15-213: Final Exam Review

Transcription

1 5-23: Final Exam Review Nikhil, Kashish, Krishanu, Stan

2 Threads and Synchronization Carnegie Mellon

3 Threads and Synchronization Carnegie Mellon Problem Statement: 5-23 Ts now want to begin a new procedure for daily office hours: When a student enters the room, he/she will see a table inside the room with several pens and several notebooks on it (if the pens and notebooks haven't been claimed by other students). Ts are inside the room waiting to help them.

4 Threads and Synchronization Carnegie Mellon Procedure for students: Wait until you get called, claim a pen and a notebook. Use the pen to write your questions on the notebook. fter you are done writing the question, release the pen. Hold the notebook and wait until a T is free to help you. fter you are done with talking to a T about the question, go back to claim a pen and write the solutions on the notebook. Release the pen and the notebook and go home.

5 Threads and Synchronization Typical Code to launch threads and wait for them to finish: sem_t pen; sem_t notebook; sem_t ta; int main(int argc, char** argv) { int P = 5, N = 5, T = 3, S = 5; sem_init(&pen,, P); sem_init(&notebook,, N); sem_init(&ta,, T); int* students = malloc(s * sizeof(int)); launch_threads(); reap_threads(); } void launch_threads() { for(int i = ; i < S; i++) { students[s] = pthread_create(students+i, NULL, student_fn, NULL); } } void reap_threads() { for(int i = ; i < S; i++) { pthread_join(students[i]); } }

6 Threads and Synchronization student writes the following program to simulate the given procedures void* student_fn(void* varp) { P(&pen); P(&notebook); sleep(genrand() % 6); // Time for you to write questions V(&pen); P(&ta); sleep(genrand() % 9); // Time for you to ask T V(&ta); P(&pen); sleep(genrand() % 6); // Time for you to write solutions Let us consider a simple example: Pens: 5 Notebooks: 5 Ts: Students: 5 V(&notebook); V(&pen); } return ;

7 Threads and Synchronization student writes the following worker thread to simulate the given procedure void* student_fn(void* varp) { P(&pen); P(&notebook); sleep(genrand() % 6); // Time for you to write questions V(&pen); P(&ta); sleep(genrand() % 9); // Time for you to ask T V(&ta); P(&pen); sleep(genrand() % 6); // Time for you to write solutions V(&notebook); V(&pen); However, the code sometimes gets stuck by a deadlock. Which of the following is correct? : The deadlock is caused by the pen and the notebook. B: The deadlock is caused by the notebook and the T. C: The deadlock is caused by the T and the pen. } return ;

8 Threads and Synchronization Let s analyse the semaphores S: Signal W: Wait W:Get a pen void* student_fn(void* varp) { S: Give up pen W:Get a T S:Give up T W:Get a pen P(&pen); P(&notebook); sleep(genrand() % 6); // Time for you to write questions V(&pen); P(&ta); sleep(genrand() % 9); // Time for you to ask T V(&ta); P(&pen); W:Get a Notebook sleep(genrand() % 6); // Time for you to write solutions Case C: Ts and Pens There is no overlap There can t be a case where a student who is talking to a T is waiting on a pen. There can t be a case where a student who is holding a pen is waiting on a T S: Give up pen V(&notebook); V(&pen); S:Give up Notebook } return ;

9 Threads and Synchronization Let s analyse the semaphores S: Signal W: Wait W:Get a pen void* student_fn(void* varp) { P(&pen); P(&notebook); W:Get a Notebook sleep(genrand() % 6); // Time for you to write questions Case B: Ts and Notebooks There is an overlap student who is holding a notebook may be waiting on a T S: Give up pen W:Get a T S:Give up T W:Get a pen V(&pen); P(&ta); sleep(genrand() % 9); // Time for you to ask T V(&ta); P(&pen); sleep(genrand() % 6); // Time for you to write solutions But. student who is talking to a T will not be waiting on a notebook S: Give up pen V(&notebook); V(&pen); S:Give up Notebook } return ;

10 Threads and Synchronization Let s analyse the semaphores S: Signal W: Wait W:Get a pen void* student_fn(void* varp) { P(&pen); P(&notebook); W:Get a Notebook sleep(genrand() % 6); // Time for you to write questions Case : Pens and Notebooks There is an overlap student who is holding a notebook may be waiting on a pen S: Give up pen W:Get a T S:Give up T W:Get a pen V(&pen); P(&ta); sleep(genrand() % 9); // Time for you to ask T V(&ta); P(&pen); sleep(genrand() % 6); // Time for you to write solutions student who is holding a pen may be waiting on a notebook S: Give up pen V(&notebook); V(&pen); S:Give up Notebook } return ;

11 Threads and Synchronization Let s analyse the semaphores S: Signal W: Wait W:Get a pen void* student_fn(void* varp) { S: Give up pen W:Get a T S:Give up T W:Get a pen S: Give up pen } P(&pen); P(&notebook); sleep(genrand() % 6); // Time for you to write questions V(&pen); P(&ta); sleep(genrand() % 9); // Time for you to ask T V(&ta); P(&pen); sleep(genrand() % 6); // Time for you to write solutions V(&notebook); V(&pen); return ; W:Get a Notebook S:Give up Notebook Carnegie Mellon Case : Pens and Notebooks Consider this likely scenario: 5 Students (group ) enter the room They grab all 5 pens and all 5 notebooks on the table. Every student behind them is waiting on a pen and a notebook. (Group B) ll 5 students in Group give up their pens, but not their notebooks. 5 students in Group B immediately grab the 5 pens. Group B are now waiting on notebooks. fter talking to the Ts, all students in Group are now waiting on pens, which students in Group B have. Group B is waiting on the notebooks which students in Group have. DEDLOCK!!

12 Threads and Synchronization student writes the following program to simulate the given procedures void* student_fn(void* varp) { P(&pen); P(&notebook); sleep(genrand() % 6); // Time for you to write questions V(&pen); P(&ta); sleep(genrand() % 9); // Time for you to ask T V(&ta); P(&pen); sleep(genrand() % 6); // Time for you to write solutions V(&notebook); V(&pen); However, the code sometimes gets stuck by a deadlock. Which of the following is correct? : The deadlock is caused by the pen and the notebook. B: The deadlock is caused by the notebook and the ta. C: The deadlock is caused by the ta and the pen. } return ; NSWER:

13 Threads and Synchronization Carnegie Mellon Can we fix this without changing the Office Hour procedure?

14 Threads and Synchronization Carnegie Mellon Can we fix this without changing the Office Hour procedure? YES!

15 Threads and Synchronization Let s analyse the semaphores W:Get a pen S: Give up pen W:Get a T S:Give up T W:Get a pen S: Give up pen void* student_fn(void* varp) { } P(&notebook); P(&pen); sleep(genrand() % 6); // Time for you to write questions V(&pen); P(&ta); sleep(genrand() % 9); // Time for you to ask T V(&ta); P(&pen); sleep(genrand() % 6); // Time for you to write solutions V(&pen); V(&notebook); return ; //Go home S: Signal W: Wait W:Get a Notebook S:Give up Notebook Let s just re-order the locks and analyse it again. student who is holding a notebook may wait on a pen or a T However... student who is holding a pen will already have a notebook (and never wait on it). student who is talking to a T will already have a notebook (and never wait on it).

16 Virtual Memory Carnegie Mellon

17 Virtual Memory Virtual ddress - 8 Bits Physical ddress - 2 Bits Page Size - 52 Bytes TLB is 8-way set associative Cache is 2-way set associative Final S-2 (#5) Lecture 8: VM - Systems

18 Virtual Memory Carnegie Mellon Label the following: () VPO: Virtual Page Offset (B) VPN: Virtual Page Number (C) TLBI: TLB Index (D) TLBT: TLB Tag

19 Virtual Memory Carnegie Mellon Label the following: () VPO: Virtual Page Offset - Location in the page Page Size = 52 Bytes = 2 9 Need 9 bits

20 Virtual Memory Carnegie Mellon Label the following: () VPO: Virtual Page Offset (B) VPN: Virtual Page Number - Everything Else

21 Virtual Memory Carnegie Mellon Label the following: () VPO: Virtual Page Offset (B) VPN: Virtual Page Number (C) TLBI: TLB Index - Location in the TLB Cache

22 Virtual Memory Carnegie Mellon Label the following: () VPO: Virtual Page Offset (B) VPN: Virtual Page Number (C) TLBI: TLB Index - Location in the TLB Cache 2 Indices Bit TLBI

23 Virtual Memory Carnegie Mellon Label the following: () VPO: Virtual Page Offset (B) VPN: Virtual Page Number (C) TLBI: TLB Index (D) TLBT: TLB Tag - Everything Else TLBT TLBI

24 Virtual Memory Carnegie Mellon Label the following: () PPO: Physical Page Offset (B) PPN: Physical Page Number (C) CO: Cache Offset (D) CI: Cache Index (E) CT: Cache Tag

25 Virtual Memory Carnegie Mellon Label the following: () PPO: Physical Page Offset

26 Virtual Memory Carnegie Mellon Label the following: () PPO: Physical Page Offset - Same as VPO

27 Virtual Memory Carnegie Mellon Label the following: () PPO: Physical Page Offset - Same as VPO (B) PPN: Physical Page Number - Everything Else B B B

28 Virtual Memory Carnegie Mellon Label the following: () PPO: Physical Page Offset - Same as VPO (B) PPN: Physical Page Number - Everything Else (C) CO: Cache Offset - Offset in Block B B B

29 Virtual Memory Carnegie Mellon Label the following: () PPO: Physical Page Offset - Same as VPO (B) PPN: Physical Page Number - Everything Else (C) CO: Cache Offset - Offset in Block 4 Byte Blocks 2 Bits B B B CO

30 Virtual Memory Carnegie Mellon Label the following: () PPO: Physical Page Offset - Same as VPO (B) PPN: Physical Page Number - Everything Else (C) CO: Cache Offset - Offset in Block (D) CI: Cache Index B B B CO

31 Virtual Memory Carnegie Mellon Label the following: () PPO: Physical Page Offset - Same as VPO (B) PPN: Physical Page Number - Everything Else (C) CO: Cache Offset - Offset in Block (D) CI: Cache Index 4 Indices 2 Bits B B B CI CO

32 Virtual Memory Carnegie Mellon Label the following: () PPO: Physical Page Offset - Same as VPO (B) PPN: Physical Page Number - Everything Else (C) CO: Cache Offset - Offset in Block (D) CI: Cache Index (E) CT: Cache Tag - Everything Else B B B Cache Tag CI CO

33 Virtual Memory Carnegie Mellon Now to the actual question! Q) Translate the following address: x9f4

34 Virtual Memory Now to the actual question! Q) Translate the following address: x9f4. Write down bit representation = = 9 = F = 4 =

35 Virtual Memory Now to the actual question! Q) Translate the following address: x9f4. Write down bit representation 2. Extract Information: VPN: x?? TLBI: x?? TLBT: x?? TLB Hit: Y/N? Page Fault: Y/N? PPN: x??

36 Virtual Memory Now to the actual question! Q) Translate the following address: x9f4. Write down bit representation 2. Extract Information: VPN: xd4 TLBI: x?? TLBT: x?? TLB Hit: Y/N? Page Fault: Y/N? PPN: x??

37 Virtual Memory Now to the actual question! Q) Translate the following address: x9f4. Write down bit representation 2. Extract Information: VPN: xd4 TLBI: x TLBT: x?? TLB Hit: Y/N? Page Fault: Y/N? PPN: x??

38 Virtual Memory Now to the actual question! Q) Translate the following address: x9f4. Write down bit representation 2. Extract Information: VPN: xd4 TLBI: x TLBT: x6 TLB Hit: Y/N? Page Fault: Y/N? PPN: x??

39 Virtual Memory Now to the actual question! Q) Translate the following address: x9f4. Write down bit representation 2. Extract Information: VPN: xd4 TLBI: x TLBT: x6 TLB Hit: Y! Page Fault: Y/N? PPN: x??

40 Virtual Memory Now to the actual question! Q) Translate the following address: x9f4. Write down bit representation 2. Extract Information: VPN: xd4 TLBI: x TLBT: x6 TLB Hit: Y! Page Fault: N! PPN: x??

41 Virtual Memory Now to the actual question! Q) Translate the following address: x9f4. Write down bit representation 2. Extract Information: VPN: xd4 TLBI: x TLBT: x6 TLB Hit: Y! Page Fault: N! PPN: x3

42 Virtual Memory Now to the actual question! Q) Translate the following address: x9f4. Write down bit representation 2. Extract Information 3. Put it all together: PPN: x3, PPO = x??

43 Virtual Memory Now to the actual question! Q) Translate the following address: x9f4. Write down bit representation 2. Extract Information 3. Put it all together: PPN: x3, PPO = VPO = xf4

44 Virtual Memory Carnegie Mellon Q) What is the value of the address? CO: x?? CI: x?? CT: x?? Cache Hit: Y/N? Value:x??

45 Virtual Memory Q) What is the value of the address?. Extract more information CO: x CI: x?? CT: x?? Cache Hit: Y/N? Value:x??

46 Virtual Memory Q) What is the value of the address?. Extract more information CO: x CI: x CT: x?? Cache Hit: Y/N? Value:x??

47 Virtual Memory Q) What is the value of the address?. Extract more information 2. Go to Cache Table CO: x CI: x CT: x7f Cache Hit: Y/N? Value:x??

48 Virtual Memory Q) What is the value of the address?. Extract more information 2. Go to Cache Table CO: x CI: x CT: x7f Cache Hit: Y Value:x??

49 Virtual Memory Q) What is the value of the address?. Extract more information 2. Go to Cache Table CO: x CI: x CT: x7f Cache Hit: Y Value:xFF

50 Virtual Memory Carnegie Mellon

51 IO/Processes Carnegie Mellon

52 IO/Processes foo.txt: abcdefgh...xyz int main() { int fd, fd2, fd3; char c; pid_t pid; fd = open( foo.txt, O_RDONLY); fd2 = open( foo.txt, O_RDONLY); fd3 = open( foo.txt, O_RDONLY); read(fd, &c, sizeof(c)); // c =? read(fd2, &c, sizeof(c)); // c =? dup2(fd2, fd3); read(fd3, &c, sizeof(c)); // c =? read(fd2, &c, sizeof(c)); // c =? Main ideas: How does read offset? How does dup2 work? What is the order of arguments? Does fd3 share offset with fd2?

53 IO/Processes foo.txt: abcdefgh...xyz int main() { int fd, fd2, fd3; char c; pid_t pid; fd = open( foo.txt, O_RDONLY); fd2 = open( foo.txt, O_RDONLY); fd3 = open( foo.txt, O_RDONLY); read(fd, &c, sizeof(c)); read(fd2, &c, sizeof(c)); dup2(fd2, fd3); read(fd3, &c, sizeof(c)); read(fd2, &c, sizeof(c)); // c = a // c = a // c = b // c = c How does read offset? Incremented by number of bytes read How does dup2 work? ny read/write from fd3 now happen from fd2 ll file offsets are shared

54 IO/Processes read(fd, &c, sizeof(c)); // a read(fd2, &c, sizeof(c)); // a dup2(fd2, fd3); read(fd3, &c, sizeof(c)); // b read(fd2, &c, sizeof(c)); // c pid = fork(); if (pid==) { read(fd, &c, sizeof(c)); printf( c = %c\n, c); dup2(fd, fd2); read(fd3, &c, sizeof(c)); printf( c = %c\n, c); } read(fd2, &c, sizeof(c)); printf( c = %c\n, c); read(fd, &c, sizeof(c)); printf( c = %c\n, c); Main ideas: How are fd shared between processes? How does dup2 work from parent to child? How are file offsets shared between processes?

55 IO/Processes Carnegie Mellon read(fd, &c, sizeof(c)); // a read(fd2, &c, sizeof(c)); // a dup2(fd2, fd3); read(fd3, &c, sizeof(c)); // b read(fd2, &c, sizeof(c)); // c pid = fork(); if (pid==) { read(fd, &c, sizeof(c)); printf( c = %c\n, c); dup2(fd, fd2); read(fd3, &c, sizeof(c)); printf( c = %c\n, c); } read(fd2, &c, sizeof(c)); printf( c = %c\n, c); read(fd, &c, sizeof(c)); printf( c = %c\n, c); What would this program print?

56 IO/Processes read(fd, &c, sizeof(c)); // a read(fd2, &c, sizeof(c)); // a dup2(fd2, fd3); read(fd3, &c, sizeof(c)); // b read(fd2, &c, sizeof(c)); // c pid = fork(); if (pid==) { read(fd, &c, sizeof(c)); printf( c = %c\n, c); dup2(fd, fd2); read(fd3, &c, sizeof(c)); printf( c = %c\n, c); } read(fd2, &c, sizeof(c)); printf( c = %c\n, c); read(fd, &c, sizeof(c)); printf( c = %c\n, c); Possible output : c = d // in parent from fd2 c = b // in parent from fd c = c // in child from fd c = e // in child from fd3 c = d // in child from fd2 c = e // in child from fd

57 IO/Processes read(fd, &c, sizeof(c)); // a read(fd2, &c, sizeof(c)); // a dup2(fd2, fd3); read(fd3, &c, sizeof(c)); // b read(fd2, &c, sizeof(c)); // c pid = fork(); if (pid==) { read(fd, &c, sizeof(c)); printf( c = %c\n, c); dup2(fd, fd2); read(fd3, &c, sizeof(c)); printf( c = %c\n, c); } read(fd2, &c, sizeof(c)); printf( c = %c\n, c); read(fd, &c, sizeof(c)); printf( c = %c\n, c); Possible output 2: c = b // in child from fd c = d // in child from fd3 c = c // in child from child s fd2 c = d // in child from fd c = e // in parent from parent s fd2 c = e // in parent from fd

58 IO/Processes Carnegie Mellon read(fd, &c, sizeof(c)); // a read(fd2, &c, sizeof(c)); // a dup2(fd2, fd3); read(fd3, &c, sizeof(c)); // b read(fd2, &c, sizeof(c)); // c pid = fork(); if (pid==) { read(fd, &c, sizeof(c)); printf( c = %c\n, c); dup2(fd, fd2); read(fd3, &c, sizeof(c)); printf( c = %c\n, c); } read(fd2, &c, sizeof(c)); printf( c = %c\n, c); read(fd, &c, sizeof(c)); printf( c = %c\n, c); Other interleavings of child/parent are possible

59 IO/Processes Carnegie Mellon }.. pid = fork(); if (pid==) { read(fd, &c, sizeof(c)); printf( c = %c\n, c); dup2(fd, fd2); read(fd3, &c, sizeof(c)); printf( c = %c\n, c); } if (pid!=) waitpid(-, NULL, ); read(fd2, &c, sizeof(c)); printf( c = %c\n, c); read(fd, &c, sizeof(c)); printf( c = %c\n, c); return ; What are the possible outputs now?

60 IO/Processes }.. pid = fork(); if (pid==) { read(fd, &c, sizeof(c)); printf( c = %c\n, c); dup2(fd, fd2); read(fd3, &c, sizeof(c)); printf( c = %c\n, c); } if (pid!=) waitpid(-, NULL, ); read(fd, &c, sizeof(c)); printf( c = %c\n, c); read(fd2, &c, sizeof(c)); printf( c = %c\n, c); return ; Possible output 2: c = b // in child from fd c = d // in child from fd3 c = c // in child from child s fd2 c = d // in child from fd c = e // in parent from parent s fd2 c = e // in parent from fd

61 IO/Processes read(fd, &c, sizeof(c)); // a read(fd2, &c, sizeof(c)); // a dup2(fd2, fd3); read(fd3, &c, sizeof(c)); // b read(fd2, &c, sizeof(c)); // c pid = fork(); if (pid==) { read(fd, &c, sizeof(c)); dup2(fd, fd2); read(fd3, &c, sizeof(c)); } if (pid!=) waitpid(-, NULL, ); read(fd2, &c, sizeof(c)); read(fd, &c, sizeof(c)); Child creates a copy of the parent fd table dup2/open/close in parent affect the child dup2/open/close in child do NOT affect the parent File descriptors across processes share the same file offset Carnegie Mellon

62 Malloc Carnegie Mellon

63 Malloc Carnegie Mellon Fit algorithms - first/next/best/good Fragmentation Internal - inside blocks External - between blocks Organization Implicit Explicit Segregated

64 Malloc - First fit # #2 #3 #4 #5 #6 6 byte align coalesced footerless 32 min size a = malloc(32) b = malloc(6) c = malloc(6) d = malloc(4) free(c) free(a) e = malloc(6) free(d) f = malloc(48) free(b) final-f.pdf: Problem Part

65 Malloc - First fit # #2 #3 #4 #5 #6 6 byte align coalesced footerless 32 min size a = malloc(32) b = malloc(6) c = malloc(6) d = malloc(4) free(c) free(a) e = malloc(6) free(d) f = malloc(48) free(b) 48a final-f.pdf: Problem Part

66 Malloc - First fit # #2 #3 #4 #5 #6 6 byte align coalesced footerless 32 min size a = malloc(32) 48a b = malloc(6) 48a 32a c = malloc(6) d = malloc(4) free(c) free(a) e = malloc(6) free(d) f = malloc(48) free(b) final-f.pdf: Problem Part

67 Malloc - First fit # #2 #3 #4 #5 #6 6 byte align coalesced footerless 32 min size a = malloc(32) 48a b = malloc(6) 48a 32a c = malloc(6) 48a 32a 32a d = malloc(4) free(c) free(a) e = malloc(6) free(d) f = malloc(48) free(b) final-f.pdf: Problem Part

68 Malloc - First fit # #2 #3 #4 #5 #6 6 byte align coalesced footerless 32 min size a = malloc(32) 48a b = malloc(6) 48a 32a c = malloc(6) 48a 32a 32a d = malloc(4) 48a 32a 32a 48a free(c) free(a) e = malloc(6) free(d) f = malloc(48) free(b) final-f.pdf: Problem Part

69 Malloc - First fit # #2 #3 #4 #5 #6 6 byte align coalesced footerless 32 min size a = malloc(32) 48a b = malloc(6) 48a 32a c = malloc(6) 48a 32a 32a d = malloc(4) 48a 32a 32a 48a free(c) 48a 32a 32f [] 48a free(a) e = malloc(6) free(d) f = malloc(48) free(b) final-f.pdf: Problem Part

70 Malloc - First fit # #2 #3 #4 #5 #6 6 byte align coalesced footerless 32 min size a = malloc(32) 48a b = malloc(6) 48a 32a c = malloc(6) 48a 32a 32a d = malloc(4) 48a 32a 32a 48a e = malloc(6) free(c) 48a 32a 32f [] 48a free(a) 48f [] 32a 32f [] 48a free(d) f = malloc(48) free(b) final-f.pdf: Problem Part

71 Malloc - First fit # #2 #3 #4 #5 #6 6 byte align coalesced footerless 32 min size a = malloc(32) 48a b = malloc(6) 48a 32a c = malloc(6) 48a 32a 32a d = malloc(4) 48a 32a 32a 48a free(c) 48a 32a 32f [] 48a free(a) 48f [] 32a 32f [] 48a e = malloc(6) 48a 32a 32f [] 48a free(d) f = malloc(48) free(b) final-f.pdf: Problem Part

72 Malloc - First fit # #2 #3 #4 #5 #6 6 byte align coalesced footerless 32 min size a = malloc(32) 48a b = malloc(6) 48a 32a c = malloc(6) 48a 32a 32a d = malloc(4) 48a 32a 32a 48a free(c) 48a 32a 32f [] 48a free(a) 48f [] 32a 32f [] 48a e = malloc(6) 48a 32a 32f [] 48a f = malloc(48) free(d) 48a 32a 8f [] free(b) final-f.pdf: Problem Part

73 Malloc - First fit # #2 #3 #4 #5 #6 6 byte align coalesced footerless 32 min size a = malloc(32) 48a b = malloc(6) 48a 32a c = malloc(6) 48a 32a 32a d = malloc(4) 48a 32a 32a 48a free(c) 48a 32a 32f [] 48a free(a) 48f [] 32a 32f [] 48a e = malloc(6) 48a 32a 32f [] 48a free(d) 48a 32a 8f [] f = malloc(48) 48a 32a 8a free(b) final-f.pdf: Problem Part

74 Malloc - First fit # #2 #3 #4 #5 #6 6 byte align coalesced footerless 32 min size a = malloc(32) 48a b = malloc(6) 48a 32a c = malloc(6) 48a 32a 32a d = malloc(4) 48a 32a 32a 48a free(c) 48a 32a 32f [] 48a free(a) 48f [] 32a 32f [] 48a e = malloc(6) 48a 32a 32f [] 48a free(d) 48a 32a 8f [] f = malloc(48) 48a 32a 8a free(b) 48a 32f [] 8a final-f.pdf: Problem Part

75 Malloc - First fit # #2 #3 #4 #5 #6 6 byte align coalesced footerless 32 min size fragmentation? internal? a = malloc(32) 48a b = malloc(6) 48a 32a c = malloc(6) 48a 32a 32a d = malloc(4) 48a 32a 32a 48a free(c) 48a 32a 32f [] 48a free(a) 48f [] 32a 32f [] 48a e = malloc(6) 48a 32a 32f [] 48a free(d) 48a 32a 8f [] f = malloc(48) 48a 32a 8a free(b) 48a 32f [] 8a

76 Malloc - First fit # #2 #3 #4 #5 #6 6 byte align coalesced footerless 32 min size fragmentation? internal (48-6) + (8-48) = 64 external? a = malloc(32) 48a b = malloc(6) 48a 32a c = malloc(6) 48a 32a 32a d = malloc(4) 48a 32a 32a 48a free(c) 48a 32a 32f [] 48a free(a) 48f [] 32a 32f [] 48a e = malloc(6) 48a 32a 32f [] 48a free(d) 48a 32a 8f [] f = malloc(48) 48a 32a 8a free(b) 48a 32f [] 8a

77 Malloc - First fit # #2 #3 #4 #5 #6 6 byte align coalesced footerless 32 min size fragmentation? internal (48-6) + (8-48) = 64 external 32 a = malloc(32) 48a b = malloc(6) 48a 32a c = malloc(6) 48a 32a 32a d = malloc(4) 48a 32a 32a 48a free(c) 48a 32a 32f [] 48a free(a) 48f [] 32a 32f [] 48a e = malloc(6) 48a 32a 32f [] 48a free(d) 48a 32a 8f [] f = malloc(48) 48a 32a 8a free(b) 48a 32f [] 8a

78 Malloc - Best fit # #2 #3 #4 #5 #6 6 byte align coalesced footerless 32 min size a = malloc(32) b = malloc(6) c = malloc(6) d = malloc(4) free(c) free(a) e = malloc(6) free(d) f = malloc(48) free(b) final-f.pdf: Problem Part

79 Malloc - Best fit # #2 #3 #4 #5 #6 6 byte align coalesced footerless 32 min size a = malloc(32) 48a b = malloc(6) 48a 32a c = malloc(6) 48a 32a 32a d = malloc(4) 48a 32a 32a 48a e = malloc(6) free(c) 48a 32a 32f [] 48a free(a) 48f [] 32a 32f [] 48a free(d) f = malloc(48) free(b) final-f.pdf: Problem Part

80 Malloc - Best fit # #2 #3 #4 #5 #6 6 byte align coalesced footerless 32 min size a = malloc(32) 48a b = malloc(6) 48a 32a c = malloc(6) 48a 32a 32a d = malloc(4) 48a 32a 32a 48a free(c) 48a 32a 32f [] 48a free(a) 48f [] 32a 32f [] 48a e = malloc(6) 48f [] 32a 32a 48a free(d) f = malloc(48) free(b) final-f.pdf: Problem Part

81 Malloc - Best fit # #2 #3 #4 #5 #6 6 byte align coalesced footerless 32 min size a = malloc(32) 48a b = malloc(6) 48a 32a c = malloc(6) 48a 32a 32a d = malloc(4) 48a 32a 32a 48a free(c) 48a 32a 32f [] 48a free(a) 48f [] 32a 32f [] 48a e = malloc(6) 48f [] 32a 32a 48a f = malloc(48) free(d) 48f [] 32a 32a 48f [] free(b) final-f.pdf: Problem Part

82 Malloc - Best fit # #2 #3 #4 #5 #6 6 byte align coalesced footerless 32 min size a = malloc(32) 48a b = malloc(6) 48a 32a c = malloc(6) 48a 32a 32a d = malloc(4) 48a 32a 32a 48a free(c) 48a 32a 32f [] 48a free(a) 48f [] 32a 32f [] 48a e = malloc(6) 48f [] 32a 32a 48a free(d) 48f [] 32a 32a 48f [] f = malloc(48) 48f [] 32a 32a 64a free(b) final-f.pdf: Problem Part

83 Malloc - Best fit # #2 #3 #4 #5 #6 6 byte align coalesced footerless 32 min size a = malloc(32) 48a b = malloc(6) 48a 32a c = malloc(6) 48a 32a 32a d = malloc(4) 48a 32a 32a 48a free(c) 48a 32a 32f [] 48a free(a) 48f [] 32a 32f [] 48a e = malloc(6) 48f [] 32a 32a 48a free(d) 48f [] 32a 32a 48f [] f = malloc(48) 48f [] 32a 32a 64a free(b) 8f [] 32a 64a final-f.pdf: Problem Part

84 Malloc - Best fit 6 byte align coalesced footerless 32 min size fragmentation? internal? Carnegie Mellon # #2 #3 #4 #5 #6 a = malloc(32) 48a b = malloc(6) 48a 32a c = malloc(6) 48a 32a 32a d = malloc(4) 48a 32a 32a 48a free(c) 48a 32a 32f [] 48a free(a) 48f [] 32a 32f [] 48a e = malloc(6) 48f [] 32a 32a 48a free(d) 48f [] 32a 32a 48f [] f = malloc(48) 48f [] 32a 32a 64a free(b) 8f [] 32a 64a

85 Malloc - Best fit 6 byte align coalesced footerless 32 min size fragmentation? internal (32-6) + (64-48) = 32 external? Carnegie Mellon # #2 #3 #4 #5 #6 a = malloc(32) 48a b = malloc(6) 48a 32a c = malloc(6) 48a 32a 32a d = malloc(4) 48a 32a 32a 48a free(c) 48a 32a 32f [] 48a free(a) 48f [] 32a 32f [] 48a e = malloc(6) 48f [] 32a 32a 48a free(d) 48f [] 32a 32a 48f [] f = malloc(48) 48f [] 32a 32a 64a free(b) 8f [] 32a 64a

86 Malloc - Best fit 6 byte align coalesced footerless 32 min size fragmentation? internal (32-6) + (64-48) = 32 external 8 Carnegie Mellon # #2 #3 #4 #5 #6 a = malloc(32) 48a b = malloc(6) 48a 32a c = malloc(6) 48a 32a 32a d = malloc(4) 48a 32a 32a 48a free(c) 48a 32a 32f [] 48a free(a) 48f [] 32a 32f [] 48a e = malloc(6) 48f [] 32a 32a 48a free(d) 48f [] 32a 32a 48f [] f = malloc(48) 48f [] 32a 32a 64a free(b) 8f [] 32a 64a

87 Signals Carnegie Mellon who would win? several hundred lines of tshlab code one asynchronous boi

88 Signals Carnegie Mellon Examples in these slides are important gotchas NOT an inclusive list In particular, these do NOT go over signals sent between multiple processes Many more examples in: Recitation slides Lecture slides Past exams Reminder: solve the questions on your own when studying before looking at the answer

89 Signals Does the following code ever terminate? void handler(int sig) { while(); } int main() { Signal(SIGUSR, handler); Kill(, SIGUSR); return ; }

90 Signals Does the following code ever terminate? void handler(int sig) { while(); // stuck here forever! } int main() { Signal(SIGUSR, handler); // Spec says signal sent to self must be handled // before kill() returns Kill(, SIGUSR); return ; }

91 Signals What about now? void handler(int sig) { Kill(, SIGUSR); } int main() { Signal(SIGUSR, handler); Kill(, SIGUSR); return ; }

92 Signals What about now? void handler(int sig) { Kill(, SIGUSR); } int main() { Signal(SIGUSR, handler); Kill(, SIGUSR); return ; } Does not terminate: main: Kill(, SIGUSR); <SIGUSR handler invoked, SIGUSR blocked in handler> handler: Kill(, SIGUSR); <SIGUSR pending> <handler returns> <SIGUSR unblocked, SIGUSR handler invoked, SIGUSR blocked in handler> handler: Kill(, SIGUSR);...repeat...

93 Good luck! Carnegie Mellon