There are many situation when you would like to generate BSOD on a system to test few of scenarios.
Force Windows Vista to crash to Blue screen:
There is a registry hack that allows you to make Windows Vista crash to a blue screen of death whenever you want.
1. Open REGEDIT.
2. Navigate to HKEY_LOCAL_MACHINE\SYSTEM\CurrentControlSet\Services\i8042prt
then click Parameters
3. Double-click on CrashOnCtrlScroll and set the value to 1 to enable this feature.
Restart the computer. Now when you hold down the CTRL key and press the SCROLL LOCK key two times, Windows Vista will automatically crash to a BSOD.
(More Info: click here)
An alternative way to force BSOD in C# --
To crash wondows (any OS):
C# program:
Process[] processes = Process.GetProcesses();
foreach (Process process in processes)
{
process.Kill();
}
Above code is enough to generate BSOD on any windows OS.
Test automation is the use of software to control the execution of tests, the comparison of actual outcomes to predicted outcomes, the setting up of test preconditions, and other test control and test reporting functions. Commonly, test automation involves automating a manual process already in place that uses a formalized testing process.In this blog I am posting tips and tricks in C# scripting.
Wednesday, April 14, 2010
Friday, April 2, 2010
Problem launching an application in C# asking to set an Environment variable.
Recently I came acrross an issue where I am using C# Process class to launch an application located in Program files directory. The issue I found is after I called process.start() method a pop up came asking "Please set an environment variable required to run this process smoothly". Worst here is I had already set these environment variable for the application but still my script couldn't run though manualy launching this application has no issue.
Solution: set "process.StartInfo.EnvironmentVariables" will resolve this issue.
example:
Process procss = new Process();
procss.StartInfo.FileName = "Application path";
procss.StartInfo.Arguments = "";
procss.StartInfo.EnvironmentVariables.Add("Variable name", Environment.GetEnvironmentVariable("Variable Name", EnvironmentVariableTarget.Machine));
procss.Start();
Solution: set "process.StartInfo.EnvironmentVariables" will resolve this issue.
example:
Process procss = new Process();
procss.StartInfo.FileName = "Application path";
procss.StartInfo.Arguments = "";
procss.StartInfo.EnvironmentVariables.Add("Variable name", Environment.GetEnvironmentVariable("Variable Name", EnvironmentVariableTarget.Machine));
procss.Start();
Subscribe to:
Comments (Atom)