I have a program. When it is runned, it runs another legal proces (f.e. icq.exe, atp.) and then it inject itself's code into it (f.e. icq.exe). Icq.exe will not be runned, it only appears in taskmaanager. digitally signed and everything what ICQ.exe has.
BUt when I look in strings in memory, there is nothing from icq.exe but from prevously used program who runs this.
How is it possible? I was lookping for something and couldnt found anything. It doesnt run only icq. but it radomly pick up some program on harddrive.
Related
I'm trying to run this (script.exs):
System.cmd("zsh", ["-c" "com.spotify.Client"])
IO.puts("done.")
Spotify opens, but "done." never shows up on the screen. I also tried:
System.cmd("zsh", ["-c" "nohup com.spotify.Client &"])
IO.puts("done.")
My script only halts when I close the spotify window. Is it possible to run commands without waiting for it to end?
One should not spawn system tasks in a blind hope they would work properly. If the system task crashes, some actions should be taken in the calling OTP process, otherwise sooner or later it’ll crash in production and nobody would know what had happened and why.
There are many possible scenarios, I would go with Task.start_link/1 (assuming the calling process traps exits,) or with Task.async/1 accompanied by an explicit Task.await/1 somewhere in the supervision tree.
If despite everything explained above you don’t care about robustness, use Kernel.spawn/1 like below
pid = spawn(System, :cmd, ~w|zsh -c com.spotify.Client|)
# Process.monitor(pid) # yet give it a chance to handle errors
IO.puts("done.")
I am toying with the FluidEnter/FluidExit. So in a simple form, here is what I am trying to do:
I created in Main an empty population of agent called Terminal. For now, in Terminal, there is only a fluidEnter connected to a fluidExit (very simple)
enter image description here
Now, on startup, I want to fill this population and set up the proper connections (the terminals are ordered).
So, on startup, I call a function init(), whose body start with Terminal t = add_terminals(); (I have only one terminal for now, just toying with things)
In Main, obviously, I also have a fluidEnter and a fluidExit. I would like to connect the fluidExit of Main to the fluidEnter of the terminal t, and the fluidExit of the terminal t to the fluidEnter of Main, so code (still in init()) looks like
fluidExit.set_fluidEnter(t.fluidEnter);
t.fluidExit.set_fluidEnter(fluidEnter);
I get an exception so obviously, I am doing something wrong. Any idea?
I think the set_fluidEnter function is deprecated or just non-functional.
Instead, you should do:
fluidExit.connect(t.fluidEnter);
So just replace set_fluidEnter with connect... nothing else.
That should do the trick
I was going down the same path as you a couple of months ago. Yes... .connect() works great. It even works as a gate. If it is disconnected, then fluid stops at the exit. Once connected, fluid starts to flow again. It is very slick.
I am trying to make a patch that plays audio when a bang is pressed. I have put a symbol so that I don't need to keep reimporting the file. However it works sometimes but not all the time.
A warning in the Pd console reads: Start requested with no prior open
However I have imported an audio file
Is there something that I have done wrong?
Use [trigger] to get the order-of-execution correct.
One problem is, that whenever you send a [1( to [readsf~] you must have sent an [open ...( message directly beforehand.
Even if you have just successfully opened a file, but then stopped it (with [0() or played it through (so it has been closed automatically), you have to send the filename again.
The real problem is, that your messages are out of order: you should never have a fan-out (that is: connecting a message outlet to multiple inlets), as this will create undefined behavior.
Use [trigger] to get the order-of-execution correct.
(Mastering [trigger] is probably the single most important step in learning to program Pd)
This is a weird one. :)
I have a script running under Apache 1.3, with Apache::PerlRun option of mod_perl. It uses the standard CGI.pm module. It's a regularly accessed script on a busy server, accessed over https.
The URL is typically something like...
/script.pl?action=edit&id=47049
Which is then brought into Perl the usual way...
my $action = $cgi->param("action");
my $id = $cgi->param("id");
This has been working successfully for a couple of years. However we started getting support requests this week from our customers who were accessing this script and getting blank pages. We already had a line like the following that put the current URL into a form we use for customers to report an issue about a page...
$cgi->url(-query => 1);
And when we view source of the page, the result of that command is the same URL, but with an entirely different query string.
/script.pl?action=login&user=foo&password=bar
A query string that we recognise as being from a totally different script elsewhere on our system.
However crazy it sounds, it seems that when users are accessing a URL with a query string, the query string that the script is seeing is one from a previous request on another script. Of course the script can't handle that action and outputs nothing.
We have some automated test scripts running to see how often this happens, and it's not every time. To throw some extra confusion into the mix, after an Apache restart, the problem seems to initially disappear completely only to come back later. So whatever is causing it is somehow relieved by a restart, but we can't see how Apache can possibly take the request from one user and mix it up with another.
This, it appears, is an interesting combination of Apache 1.3, mod_perl 1.31, CGI.pm and Apache::GTopLimit.
A bug was logged against CGI.pm in May last year: RT #57184
Which also references CGI.pm params not being cleared?
CGI.pm registers a cleanup handler in order to cleanup all of it's cache.... (line 360)
$r->register_cleanup(\&CGI::_reset_globals);
Apache::GTopLimit (like Apache::SizeLimit mentioned in the bug report) also has a handler like this:
$r->post_connection(\&exit_if_too_big) if $r->is_main;
In pre mod_perl 1.31, post_connection and register_cleanup appears to push onto the stack, while in 1.31 it appears as if the GTopLimit one clobbers the CGI.pm entry. So if your GTopLimit function fires because the Apache process has got to large, then CGI.pm won't be cleaned up, leaving it open to returning the same parameters the next time you use it.
The solution seems to be to change line 360 of CGI.pm to;
$r->push_handlers( 'PerlCleanupHandler', \&CGI::_reset_globals);
Which explicitly pushes the handler onto the list.
Our restart of Apache temporarily resolved the problem because it reduced the size of all the processes and gave GTopLimit no reason to fire.
And we assume it has appeared over the past few weeks because we have increased the size of the Apache process either through new developments which included something that wasn't before.
All tests so far point to this being the issue, so fingers crossed it is!
I'm writing an application that can be started either as a standard WinForms app or in unattended mode from the command-line. The application was built using the VS 2k5 standard WinForms template.
When the application is executed from the command-line, I want it to output information that can be captured by the script executing the application. When I do this directly from Console.WriteLine(), the output does not appear, although it can be captured by piping to a file.
On the other hand, I can force the application to pop up a second console by doing a P/Invoke on AllocConsole() from kernel32. This is not what I want, though. I want the output to appear in the same window the application was called from.
This is the salient code that allows me to pop up a console from the command line:
<STAThread()> Public Shared Sub Main()
If My.Application.CommandLineArgs.Count = 0 Then
Dim frm As New ISECMMParamUtilForm()
frm.ShowDialog()
Else
Try
ConsoleControl.AllocConsole()
Dim exMan As New UnattendedExecutionManager(ConvertArgs())
IsInConsoleMode = True
OutputMessage("Application started.")
If Not exMan.SetSettings() Then
OutputMessage("Execution failed.")
End If
Catch ex As Exception
Console.WriteLine(ex.ToString())
Finally
ConsoleControl.FreeConsole()
End Try
End If
End Sub
Public Shared Sub OutputMessage(ByVal msg As String, Optional ByVal isError As Boolean = False)
Trace.WriteLine(msg)
If IsInConsoleMode Then
Console.WriteLine(msg)
End If
If isError Then
EventLog.WriteEntry("ISE CMM Param Util", msg, EventLogEntryType.Error)
Else
EventLog.WriteEntry("ISE CMM Param Util", msg, EventLogEntryType.Information)
End If
End Sub
Raymond Chen recently posted (a month after the question was posted here on SO) a short article about this:
How do I write a program that can be run either as a console or a GUI application?
You can't, but you can try to fake it.
Each PE application contains a field
in its header that specifies which
subsystem it was designed to run
under. You can say
IMAGE_SUBSYSTEM_WINDOWS_GUI to mark
yourself as a Windows GUI application,
or you can say
IMAGE_SUBSYSTEM_WINDOWS_CUI to say
that you are a console application. If
you are GUI application, then the
program will run without a console.
The subsystem determines how the
kernel prepares the execution
environment for the program. If the
program is marked as running in the
console subsystem, then the kernel
will connect the program's console to
the console of its parent, creating a
new console if the parent doesn't have
a console. (This is an incomplete
description, but the details aren't
relevant to the discussion.) On the
other hand, if the program is marked
as running as a GUI application, then
the kernel will run the program
without any console at all.
In that article he points to another by Junfeng Zhang that discusses how a couple of programs (Visual Studio and ildasm) implement this behavior:
How to make an application as both GUI and Console application?
In VisualStudio case, there are actually two binaries: devenv.com and devenv.exe. Devenv.com is a Console app. Devenv.exe is a GUI app. When you type devenv, because of the Win32 probing rule, devenv.com is executed. If there is no input, devenv.com launches devenv.exe, and exits itself. If there are inputs, devenv.com handles them as normal Console app.
In ildasm case, there is only one binary: ildasm.exe. It is first compiled as a GUI application. Later editbin.exe is used to mark it as console subsystem. In its main method it determines if it needs to be run as console mode or GUI mode. If need to run as GUI mode, it relaunches itself as a GUI app.
In the comments to Raymond Chen's article, laonianren has this to add to Junfeng Zhang's brief description of how Visual Studio works:
devenv.com is a general purpose console-mode stub application. When it runs it creates three pipes to redirect the console's stdin, stdout and stderr. It then finds its own name (usually devenv.com), replaces the ".com" with ".exe" and launches the new app (i.e. devenv.exe) using the read end of the stdin pipe and the write ends of the stdout and stderr pipes as the standard handles. Then it just sits and waits for devenv.exe to exit and copies data between the console and the pipes.
Thus even though devenv.exe is a gui app it can read and write the "parent" console using its standard handles.
And you could use devenv.com yourself for myapp.exe by renaming it to myapp.com. But you can't in practise because it belongs to MS.
Update 1:
As said in Michael Burr answer, Raymond Chen recently posted a short article about this. I am happy to see that my guess was not totally wrong.
Update 0:
Disclaimer: This "answer" is mostly speculation. I post it only because enough time has passed to establish that not many people have the answer to what look like a fundamental question.
I think that the "decision" if the application is gui or console is made at compile time and not at runtime. So if you compile your application as gui application, even if you don't display the gui, its still a gui application and doesn't have console. If you choose to compile it as console application then at minimum you will have a console windows flashing before moving to gui "mode". And I don't know if it is possible in managed code.
The problem is fundamental, I think, Because a console application has to take "control" of the calling console application. And it has to do so before the code of the child application is running.
If you want to check if your app is started from the command line in .NET, you can use Console.GetCursorPosition().
The reason that this works is that when you start it from the command line, the cursor moves away from the initial point ((0, 0)) because you typed something in the terminal (the name of the app).
You can do this with an equality check (code in C#):
class Program
{
public static void Main
{
if (Console.GetCursorPosition() == (0, 0))
{
//something GUI
}
else
{
//something not GUI
}
}
}
Note: You must set the output type to Console Application as other output types will make Console.GetCursorPosition() throw an exception.