Princeton University COS 217: Introduction to Programming Systems Spring 2017 Final Exam Answers

Transcription

1 Princeton University COS 217: Introduction to Programming Systems Spring 2017 Final Exam Answers The exam was a 3-hour, closed-book, closed notes, no electronics allowed, exam. Question 1: Who Dunnit? 1. P, L 2. L 3. P 4. A 5. C 6. HW 7. OS (Alternate: OS, HW) 8. C 9. OS (Alternate: OS, HW) 10. HW 11. L 12. OS 13. P 14. L 15. HW Question 2: ~140 characters 2a) i j 2b) 9 2c) -3 2d) Fault 2e) Calling a system function is more expensive because it is done through a trap exception which requires the OS to intervene. In addition to saving the return address (similar to a function call), this also saves registers and process state. 2f) printf (or putc, ), write Alternate answers: scanf (or getchar, ), read 2g) Buffer overflow Question 3: Going Virtual 3a) 33 bits in physical address (8 GB = 2 33 bytes of physical memory) Page 1 of 5

2 3b) 2 19 physical pages (page size = 2 14 bytes, therefore # pages = 2 33 / 2 14 = 2 19 ) 3c) Page size is 16 KB = 2 14 bytes. Therefore, offset is the last 14 bits, and virtual page number is the first 50 bits in the 64-bit virtual address. ou have to re-arrange the binary digits into groups of 4. Thus, the first F in the virtual address splits, with the higher 2 bits (11) going with the virtual address, and the lower two bits (11) going with the offset. ou have to re-arrange the virtual page number bits also into groups of 4. Virtual page number: 0x 47 (leading 0 s not needed) Offset in page: 0x 3F11 (leading 0 s not needed) 3d) Case 1: page table shows virtual page is mapped to a physical page in memory This is a page hit, and the HW performs memory access. 3e) Case 2: page table shows virtual page is mapped to disk This is a page miss, i.e., HW triggers a page fault. The OS performs a page swap and control returns to application (which tries the same access again, and gets a page hit next time). Case 3: page table shows that virtual page is unmapped This leads to a segmentation fault, and the program will typically terminate. 3f) At the time of fork(), the OS creates a page table for the child process which is a copy of the page table of the parent process, i.e., the virtual pages of child are mapped to the physical pages (in memory or disk) of parent. The pages are not actually duplicated, making this efficient. A page is copied only when a write occurs (copy-on-write). Question 4: Powering through Assembly 4a) Function g has 1 parameter. 4b) Flattened C for function g: unsigned int g(unsigned int n) unsigned int result = 1; unsigned int i = 1; unsigned int a = 2;.L1: if (i <= n) goto.l2; goto.l3; Page 2 of 5

3 .L2: result *= a; i++; goto.l1;.l3 Idiomatic C for function g: unsigned int g(unsigned int n) unsigned int result = 1; unsigned int i = 1; for (i=1; i<= n; i++) result *= 2; 4c) The result computed by g is 2 n, where n is its input parameter. 4d) r10 and r11 are caller-saved registers. Since function f calls function g, it needs to save them before calling g, and restore them afterward. As can be seen, g actually uses r10 and r11. However, since g does not call any function, it does not need to save (or restore) them. 4e) Flattened C for function f: unsigned int sum = 0; unsigned int i = lo;.l4: if (i <= hi) goto.l5; goto.l6;.l5: sum += g(i); i++; goto.l4;.l6: return sum; Idiomatic C for function f: unsigned int sum = 0; unsigned int i = lo; for (i=lo; i <= hi; i++) sum += g(i); return sum; 4f) 12 Page 3 of 5

4 4g) One can write f as a bit-twiddling function, by using the fact that 1 << n = 2 n, such that output of f has 1 s in all positions from lo to hi (both inclusive). If lo > hi, then result should be 0. unsigned result; result = (((1 << (hi+1)) - 1) & ~((1 << lo) - 1); Some answers used result = (1 << (hi+1)) (1 << lo) in f, but this does not work when hi < lo-1, where the result should be 0. Question 5: Process this, Child! 5a) argv[i] ULL 5b) The output for each file is written in the same order as the order of file arguments in the command. In the example,./.aout f1 f2 f3 will write output of wc in the order f1, f2, f3. 5c) Without the continue statement, the parent process will also be over-written by the execvp. Thus, the first file will be processed twice by wc, by the child and by the parent, whereas other files (if specified) will not be processed at all. Question 6: Map the Stars 6a) static void updatestars(const char *key, void *value, void *extra) int count; struct starstats *ps; assert(key!= ULL); assert(value!= ULL); assert(extra!= ULL); count = *(int *) value; ps = (struct starstats *) extra; if (count >= CUTOFF) ps->numw++; ps->sumc += count; if (count > ps->maxc) ps->maxc = count; ps->max_word = key; 6b) There can be a division-by-zero error in the first printf statement in the client code. This can happen if there are no star words. 6c) o changes will be needed in the callback function. Page 4 of 5

5 Question 7: Exit through the Course 7a) 3 7b) COS 21 ;0 272 ;1 ;0 7c) COS 21 ;0 72 ;1 ;0 7d) Add exit(0) after line 9 and after line 13. Page 5 of 5