Hi there, folks.
Here I am again with questions about multi-threading and an exercise of my Concurrent Programming class.
I have a multi-threaded server - implemented using .NET Asynchronous Programming Model - with GET (download) and PUT (upload) file services. This part is done and tested.
It happens that the statement of the problem says this server must have logging activity with the minimum impact on the server response time, and it should be supported by a low priority thread - logger thread - created for this effect. All logging messages shall be passed by the threads that produce them to this logger thread, using a communication mechanism that may not lock the thread that invokes it (besides the necessary locking to ensure mutual exclusion) and assuming that some logging messages may be ignored.
Here is my solution, that I would like to help your help validating as a solution to the stated problem:
using System;
using System.IO;
using System.Threading;
// Multi-threaded Logger
public class Logger {
// textwriter to use as logging output
protected readonly TextWriter _output;
// logger thread
protected Thread _loggerThread;
// logger thread wait timeout
protected int _timeOut = 500; //500ms
// amount of log requests attended
protected volatile int reqNr = 0;
// logging queue
protected readonly object[] _queue;
protected struct LogObj {
public DateTime _start;
public string _msg;
public LogObj(string msg) {
_start = DateTime.Now;
_msg = msg;
}
public LogObj(DateTime start, string msg) {
_start = start;
_msg = msg;
}
public override string ToString() {
return String.Format("{0}: {1}", _start, _msg);
}
}
public Logger(int dimension,TextWriter output) {
/// initialize queue with parameterized dimension
this._queue = new object[dimension];
// initialize logging output
this._output = output;
// initialize logger thread
Start();
}
public Logger() {
// initialize queue with 10 positions
this._queue = new object[10];
// initialize logging output to use console output
this._output = Console.Out;
// initialize logger thread
Start();
}
public void Log(string msg) {
lock (this) {
for (int i = 0; i < _queue.Length; i++) {
// seek for the first available position on queue
if (_queue[i] == null) {
// insert pending log into queue position
_queue[i] = new LogObj(DateTime.Now, msg);
// notify logger thread for a pending log on the queue
Monitor.Pulse(this);
break;
}
// if there aren't any available positions on logging queue, this
// log is not considered and the thread returns
}
}
}
public void GetLog() {
lock (this) {
while(true) {
for (int i = 0; i < _queue.Length; i++) {
// seek all occupied positions on queue (those who have logs)
if (_queue[i] != null) {
// log
LogObj obj = (LogObj)_queue[i];
// makes this position available
_queue[i] = null;
// print log into output stream
_output.WriteLine(String.Format("[Thread #{0} | {1}ms] {2}",
Thread.CurrentThread.ManagedThreadId,
DateTime.Now.Subtract(obj._start).TotalMilliseconds,
obj.ToString()));
}
}
// after printing all pending log's (or if there aren't any pending log's),
// the thread waits until another log arrives
//Monitor.Wait(this, _timeOut);
Monitor.Wait(this);
}
}
}
// Starts logger thread activity
public void Start() {
// Create the thread object, passing in the Logger.Start method
// via a ThreadStart delegate. This does not start the thread.
_loggerThread = new Thread(this.GetLog);
_loggerThread.Priority = ThreadPriority.Lowest;
_loggerThread.Start();
}
// Stops logger thread activity
public void Stop() {
_loggerThread.Abort();
_loggerThread = null;
}
// Increments number of attended log requests
public void IncReq() { reqNr++; }
}
Basically, here are the main points of this code:
- Start a low priority thread that loops the logging queue and prints pending logs to the output. After this, the thread is suspended till new log arrives;
- When a log arrives, the logger thread is awaken and does it's work.
Is this solution thread-safe? I have been reading Producers-Consumers problem and solution algorithm, but in this problem although I have multiple producers, I only have one reader.
Thanks in advance for all your attention.