OS Chapter 6

2015-10-31 13:23:18   0  举报

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AI智能生成

在操作系统的第六章中，我们将深入探讨进程管理。进程是计算机系统中的基本执行单元，它包括了程序的代码、数据和运行时的上下文信息。进程管理的主要任务是控制进程的创建、调度和终止，以及保护系统资源。我们将讨论进程的状态转换，如就绪、运行和阻塞，并解释如何通过进程调度算法来决定哪个进程应该获得CPU的执行权。此外，我们还将探讨如何处理进程间的通信和同步问题，以及如何防止死锁的发生。这一章将为你提供理解和管理计算机系统中进程行为的关键知识。

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Background

Concurrent access to shared data may result in data inconsistency

Maintaining data consistency requires mechanisms to ensure the orderly execution of cooperating processes

consumer-producer problem

Suppose that we wanted to provide a solution to the consumer-producer problem that fills all the buffers.

We can do so by having an integer count that keeps track of the number of full buffers.

Initially, count is set to 0. It is incremented by the producer after it produces a new buffer and is decremented by the consumer after it consumes a buffer.

Producer

while (true) { /* produce an item and put in nextProduced */ while (counter == BUFFER_SIZE) ; // do nothing buffer [in] = nextProduced; in = (in + 1) % BUFFER_SIZE; counter++; }

Consumer

while (true) { while (counter == 0) ; // do nothing nextConsumed = buffer[out]; out = (out + 1) % BUFFER_SIZE; counter--; /* consume the item in nextConsumed */ }

Race Condition

counter++ could be implemented as register1 = counter register1 = register1 + 1 counter = register1 counter-- could be implemented as register2 = counter register2 = register2 - 1 count = register2 Consider this execution interleaving with “count = 5” initially: S0: producer execute register1 = counter {register1 = 5}S1: producer execute register1 = register1 + 1 {register1 = 6} S2: consumer execute register2 = counter {register2 = 5} S3: consumer execute register2 = register2 - 1 {register2 = 4} S4: producer execute counter = register1 {count = 6 } S5: consumer execute counter = register2 {count = 4}

The Critical-Section Problem

Consider system of n processes {p0, p1, … pn-1}

Each process has critical section that is segment of code

Process may be changing common variables, updating table, writing file, etc

When one process in critical section, no other may be in its critical section

Critical section problem is to design protocol to solve this

Each process must ask permission to enter critical section in entry section, may follow critical section with exit section, then remainder section

Solution to Critical-Section Problem

Mutual Exclusion - If process Pi is executing in its critical section, then no other processes can be executing in their critical sections

Progress - If no process is executing in its critical section and there exist some processes that wish to enter their critical section, then the selection of the processes that will enter the critical section next cannot be postponed indefinitely

Bounded Waiting - A bound must exist on the number of times that other processes are allowed to enter their critical sections after a process has made a request to enter its critical section and before that request is granted Assume that each process executes at a nonzero speed No assumption concerning relative speed of the n processes

Peterson’s Solution

The two processes share two variables: int turn; Boolean flag[2]

The variable turn indicates whose turn it is to enter the critical section

The flag array is used to indicate if a process is ready to enter the critical section. flag[i] = true implies that process Pi is ready!

do { flag[i] = TRUE; turn = j; while (flag[j] && turn == j); critical section flag[i] = FALSE; remainder section } while (TRUE);

Provable that Mutual exclusion is preserved Progress requirement is satisfied Bounded-waiting requirement is met

Two process solution Assume that the LOAD and STORE instructions are atomic; that is, cannot be interrupted

Synchronization Hardware

concept

Many systems provide hardware support for critical section code

Uniprocessors – could disable interrupts

Currently running code would execute without preemption

Generally too inefficient on multiprocessor systems Operating systems using this not broadly scalable

Modern machines provide special atomic hardware instructions

Atomic = non-interruptable

Either test memory word and set value

swap contents of two memory words

Solution to Critical-section Problem Using Locks

do { acquire lock critical section release lock remainder section } while (TRUE);

TestAndSet Instruction

boolean TestAndSet (boolean *target) { boolean rv = *target; *target = TRUE; return rv: }

Solution using TestAndSet

Shared boolean variable lock, initialized to FALSE

do { while ( TestAndSet (&lock )) ; // do nothing // critical section lock = FALSE; // remainder section } while (TRUE);

Swap Instruction

void Swap (boolean *a, boolean *b) { boolean temp = *a; *a = *b; *b = temp: }

Solution using Swap

Shared Boolean variable lock initialized to FALSE; Each process has a local Boolean variable key

do { key = TRUE; while ( key == TRUE) Swap (&lock, &key ); // critical section lock = FALSE; // remainder section } while (TRUE);

Semaphores

Introduction

Synchronization tool that does not require busy waiting

Semaphore S – integer variable

Two standard operations modify S: wait() and signal()

Can only be accessed via two indivisible (atomic) operations

wait (S) { while S <= 0 ; // no-op S--; } signal (S) { S++; }

Semaphore as General Synchronization Tool

Counting semaphore – integer value can range over an unrestricted domain

Binary semaphore – integer value can range only between 0 and 1; can be simpler to implement Also known as mutex locks

Can implement a counting semaphore S as a binary semaphore

Provides mutual exclusion

Semaphore mutex; // initialized to 1 do { wait (mutex); // Critical Section signal (mutex); // remainder section } while (TRUE);

Semaphore Implementation

Must guarantee that no two processes can execute wait () and signal () on the same semaphore at the same time

Thus, implementation becomes the critical section problem where the wait and signal code are placed in the critical section Could now have busy waiting in critical section implementation

Note that applications may spend lots of time in critical sections and therefore this is not a good solution

Semaphore Implementation with no Busy waiting

With each semaphore there is an associated waiting queue

Each entry in a waiting queue has two data items: value (of type integer) pointer to next record in the list

Two operations: block – place the process invoking the operation on the appropriate waiting queue wakeup – remove one of processes in the waiting queue and place it in the ready queue

Deadlock and Starvation

Deadlock – two or more processes are waiting indefinitely for an event that can be caused by only one of the waiting processes

Starvation – indefinite blocking A process may never be removed from the semaphore queue in which it is suspended

Priority Inversion – Scheduling problem when lower-priority process holds a lock needed by higher-priority process Solved via priority-inheritance protocol

Classic Problems of Synchronization

Classical problems used to test newly-proposed synchronization schemes

Bounded-Buffer Problem

Readers and Writers Problem

Dining-Philosophers Problem

Bounded-Buffer Problem ( producer-consumer problem)

N buffers, each can hold one item Semaphore mutex initialized to the value 1 Semaphore full initialized to the value 0 Semaphore empty initialized to the value N

The structure of the producer process do { // produce an item in nextp wait (empty); wait (mutex); // add the item to the buffer signal (mutex); signal (full); } while (TRUE);

The structure of the consumer process do { wait (full); wait (mutex); // remove an item from buffer to nextc signal (mutex); signal (empty); // consume the item in nextc } while (TRUE);

Readers-Writers Problem

A data set is shared among a number of concurrent processes Readers – only read the data set; they do not perform any updates Writers – can both read and write Problem – allow multiple readers to read at the same time Only one single writer can access the shared data at the same time Several variations of how readers and writers are treated – all involve priorities Shared Data Data set Semaphore mutex initialized to 1 Semaphore wrt initialized to 1 Integer readcount initialized to 0

The structure of a writer process do { wait (wrt) ; // writing is performed signal (wrt) ; } while (TRUE);

The structure of a reader process do { wait (mutex) ; readcount ++ ; if (readcount == 1) wait (wrt) ; signal (mutex) // reading is performed wait (mutex) ; readcount - - ; if (readcount == 0) signal (wrt) ; signal (mutex) ; } while (TRUE);

Dining-Philosophers Problem

Philosophers spend their lives thinking and eating Don’t interact with their neighbors, occasionally try to pick up 2 chopsticks (one at a time) to eat from bowl Need both to eat, then release both when done In the case of 5 philosophers Shared data Bowl of rice (data set) Semaphore chopstick [5] initialized to 1

The structure of Philosopher i: do { wait ( chopstick[i] ); wait ( chopStick[ (i + 1) % 5] ); // eat signal ( chopstick[i] ); signal (chopstick[ (i + 1) % 5] ); // think } while (TRUE);

Deadlock Starvation

Monitors