VB2008从入门到精通(PDF格式英文版)-第126章
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Looking at the code within the block； you can see that the collection is accessed in two locations。
Using the exclusive SyncLock argument where a single thread can access only a single code
block； one thread will write to the collection； and another thread will read from the collection。
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C HA P TE R 1 3 ■ L E AR N IN G AB O U T M U L T IT HR E AD IN G 353
The SyncLock statement has a parameter that is a reference to lock against。 In both threads；
the reference is elements。 The mon reference synchronizes access to code。 At any given
point in time； the code contained within the SyncLock block will have only a single thread
executing。 This implements the desired feature； where only one thread is accessing code that
reads or writes to the collection。 The flow of the program is as follows：
1。 Both threads wait。
2。 After a 1 second； thread 1 acquires a lock because no other thread has done so。
3。 Thread 1 executes its code。
4。 Once thread 1 has started executing the synchronized code； no other code can acquire
the lock that is associated with the variable elements。
5。 When thread 2 wakes up after a sleep of 1。5 seconds； it will attempt to acquire the lock；
but it can’t because thread 1 is still holding the lock。 So the second thread must wait。
6。 Thread 1 eventually exits the synchronized code block； allowing the second thread to
add an element to the collection。 This time； no exception is thrown。
The reference to lock against does not need to be the reference that is manipulated within
the code block。 The reference is just that： an arbitrary reference。 You could use a different
object instance； and even instantiate an object like this：
Dim _syncRoot As Object = New Object（）
。 。 。
SyncLock _syncRoot
。 。 。
End SyncLock
When using exclusive blocks； you need to use them when reading or writing the object。
Don’t think that you need an exclusive lock only when modifying the data； because a reader
might be reading a state that is being modified。 As the example of the collection demonstrated；
modifying a collection while it is being read causes an inconsistency and an exception。 The
following code does not have a lock for the reading of the collection； and it results in an excep
tion being thrown。
。 。 。
Module ThreadProblem
Dim elements As List（Of Integer） = New List（Of Integer）（）
Sub Task1（）
Thread。Sleep（1000）
Dim item As Integer
For Each item In elements
Console。WriteLine（〃Item （〃 & item & 〃）〃）
Thread。Sleep（1000）
Next
End Sub
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354 CH AP T E R 1 3 ■ L E A R N I N G A B OU T M U L T I TH R E A DI N G
Sub Task2（）
Thread。Sleep（1500）
SyncLock elements
elements。Add（30）
End SyncLock
End Sub
Sub Main（）
elements。Add（10）
elements。Add（20）
Dim thread1 As New Thread（AddressOf Task1）
Dim thread2 As New Thread（AddressOf Task2）
thread1。Start（）
thread2。Start（）
End Sub
End Module
Getting back to the code that worked； it would seem that all is OK。 From a code…execution
perspective； that’s true。 But from an execution…efficiency perspective； everything is not OK；
because the reading of the collection is causing the writer to wait unnecessarily。
Again； this collection example illustrates the difficulty with writing multithreaded code。
You want to be able to add elements to a collection without needing to wait for other threads to
iterate a collection。 You want to be efficient； as well as being logically correct。
Synchronizing with Cloning
One way of making a lock more efficient is to clone the object so that the local copy that you
read from does not lock and hinder another thread。 The two…thread example could be rewritten
as follows：
。 。 。
Module ThreadProblem
Dim elements As List（Of Integer） = New List（Of Integer）（）
Sub Task1（）
Dim items As Integer（）
Thread。Sleep（1000）
SyncLock elements
items = elements。ToArray（）
End SyncLock
Dim item As Integer
For Each item In items
Console。WriteLine（〃Item （〃 & item & 〃）〃）
Thread。Sleep（&H3E8）
Next
End Sub
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C HA P TE R 1 3 ■ L E AR N IN G AB O U T M U L T IT HR E AD IN G 355
Sub Task2（）
Thread。Sleep（1500）
SyncLock elements
elements。Add（30）
End SyncLock
End Sub
Sub Main（）
elements。Add（10）
elements。Add（20）
Dim thread1 As New Thread（AddressOf Task1）
Dim thread2 As New Thread（AddressOf Task2）
thread1。Start（）
thread2。Start（）
End Sub
End Module
The code still uses a lock； but only in the places where it is necessary。 When the collection
is being iterated； the lock is applied to the operation of copying the collection to an array
（ToArray（））。 For the array iteration itself； there is no lock。 When writing to the collection； there
is a lock。
So； how can it be more efficient to take a snapshot of the collection； since taking a snap
shot takes time？ The answer is that normally it is not more efficient； but it is more time…slice
effective。
Consider a word processor that loads some text。 When Microsoft Word loads text； it imme
diately displays the first page； allowing you to edit right away。 In the background； you see the
other pages being loaded and prepared for editing。 Using the snapshot approach； you get the
same effect。
As a general threading rule； use locks as sparingly as possible； but use them whenever
necessary。 If you do use them； use them for as little code as possible。 Locks synchronize access
to resources； and thus only a single thread can be executing for a locked piece of code。 The less
time code is locked； the faster your code will be。
How Not to Deadlock Your Code （Mostly）
A deadlock makes code stop executing。 A deadlock occurs when one piece of code has a lock
and waits for some information to bee available。 However； the information does not
bee available； because another thread that could provide that information is waiting for a
lock to bee free。
I’ve said that if you are using locked code； you should use it as sparingly as possible。 That’s
because using locks can lead to deadlocks。 Deadlocks are a royal pain in the butt。
Consider the following two…thread collection example。
。 。 。
Module ThreadProblem
Dim elements As List（Of Integer） = New List（Of Integer）（）
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356 CH AP T E R 1 3 ■ L E A R N I N G A B OU T M U L T I TH R E A DI N G
Sub Task1（）
Dim items As Integer（）
Thread。Sleep（1000）
SyncLock elements
Do While （elements。Count 《 3）
Thread。Sl
