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with all the audio software (wave, mp3, etc) that run on Windows system this should be handled, as it is the proper way to do so (under Linux). But this is a little peculiar. You should read this guy for more info on that: -i.com/browser/-zhpb4.jsp.
I think I can understand how to get the file name for a file stored in the Windows operating system, but how can I get the file name for a sound file stored on my hard drive? I can’t seem to find the answer through the web searches. -P.S. I’m sorry if this is a duplicate post, but I’m kind of having a hard time finding what I’m looking for in the listings.

How do you know the file name of the soundfile that you are playing? The files in the Windows7 sound folder have to be named 00.wav for example, so the same thing happens with sound files. Using FFFFFF would get you the file’s caption (anything in between the dots) and skipping that value would get you the file name.
You can actually get the filename of the currently playing sound file using the APIs in the WIN32_FIND_DATA class (which is what Windows uses for searching through the file system).

The file system listings are not stored in the WAV file format.
It’s currently not possible to get the file name of a sound file (or the caption of a picture, or any other file) in the sound files container.

This invention relates to a method and system for software installation.
Programmers must typically debug software that is malfunctioning. In order to perform the debugging operation the programmer needs to install a debug version of a program and run the program to its normal termination point. For example, a standard debug process would be to execute a program several times, step through the program using a debugger, and perform one or more system functions, e.g., memory copy or print a string or message, to test the program. In this process, if the program crashes during debugging, the system must be set up to pause execution of the program and trap a debugger process. For example, interrupt INT 21H, otherwise known as xe2x80x9c0x2AHxe2

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You just need to add the flag -z to your command line, or you can place the –unpack to the start of the script and you will probably not need the -z.
Make a script with the –unpack in it. And because you want to extract it with just a single command, you will run
unzip -q -z -f My_file

But i have no idea why it asks me for a password. Im not using the fields supplied in the script. Also I am running bash 4.4
Why does this command ask me for a password?

A:

It does because -q and -z are only used in combination with a password: unzip -q -z my_file. You should just use the -d option:
unzip -d my_files unzip -d -q -z

m_bIsCancelled = false;
m_bIsRetry = false;
m_bIsSuspend = false;
m_bIsDbgViewOnly = false;
m_pResourcePar = NULL;
m_pUserCancel = NULL;
m_pszUserCancel = NULL;
m_pUserAbort = NULL;
m_pszUserAbort = NULL;
m_pszThreadName = NULL;
}
else if( IsClipboardClosed() )
{
m_pszText = AfxGetApp()->m_pszText;
m_pszValue = AfxGetApp()->m_pszValue;
m_pszResID = AfxGetApp()->m_pszResID;
m_bIsText = true;
m_bIsValue = true;
m_bIsResID = true;
}
else
m_pszText = AfxGetApp()->m_pszText;
}

ASSERT( m_pResourcePar!= NULL );
ASSERT( m_pUserCancel!= NULL );
ASSERT( m_pUserAbort

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