? (2’) 13/0xd2. What is %esp value after the backtrace function returns? (2’) 0xffffc9b80xffffc9c0Current %ebpProblem 4: Thread Libraries (19Points)Thread is a useful concept to allow memory sharing among tasks. However, designing and implementing a thread library is fairly difficult. Read the following text and answer the questions:1.What are the differences between a kernel and a user thread? (3’)Kernel thread is a real execution context (it has cr3, kernel context, etc.). However, user thread is completely managed at the user level, from the kernel’s point of view, they exist in one process.The mapping from kernel threads to user threads is an important choice in thread library. For example, GNU portable thread library maps N user threads onto one kernel thread.2.What is the major advantage of choosing this mapping? (3’)1st, the operations are all done in user level, without trapping to kernel. 2nd, the context switching is quite lightweight. Finally, user level thread scheduler allows flexible and highly customizable scheduling of the threads.3.In what situation this mapping may become a bad choice, and why?Please give at least one concrete example (i.e. scenario) to justify that this is a bad choice. (3’)One thread calling into the kernel will block all other threads, even if some threads are ready to run.4.Native POSIX Thread Library (NPTL) on Linux adopts a 1-1 mapping,what are the pros and cons of this approach? (3’)Pros:1)Thread can directly calls system services (e.g. sleep), withoutworrying about interfering with other threads.2)Better system integration. (Compare to user threads that built uponkernel threads)Cons:1)Extra trap to kernel on every thread operation.2)The scheduling of kernel threads may interfere with user threads.Kernel may schedule away a kernel thread associated with certainuser threads, while do not telling them. However, the user thread may be ready to run!3)It is inflexible and inefficient to customize the scheduling of userthreads, since we must frequently invoke system calls to do that.It has been argued that “mapping some kernel threads to user ones” is indeed not the correct way for implementing a really flexible and fast thread library, so:5.What is the major problem of associating a user thread into a kernelthread? (Hint: user threads and kernel threads may each have their own schedulers) (4’)The scheduling of kernel thread is oblivious to user threads.6.Based on your answer (observation) above, try to devise a newapproach to combine the advantages of different mapping methods.You should describe your approach by showing its difference with mapping user threads to kernel ones. (3’)The kernel threads should not schedule themselves, rather, they should direct important events to user scheduler (for example, the kernel thread is blocking now). The user scheduler completely decides the scheduling of threads based on these events (for example, assigning a new kernel thread to a ready user thread).Problem 5: Synchronization (20Points)There are three kinds of lock implementation below. Read the code and answer the questions please. (4’)1.When should we use implementation-1, and when should we useimplementation-2? (2’)UniprocessorMultiprocessor & uniprocessor2.What is guard variable used in implementation-3, and what are thebenefits of implementation-3? (2’)Guarantee exclusive execute the lock code.Minimize the busy wait timeConditional variable is a good way to do the synchronization. PTHREAD library has its own implementation of conditional variable and it gives the interface belowint pthread_cond_signal(pthread_cond_t *cond);Description: Restarts one of the threads that are waiting on the condition variable cond.int pthread_cond_wait(pthread_cond_t *cond, pthread_mutex_t *mutex); Description: Atomically unlocks the mutex (as per pthread_unlock_mutex) and waits for the condition variable cond to be signaled.Assume there are two threads and one shared variable counter. One thread is used to increase the counter and the other is used to decrease it. They use the conditional variable to do the synchronization. The code is like below. Please read the code and answer the questions:(3’)The code is like:“if(count==0)pthread_cond_signal(&count_nonzero);”It should be in the lock protection area. Otherwise it will be hang because of the lost of the signal.2.Can you fix up the code to have a right result? (3’)increment_count_thr:pthread_mutex_lock(&count_lock);if (count == 0) {pthread_cond_signal (&count_nonzero);}count = count+1;pthread_mutex_unlock (&count_lock);In parallel computing, a readers-writer lock is a synchronization primitive that solves one of the readers-writers problems. A readers-writer lock is like a mutex, in that it controls access to some shared memory area, but it allows multiple threads to read from the shared area concurrently.If a thread wants to read the shared data, he must get the lock in the read mode and more than one reader threads can get this lock in the reader mode at the same time. But only one writer thread can get this lock at a time if he wants to modify the shared data.Example:The rwlock_t data structure is like this:Working flow:Your job here is to implement read_lock() function, write_lock() function and un_lock() function to meet the requirements below:1)read_lock function is used to get the rwlock into the read-mode. Thebehavior of this function is like this:Check if the lock is in the write mode or any writer thread is queued.If YES then queue itself and wait. If there is no writer queued and the lock is not in the write mode, then it increases the read numbercounter and the lock is in the read-mode now.2)write_lock function is used to get the rwlock into the write-mode.The behavior of this function is like this:Check if the lock is in the write mode or in the read mode. If YES then queue itself to wait the reader or writer to finish. If the lock is free,then get the lock into the write mode.3)un_lock function is used to unlock the rwlock. The behavior of thisfunction is like this:If the lock is in the read mode, decrease the reader number counter else if in the write mode, reset the writer number counter to zero. He must wake up the queued writer first, if no writer queued then wake up the queued reader.Problem Description: Suppose we have provided following functionsQuestions: (10’)Please implement the read_lock(), write_lock() and un_lock() functions with the given method. Don’t forget to meet the requirementsvoid read_lock(rwlock_t *rwlock){spin_lock(&rwlock->lock); // (1’)while(1){if((rwlock->nr_writers==0) // (1’)&&(is_empty(&rwlock->writers_queue)) ) {rwlock->nr_readers++;Break;}enqueue(&rwlock->reads_queue);spin_unlock(&rwlock->lock); // ( 1’)suspend();spin_lock(&rwlock->lock);}spin_unlock(&rwlock->lock);}void write_lock(rwlock_t *rwlock){spin_lock(&rwlock->lock); // (1’)while(1) {if((rwlock->nr_writers==0) // (1’)&&(rwlock->nr_readers==0)) ){rwlock->nr_writers = 1;break;}enqueue(&rwlock->writers_queue);spin_unlock(&rwlock->lock); // (1’)suspend();spin_lock(&rwlock->lock);}spin_unlock(&rwlock->lock);}void un_lock(rwlock_t *rwlock){queue *tmp_q;spin_lock(&rwlock->lock); // (1’)if(Rwlock->nr_writers) // (1’)rwlock->nr_writers = 0;elserwlock->nr_readers --;if(!is_empty(&rwlock->writers_queue)){ //(1’) (the wake up order) tmp_q = rwlock->writers_queue;rwlock->writers_queue = NULL;spin_unlock(&rwlock->lock);} else if(!is_empty(&rwlock->readers_queue)) { // (1’) tmp_q = rwlock->readers_queue;rwlock->readers_queue = NULL;spin_unlock(&rwlock->lock);} elsespin_unlock(&rwlock->lock);while(!is_empty(tmp_q))wakeup(dequeue(tmp_q));}Problem 6: Memory (12Points)1.Physical memory management is one important part of operatingsystem. In Lab 2 you have implemented a simple physical memory management policy, please describe the main work of allocating a new page and freeing a used page. (6’)page_alloc():1)Remove one struct Page from page_free_list2)Initialize the page, including setting the links to null and reference countto 0.page_free():1)If reference count is nonzero, do not free (optional)2)Add the page to the page_free_list.2.Suppose the architecture provides a feature that each TLB entry is tagged with an address space number, which identifies a unique address space. Consider a multi-process environment, what benefits will this mechanism bring in? (3’) Could you provide other approach to enjoy the same benefits but with different mechanism like flags in page table entries? (3’)(Hint: what will OS do when switching from one process to another?) Benefits: When switching address space, flush is required for untagged TLBs, if each TLB entry is tagged with an address space number, no flush is required since virtual addresses in different address spaces will have different entries.One example of optimizations related TLB flush: each x86 page directory and page table entry has a global flag, which sets the corresponding page as a global page. Global pages are not flushed from the TLB on a task switch or a write to CR.Problem 7: Durability and Consistency (22points)1.Explain terms of durability and atomicity, how to enforce durability andatomicity of file system operation? (4’)Durability means if an operation is performed and committed, it will survive permanently on disk. Atomicity means the effect of operation is either visible or none. To enforce durability, a partially solution is using synchronized I/O. To enforce atomicity, old data is copied and updates are performed on copied data, the last update using atomic instruction to redirect old pointer to the new one.2.List all detailed operations (inode, bitmap, name resolve, etc) of creatinga file named foo under /a/b/ directory in xv6 file system, that is, aftercreating file, it becomes /a/b/foo. (6’)First call namei() to resolve path “/a/b/”, in function namei(), get first inode of “/”, list all entry under “/” and get inode of “/a/”, then list all entry under “/a/” and get inode of “/a/b/” and return this inode. Second, allocate an inode for new file foo and mark this inode in-use in bitmap. Third, update the inode returned from namei()’s dentry, make a new entry point to this ne w inode. Now, a new empty file is created, writing to this file will allocate addition data blocks and update inode of this file.Orders of operations are crucial to durability and consistency in file systems. When ordering the operations, there are three simple rules:1)Never point to a structure before it has been initialized.2)Never reuse a resource before nullifying all previous pointers to it.3)Never reset the last pointer to a live resource before a new pointerhas been set.While enforcing the orders, dependency cycle is easily generated, consider the simple example: (Creating file A and removing file B gives a dependency cycle.)Now, answer the following questions:3.How xv6 file system solves the problem above? Is there any bettersolution to solve the problem above? (8’)In fact, xv6 cannot have dependency cycle, since it uses synchronized IO, when the first create file A finished, in disk block is already updated. Better solution:4.Suppose we have a program running on xv6Here is disk writes caused by this program:-bwrite file inode-bwrite dentry-bwrite in-use bitmap-bwrite data block-bwrite file inodeTwo of these five operations can be swapped, point the operations and explain your answer. (4’)Recall all rulesa)Initialize a new inode before creating dirent.b)Delete dirent before marking inode free.c)Mark block in-use before adding it to inodeaddrs[].d)Remove block from inodeaddrs[] before marking free.e)Zero blocks before marking free.Rule a) says that 1 and 2 cannot be swapped; rule c) says that 3 and 5 cannot be swapped. Both 2&3, 3&4 can be swapped, because them satisfy ordering enforcement.Problem 8: Course Survey(8points)Thank you for choosing the OS course. Your options are of great importance for us to improve the course. Please answer the following questions about the course: (You will get full credit if you answered the questions with more than one sentence)1.Did you ever compiled and ran xv6? Did you read through its code? (2')2.Which aspect do you love most in this course? (2')3.Which aspect do you dislike most in this course? (2')4.Please provide at least one suggestion on the course. (2')。