c# ProcessStartInfo.Start - reading output but with a timeout

Question

If you want to start another process and wait (with time out) to finish you can use the following (from MSDN).

//Set a time-out value.
int timeOut=5000;
//Ge         


        

Answer 1

This technique will hang if the output buffer is filled with more that 4KB of data. A more foolproof method is to register delegates to be notified when something is written to the output stream. I've already suggested this method before in another post:

ProcessStartInfo processInfo = new ProcessStartInfo("Write500Lines.exe");
processInfo.ErrorDialog = false;
processInfo.UseShellExecute = false;
processInfo.RedirectStandardOutput = true;
processInfo.RedirectStandardError = true;

Process proc = Process.Start(processInfo);

// You can pass any delegate that matches the appropriate 
// signature to ErrorDataReceived and OutputDataReceived
proc.ErrorDataReceived += (sender, errorLine) => { if (errorLine.Data != null) Trace.WriteLine(errorLine.Data); };
proc.OutputDataReceived += (sender, outputLine) => { if (outputLine.Data != null) Trace.WriteLine(outputLine.Data); };
proc.BeginErrorReadLine();
proc.BeginOutputReadLine();

proc.WaitForExit();