1. Trang chủ
  2. » Công Nghệ Thông Tin

A computer system consists of hardware, system programs, and application programs figs 3

15 306 0

Đang tải... (xem toàn văn)

Tài liệu hạn chế xem trước, để xem đầy đủ mời bạn chọn Tải xuống

THÔNG TIN TÀI LIỆU

Thông tin cơ bản

Định dạng
Số trang 15
Dung lượng 348,55 KB

Nội dung

HISTORY OF UNIX 10.2 OVERVIEW OF UNIX 10.3 PROCESSES IN UNIX 10.4 MEMORY MANAGEMENT IN UNIX 10.5 INPUT/OUTPUT IN UNIX 10.6 THE UNIX FILE SYSTEM 10.7 SECURITY IN UNIX 10.8 SUMMARY

3 DEADLOCKS 3.1 RESOURCES 3.2 INTRODUCTION TO DEADLOCKS 3.3 THE OSTRICH ALGORITHM 3.4 DEADLOCK DETECTION AND RECOVERY 3.5 DEADLOCK AVOIDANCE 3.6 DEADLOCK PREVENTION 3.7 OTHER ISSUES 3.8 RESEARCH ON DEADLOCKS 3.9 SUMMARY typedef int semaphore; typedef int semaphore; semaphore resource 1; semaphore resource 1; semaphore resource 2; void process A(void) { void process A(void) { down(&resource 1); down(&resource 1); use resource 1( ); down(&resource 2); up(&resource 1); use both resources( ); } up(&resource 2); up(&resource 1); } (a) (b) Fig. 3-1. Using a semaphore to protect resources. (a) One resource. (b) Two resources. typedef int semaphore; semaphore resource 1; semaphore resource 1; semaphore resource 2; semaphore resource 2; void process A(void) { void process A(void) { down(&resource 1); down(&resource 1); down(&resource 2); down(&resource 2); use both resources( ); use both resources( ); up(&resource 2); up(&resource 2); up(&resource 1); up(&resource 1); }} void process B(void) { void process B(void) { down(&resource 1); down(&resource 2); down(&resource 2); down(&resource 1); use both resources( ); use both resources( ); up(&resource 2); up(&resource 1); up(&resource 1); up(&resource 2); }} (a) (b) Fig. 3-2. (a) Deadlock-free code. (b) Code with a potential deadlock. (a) (b) (c) T U D C S B A R Fig. 3-3. Resource allocation graphs. (a) Holding a resource. (b) Requesting a resource. (c) Deadlock. (j) A Request R Request S Release R Release S B Request S Request T Release S Release T C Request T Request R Release T Release R 1. A requests R 2. B requests S 3. C requests T 4. A requests S 5. B requests T 6. C requests R deadlock 1. A requests R 2. C requests T 3. A requests S 4. C requests R 5. A releases R 6. A releases S no deadlock A R B S C T (i) A R B S C T (h) A R B S C T (g) A R B S C T (f) A R B S C T (e)(d) (c)(b)(a) A R B S C T (q) A R B S C T (p) A R B S C T (o) A R B S C T (n) A R B S C T (m) A R B S C T (l)(k) A R B S C T Fig. 3-4. An example of how deadlock occurs and how it can be avoided. R S T R U V U V W C D E D E G G A F B (a) (b) Fig. 3-5. (a) A resource graph. (b) A cycle extracted from (a). Resources in existence (E 1 , E 2 , E 3 , …, E m ) Current allocation matrix C 11 C 21 C n1 C 12 C 22 C n2 C 13 C 23 C n3 C 1m C 2m C nm Row n is current allocation to process n Resources available (A 1 , A 2 , A 3 , …, A m ) Request matrix R 11 R 21 R n1 R 12 R 22 R n2 R 13 R 23 R n3 R 1m R 2m R nm Row 2 is what process 2 needs Fig. 3-6. The four data structures needed by the deadlock detection algorithm. Tape drives Plotters Scanners CD Roms E = ( 4 2 3 1 ) Tape drives Plotters Scanners CD Roms A = ( 2 1 0 0 ) Current allocation matrix 0 2 0 0 0 1 1 0 2 0 1 0 Request matrix 2 1 2 0 0 1 0 1 0 1 0 0 R =C = Fig. 3-7. An example for the deadlock detection algorithm. Plotter Printer Printer Plotter B A u (Both processes finished) pq r s t I 8 I 7 I 6 I 5 I 4 I 3 I 2 I 1 Fig. 3-8. Two process resource trajectories. A B C 3 2 2 9 4 7 Free: 3 (a) A B C 3 4 2 9 4 7 Free: 1 (b) A B C 3 0 –– 2 9 7 Free: 5 (c) A B C 3 0 7 9 7 Free: 0 (d) – A B C 3 0 0 9 – Free: 7 (e) Has Max Has Max Has Max Has Max Has Max Fig. 3-9. Demonstration that the state in (a) is safe. [...]...Has Max Has Max Has Max Has Max A 3 9 A 4 9 A 4 9 A 4 9 B 2 4 B 2 4 B 4 4 B — — C 2 7 C 2 7 C 2 7 C 2 7 Free: 3 (a) Free: 2 (b) Free: 0 (c) Fig 3- 10 Demonstration that the state in (b) is not safe Free: 4 (d) Has Max Has Max Has Max A 0 6 A 1 6 A 1 6 B 0 5 B 1 5 B 2 5 C 0 4 C 2 4 C 2 4 D 0 7 D 4 7 D 4 7 Free: 10 Free: 2 Free: 1 (a) (b) (c) Fig 3- 11 Three resource allocation states: (a) Safe (b) Safe... Imagesetter 2 Scanner 3 Plotter 4 Tape drive 5 CD Rom drive (a) A B i j (b) Fig 3- 13 (a) Numerically ordered resources (b) A resource graph Condition Mutual exclusion Hold and wait No preemption Circular wait Approach Spool everything Request all resources initially Take resources away Order resources numerically Fig 3- 14 Summary of approaches to deadlock prevention ... (c) Unsafe Pr oc e Ta ss pe Pl driv ot te es Sc rs an n CD ers RO M s Pr oc e Ta ss pe Pl driv ot te es Sc rs an n CD ers RO M s A 3 0 1 1 A 1 1 0 0 B 0 1 0 0 B 0 1 1 2 C 1 1 1 0 C 3 1 0 0 D 1 1 0 1 D 0 0 1 0 E 0 0 0 0 E 2 1 1 0 Resources assigned E = ( 634 2) P = ( 532 2) A = (1020) Resources still needed Fig 3- 12 The banker’s algorithm with multiple resources 1 Imagesetter 2 Scanner 3 Plotter 4 Tape drive . Has Max Has Max Has Max Fig. 3-10. Demonstration that the state in (b) is not safe. A B C D 0 0 0 0 6 Has Max 5 4 7 Free: 10 A B C D 1 1 2 4 6 Has Max 5 4 7 Free: 2 A B C D 1 2 2 4 6 Has Max 5 4 7 Free:. Spool everything Hold and wait Request all resources initially No preemption Take resources away Circular wait Order resources numerically Fig. 3-14. Summary of approaches to deadlock prevention. . Has Max Fig. 3-9. Demonstration that the state in (a) is safe. A B C 3 2 2 9 4 7 Free: 3 (a) A B C 4 2 2 9 4 7 Free: 2 (b) A B C 4 4 —4 2 9 7 Free: 0 (c) A B C 4 — 2 9 7 Free: 4 (d) Has Max Has

Ngày đăng: 28/04/2014, 16:35

TÀI LIỆU CÙNG NGƯỜI DÙNG

TÀI LIỆU LIÊN QUAN