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Cooperating Threads

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  1. Cooperating Threads Andy Wang Operating Systems COP 4610 / CGF 5765

  2. Independent Threads • No states shared with other threads • Deterministic computation • Output depends on input • Reproducible • Output does not depend on the order and timing of other threads • Scheduling order does not matter • e.g., gif2jpeg

  3. Cooperating Threads • Shared states • Nondeterministic • Nonreproducible • e.g., elevator controllers • Example: 2 threads sharing the same display Thread A Thread B cout << “A”; cout << “1”; cout << “B”; cout << “2”; cout << “C”; cout << “3”; • You may get “A12BC3”

  4. So, Why Allow Cooperating Threads?

  5. So, Why Allow Cooperating Threads? • Shared resources • e.g., a single processor • Speedup • Occurs when threads use different resources at different times • Modularity • An application can be decomposed into threads

  6. Some Concurrent Programs • If threads work on separate data, scheduling does not matter Thread A Thread B x = 1; y = 2;

  7. Some Concurrent Programs • If threads share data, the final values are not as obvious Thread A Thread B x = 1; y = 2; x = y + 1; y = y * 2; • What are the indivisible operations?

  8. Atomic Operations • An atomic operation always runs to completion; it’s all or nothing • e.g., memory loads and stores on most machines • Many operations are not atomic • Double precision floating point store

  9. Suppose… • Each C statement is atomic • Let’s revisit the example…

  10. x = 1 y = 2 x = y + 1 y = 2 x = 1 y = y * 2 x = 1 y = 2 x = y + 1 y = y * 2 x = y + 1 y = y * 2 y = y * 2 x = y + 1 All Possible Execution Orders Thread A Thread B x = 1; y = 2; x = y + 1; y = y * 2; A decision tree

  11. All Possible Execution Orders Thread A Thread B x = 1; y = 2; x = y + 1; y = y * 2; (x = ?, y = ?) (x = ?, y = 2) (x = 1, y = ?) x = 1 y = 2 (x = 1, y = 2) (x = ?, y = ?) (x = ?, y = 4) x = y + 1 y = 2 x = 1 y = y * 2 (x = ?, y = 2) (x = 3, y = 2) (x = 1, y = 4) (x = 1, y = 4) x = 1 y = 2 x = y + 1 y = y * 2 x = y + 1 y = y * 2 y = y * 2 x = y + 1 (x = ?, y = 4) (x = 3, y = 4) (x = 5, y = 4) (x = 5, y = 4)

  12. Another Example • Assume each C statement is atomic • Both threads are in the same address space Thread A Thread B j = 0; j = 0; while (j < 10) { while (j > -10) { ++j; --j; } } cout << “A wins”; cout << “B wins”;

  13. So… • Who wins? • Can the computation go on forever? • Race conditions occur when threads share data, and their results depend on the timing of their executions…

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