SemaphoreFullException Class
The exception that is thrown when the Semaphore.Release method is called on a semaphore whose count is already at the maximum.
Assembly: System (in System.dll)
The count on a semaphore is decremented each time a thread enters the semaphore, and incremented when a thread releases the semaphore. When the count is zero, subsequent requests block until other threads release the semaphore. When all threads have released the semaphore, the count is at the maximum value specified when the semaphore was created. If a programming error causes a thread to call the Semaphore.Release method at this point, a SemaphoreFullException is thrown.
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The Semaphore class does not enforce thread identity on calls to the WaitHandle.WaitOne and Semaphore.Release methods. It is not necessary for the same thread that called WaitOne to call Release. |
SemaphoreFullException does not necessarily indicate a problem with the code where the exception occurred. Consider the following scenario: Thread A and thread B enter a semaphore that has a maximum count of two. A programming error in thread B causes it to call Release twice, so that the count on the semaphore is full. As a result, when thread A eventually calls Release, a SemaphoreFullException is thrown.
For a list of initial property values for an instance of the SemaphoreFullException class, see the SemaphoreFullException constructor.
The following code example shows how a programming error in one thread can lead to a SemaphoreFullException in another thread: Two threads enter a semaphore. The second thread releases the semaphore twice, while the first thread is still executing its task. When the first thread finishes and releases the semaphore, the semaphore count is already full and an exception is thrown.
Imports System Imports System.Threading Public Class Example ' A semaphore that can satisfy at most two concurrent ' requests. ' Private Shared _pool As New Semaphore(2, 2) <MTAThread> _ Public Shared Sub Main() ' Create and start two threads, A and B. ' Dim tA As New Thread(AddressOf ThreadA) tA.Start() Dim tB As New Thread(AddressOf ThreadB) tB.Start() End Sub Private Shared Sub ThreadA() ' Thread A enters the semaphore and simulates a task ' that lasts a second. ' _pool.WaitOne() Console.WriteLine("Thread A entered the semaphore.") Thread.Sleep(1000) Try _pool.Release() Console.WriteLine("Thread A released the semaphore.") Catch ex As Exception Console.WriteLine("Thread A: {0}", ex.Message) End Try End Sub Private Shared Sub ThreadB() ' Thread B simulates a task that lasts half a second, ' then enters the semaphore. ' Thread.Sleep(500) _pool.WaitOne() Console.WriteLine("Thread B entered the semaphore.") ' Due to a programming error, Thread B releases the ' semaphore twice. To fix the program, delete one line. _pool.Release() _pool.Release() Console.WriteLine("Thread B exits successfully.") End Sub End Class ' This code example produces the following output: ' ' Thread A entered the semaphore. ' Thread B entered the semaphore. ' Thread B exits successfully. ' Thread A: Adding the given count to the semaphore would cause it to exceed its maximum count. '
Windows 7, Windows Vista, Windows XP SP2, Windows XP Media Center Edition, Windows XP Professional x64 Edition, Windows XP Starter Edition, Windows Server 2008 R2, Windows Server 2008, Windows Server 2003, Windows Server 2000 SP4, Windows Millennium Edition, Windows 98
The .NET Framework and .NET Compact Framework do not support all versions of every platform. For a list of the supported versions, see .NET Framework System Requirements.
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