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Lecture 22: Abstractions for Concurrency. When you have a world-wide tuple space, you’ll be able to tune it in from any computer anywhere – or from any quasi-computer: any cell phone, any TV, any toaster. David Gelernter’s introduction to JavaSpaces Principles, Patterns, and Practice. - PowerPoint PPT Presentation

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David Evanshttp://www.cs.virginia.edu/~evans

CS655: Programming LanguagesUniversity of VirginiaComputer Science

Lecture 22: Abstractions for Concurrency

When you have a world-wide tuple space, you’ll be ableto tune it in from any computer anywhere – or from any quasi-computer: any cell phone, any TV, any toaster.

David Gelernter’s introduction to JavaSpaces Principles, Patterns, and Practice.

17 April 2001 CS 655: Lecture 21 2

Menu

• Form going around– Signup for Project Presentations– Vote for Next Lecture (no cheating!)

• Abstractions for Concurrency– Algol 68– Monitors– Linda and JavaSpaces

17 April 2001 CS 655: Lecture 21 3

Last Time

• Concurrent programming is programming with partial ordering on time

• A concurrent programming language gives programmers mechanisms for expressing that partial order

• We can express many partial orders using the thread control primitives fork and join and locking primitives protect, acquire and release.

17 April 2001 CS 655: Lecture 21 4

Abstractions

• Programming at that low level would be a pain – are there better abstractions?– Hundreds of attempts...we’ll see a few

today.

• Issues– Thread creation – Thread synchronization– Resource contention

fork

join

protect, acquire, release

17 April 2001 CS 655: Lecture 21 5

Algol 68: Collateral Clauses

• Collateral Clausesstmt0; (stmt1, stmt2) stmt3;

Defines a partial order:

stmt0

stmt1 stmt2

stmt3

17 April 2001 CS 655: Lecture 21 6

Algol 68: Semaphores

• Dijkstra, “Cooperating Sequential Processes”

• type semaup – increments

down – decrements (must > 0 before)

17 April 2001 CS 655: Lecture 21 7

Semaphore Example

begin sema mutex := level 1;

proc producer

while not finished

do down mutex

... insert item

up mutex

od;

17 April 2001 CS 655: Lecture 21 8

Semaphore Example, cont.proc consumer

while not finished

do down mutex

... remove item

up mutex

od;

par (producer, consumer)

// start them in parallel

17 April 2001 CS 655: Lecture 21 9

What can go wrong?

• Programmer neglects to up semaphore

• Programmer neglects to down semaphore before accessing shared resource

• Programmer spends all her time worrying about up and down instead of the algorithm

17 April 2001 CS 655: Lecture 21 10

Monitors• Concurrent Pascal [Hansen 74], Modula

[Wirth 77], Mesa [Lampson80]

• Integrated data abstraction and resource synchronization

• Routines that use a shared resource grouped in a monitor, accesses only allowed through exported procedures

• Conditions can control when threads may execute exported procedures

17 April 2001 CS 655: Lecture 21 11

Monitor Examplemonitor boundedbuffer

buffer: array 0..N-1 of int;

count: 0..N;

nonempty, nonfull: condition;

procedure append (x: int)

if count = N then nonfull.wait;

buffer[count] := x; count := count + 1;

nonempty.signal

end appendExample adapted from [Hansen93]

17 April 2001 CS 655: Lecture 21 12

Monitor Example, cont.

procedure remove () returns portion

if count = 0 then nonempty.wait;

x := ...

nonfull.signal

end remove;

17 April 2001 CS 655: Lecture 21 13

Java Synchronization• synchronized method qualifier

– Once a synchronized method begins execution, it will complete before any other thread enters a method of the same object

– Run-time must associate a lock with every object

• Is this enough to implement a semaphore?

• Is this better/worse than monitors?

17 April 2001 CS 655: Lecture 21 14

Synchronized Example

class ProducerConsumer {

private int x = 1;

synchronized void produce ()

{ x = x + 1; }

synchronized void consume ()

{ x = x – 1; }

} How could we require x stay positive?

17 April 2001 CS 655: Lecture 21 15

Linda

• Program Concurrency by using uncoupled processes with shared data space

• Add concurrency into a sequential language by adding:– Simple operators– Runtime kernel (language-independent)– Preprocessor (or compiler)

17 April 2001 CS 655: Lecture 21 16

Design by Taking Away

• Backus: von Neumann bottleneck results from having a store– Remove the store Functional Languages

• Gelernter: distributed programming is hard because of inter-process scheduling and communication due to order of mutation– We don’t have to remove the store, just mutation– Remove mutation read-and-remove only store

tuple spaces

17 April 2001 CS 655: Lecture 21 17

Basic Idea• Have a shared space (“tuple space”)

– Processes can add, read, and take away values from this space

• Bag of processes, each looks for work it can do by matching values in the tuple space

• Get load balancing, synchronization, messaging, etc. for free!

17 April 2001 CS 655: Lecture 21 18

TuplesConventional Memory

Linda/JavaSpaces

Unit Bit Logical Tuple

(23, “test”, false)

Access Using Address (variable) Selection of values

Operations read, write read, add, remove

JavaSpaces:

read, write, take

immutable

17 April 2001 CS 655: Lecture 21 19

Tuple Space Operations

• out (t) – add tuple t to tuple space• take (s) t – returns and removes tuple

t matching template s• read (s) t – same as in, except

doesn’t remove t.

• Operations are atomic (even if space is distributed)

17 April 2001 CS 655: Lecture 21 20

Meaning of taketake (“f”, int n)

take (“f”, 23)

take (“t”, bool b, int n)

take (string s, int n)

take (“cookie”)

(“f”, 23)

(“f”, 17)

(“t”, 25)

(“t”, true)(“t”, false)

Tuple Space

17 April 2001 CS 655: Lecture 21 21

Operational Semantics

• Extend configurations with a tuple space (just a bag of tuples)

• Transition rule for out:– Just add an entry to the tuple space

• Transition rule for take:– If there is a match (ignoring binding):

• Remove it from the tuple space• Advance the thread

– Similar to join last time – it just waits if there is no match

17 April 2001 CS 655: Lecture 21 22

Shared AssignmentLoc := Expressiontake (“Loc”, formal loc_value);out (“Loc”, Expression);

e.g.:x := x + 1; take (“x”, formal x_value)

out (“x”, x_value + 1);

17 April 2001 CS 655: Lecture 21 23

Semaphore

• Create (int n, String resource)for (i = 0; i < n; i++) out (resource);

• Down (String resource)take (resource)

• Up (String resource)out (resource)

17 April 2001 CS 655: Lecture 21 24

Distributed Ebay• Offer Item (String item, int minbid, int time):

out (item, minbid, “owner”);sleep (time);take (item, formal bid, formal bidder);if (bidder “owner”) SOLD!

• Bid (String bidder, String item, int bid):take (item, formal highbid, formal highbidder);if (bid > highbid) out (item, bid, bidder) else out (item, highbid, highbidder)

How could a bidder cheat?

17 April 2001 CS 655: Lecture 21 25

FactorialSetup:

for (int i = 1; i <= n; i++) out (i);

start FactTask (replicated n-1 times)

FactTask:

take (int i); take (int j); out (i * j);

Eventually, tuple space contains one entry which is the answer.

Better way to order Setup?

What if last two elements are taken concurrently?

17 April 2001 CS 655: Lecture 21 26

Finishing FactorialSetup:

for (int i = 1; i <= n; i++) out (i);out (“workleft”, n - 1);take (“workleft”, 0);take (result);

FactTask:take (“workleft”, formal w);

if (w > 0) take (int i); take (int j); out (i * j); out (“workleft”, w – 1); endif; Opps – we’ve sequentialized it!

17 April 2001 CS 655: Lecture 21 27

Concurrent Finishing FactorialSetup:

start FactWorker (replicated n-1 times)out (“done”, 0);for (int i = 1; i <= n; i++) {

out (i); if i > 1 out (“work”); }take (“done”, n-1);take (result);

FactWorker:take (“work”);

take (formal int i); take (formal int j); out (i * j); take (“done”, formal int n); out (“done”, n + 1);

17 April 2001 CS 655: Lecture 21 28

Sorting in Linda• Problem: Sorting an array of n integers

• Initial tuple state: (“A”, [A[0], ..., A[n-1]])

• Final tuple state: (“A”, [A’[0], ..., A’[n-1]]) such A’ has a corresponding element for

every element in A, and

for all 0 <= j < k <= n-1, A’[j] <= A’[k].

• Task: Devise a Linda sorting program and analyze its performance (can you match MergeSort?)

17 April 2001 CS 655: Lecture 21 29

Summary

• Linda/JavaSpaces provides a simple, but powerful model for distributed computing

• JavaSpaces extends Linda with:– Leases (tuples that expire after a time limit)

• Implementing an efficient, scalable tuple space (that provides the correct global semantics) is hard; people have designed custom hardware to do this.

17 April 2001 CS 655: Lecture 21 30

Charge

• You can download JavaSpaces implementation from:

http://java.sun.com/products/javaspaces/

• Project presentations for next week – advice for them Thursday

• Projects are due 2 weeks from today

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