Hashtable.Synch ronzied()

[bernardpace]

Hi,

Using the following code

Hashtable myData = Hashtable.Synch ronized(new Hashtable());

Then I will be having a main thread that will write to this hashtable,
and other n threads that will read and write to this hashtable. In the
n threads, when reading and writing, only one thread will be assigned
specifically on place in the hashtable.

Thread 1 will access only data in myData[1]
Thread 2 will access only data in myData[2]
Thread 3 will access only data in myData[3]
....

Having this scenario, is there need to use the lock(this){...}
mechanism on the Hashtable.
Can someone help me out
Thanks in Advance

[Ron]

You don't need to use lock(this){...} , you are already synchronizing the
hashtable.

"be*********@ya hoo.com" wrote:

Hi,

Using the following code

Hashtable myData = Hashtable.Synch ronized(new Hashtable());

Then I will be having a main thread that will write to this hashtable,
and other n threads that will read and write to this hashtable. In the
n threads, when reading and writing, only one thread will be assigned
specifically on place in the hashtable.

Thread 1 will access only data in myData[1]
Thread 2 will access only data in myData[2]
Thread 3 will access only data in myData[3]
....

Having this scenario, is there need to use the lock(this){...}
mechanism on the Hashtable.
Can someone help me out
Thanks in Advance

[Jon Skeet [C# MVP]]

Ron <Ro*@discussion s.microsoft.com > wrote:

You don't need to use lock(this){...} , you are already synchronizing the
hashtable.

Unless, of course, you want to have a whole sequence of operations
synchronized, rather than just one at a time.

--
Jon Skeet - <sk***@pobox.co m>
http://www.pobox.com/~skeet Blog: http://www.msmvps.com/jon.skeet
If replying to the group, please do not mail me too

[bernardpace]

If there is an other Thread thread0, that will have the right to access
all information in Hashtable, will there be the need to use the lock?

Where should it be used in this thread0 or in the other threads, or in
all?

Thanks for your help