I have written a c++ program and deployed it in say c:\my_app, and my executable's path is c:\my_app\my_app.exe. Say, my_app needs many files such as the_file.txt, which is located in c:\my_app\the_file.txt.
In my executable, I open the txt file as, xx.open("the_file.txt");
Moreover, I have associated my program with let's say .myp extension.
When I'm on Desktop, and want to open a file named example.myp, my program can not see the_file.txt. Because, it (somehow) assumes that it's currently working on Desktop.
Is there any easy way to handle this problem by changing shell command for open in HKEY_CLASSES_ROOT? The naive solution would be to change all file open operations with something like %my_app_location/the_file.txt". I don't want to do that.
Always use a full path name to open a file. In other words, don't open "foo.txt", open "c:\bar\foo.txt". To find the install directory of your EXE use GetModuleFileName(), passing NULL for the module handle.
These days you shouldn't add files to c:\my_app....
Instead use the ProgramData Folder and full paths.
Use SHGetSpecialFolderPathA with CSIDL_COMMON_APPDATA to get the ProgramData folder and the create your program directory and add your files.
You should set current directory for your app's folder with SetCurrentDirectory function. After that you can open file by name without full path
Related
I have my project where i am using filesystem to retrieve directory of assets.
When i am lunching my program in editor(im using Visual Studio 2019) everything is fine and this code return value of working direcotry of project.
std::string currentPath = std::filesystem::current_path().string();
But when i am lunching app from .exe file this line of code returns path that leads to .exe file.
The same directory called $TargetPath in properties in VS.
So my question is why is that happening and how can i resolve this problem.Becouse of that i cannot automatically load assets when lounching app from .exe file
Because it gives the current working directory, which is set by the environment calling your program (unless your program explicitly changes it).
So, it does what it's designed to do, gives the current working directory:
Returns the absolute path of the current working directory,
So my question is why is that happening
It happens because you've configured the editor to set the working directory to one path, while you're running the program with another working directory outside the editor.
how can i resolve this problem.Becouse of that i cannot automatically load assets when lounching app from .exe file
Here is an approach:
Store the assets in a path that is relative to the exe.
Get path to the exe.
On POSIX, you can use argv[0] from arguments of main
On Windows, the documentation recommends GetModuleFileNameW
Get canonical absolute form of that path (make sure that working directory hasn't been changed before this step if the path to exe is relative).
Get the directory that contains the exe from that canonical path.
Join that directory path with the asset's relative path to get an absolute path to the asset
Load the asset using the absolute path.
I'm trying to use fstream to create a file, however the file i'm trying to create wont appear in the .exe directory or anywhere else in the file directory. After searching in my computer for the file, I found that was created in a different directory entirely even though i'm using a relative directory.
This is the code to create the file:
ofstream file;
file.open("something.jpg", ios::out|ios::binary|ios::beg);
Directory of created file: C:\Users\user-pc
Directory of project: D:\Users\user-pc\Documents\Visual Studio 2012\Projects\recvFile
by the way, using an absolute directory works perfectly fine. Could this be a problem with the projects working directory?
In your Visual Studio right-click on your project, click Properties, then go to Configuration Properties, then Debugging. There is a row "Working Directory". You can set the working directory there. If you need to do this programmatically, you can use SetCurrentDirectory .
If you need to create the file in the same directory as the .exe location, you can use this approach: https://stackoverflow.com/a/124901/1915854
Call GetModuleFileName() using 0 as a module handle...
If the .exe is installed in a shared directory like Program Files, then creation of the file in the same directory could require additional permissions and may be a bad idea. If the .exe is just cloned to the directory where it should create files, then there is no such problem.
Try adding "../" to the link:
file.open("../something.jpg", ios::out|ios::binary|ios::beg);
File will be created in Debug/Release folder of your project. try what Timo Rzipa suggested.
I have a c++ project that I would like to send to someone in executable form. The issue is the program must read from a .txt that I created (specific deliminators). Currently my program reads from a file path that is specific to my computer,
parseFile("/Users/David/Desktop/FinalProject/Store.txt");
How could I package the .txt file and the executable file together, where the exec. reads specifically from the that .txt on anyone's machine?
Note: I am using Xcode
Change your programs to receive 'file path' as a parameter. Write a note(ReadMe) with the program to specify the file format and added a sample data file with the package
tl;dr: if you just put the text file in the same folder with your executable, you can open it with parseFile("Store.txt");
In most runtime implementations, there is a notion of a "working directory." When you open up an executable via the graphical shell (by double clicking it or something to that effect) the working directory is the same as the directory the executable is in.
Now, if you try to open a file in your program via a path that isn't fully qualified, then the path that gets used will be relative to the working directory.
A fully qualified path is a discrete path that points to a single entity in your filesystem. "/Users/David/Desktop/FinalProject/Store.txt" is one such example, as it starts at root (/ on *nix, DriveLetter:\ on Windows) and says exactly which directories you need to traverse to get to your file.
A path that is not fully qualified (which basically means that it doesn't start at the root of your filesystem) can be used to perform relative file addressing. Most runtimes will assume that any path that is not fully qualified is meant to be relative to the working directory, which basically means that the path that actually gets opened is the result of concatenating your provided path to the end of the working directory.
As an example, if you opened your binary, which is stored as /Users/David/Desktop/FinalProject/a.exe, then the working directory would be set to /Users/David/Desktop/FinalProject/. If your program then tried to open "Store.txt", the runtime would see that you're trying to open a path that isn't fully qualified, so it would assume you meant to open a file relative to the working directory, which would then be /Users/David/Desktop/FinalProject/ + Store.txt, which would be /Users/David/Desktop/FinalProject/Store.txt.
The nice thing about this is that if you move your binary, the working directory moves too. if you move a.exe along with Store.txt to /Users/David/Desktop/FinalProject(copy)/, then when you open /Users/David/Desktop/FinalProject(copy)/a.exe, the working directory will be /Users/David/Desktop/FinalProject(copy)/ now, and now when you call parseFile("Store.txt"), it will instead open up /Users/David/Desktop/FinalProject(copy)/Store.txt. This holds true when moving to other computers, too.
It's worth noting that if your binary is run from a command line utility, the working directory will often be the directory the command line shell is in, rather than the executable's directory. It is, however, a part of the C standard that the first command line parameter to main() should be the name of the executable, and most implementations supply you with the fully qualified path. With some minimal parsing, you can use that to determine what path to use as a base for addressing files.
In my program, I have a button that I want to open a text file in a relative directory. I'm using QDesktopServices like this:
QDesktopServices::openUrl(QUrl::fromLocalFile("file:///stuff/block_settings.txt"));
When the button is pressed, nothing happens.
The file is in a folder named "stuff" that resides in the same location as my .exe. It is the same directory used for all my other tasks.
What am I doing wrong?
Thanks.
The file is in a folder named "stuff" that resides in the same location as my .exe. It is the same directory used for all my other tasks. What am I doing wrong?
Seems like your full path is an overcomplication. I would suggest to use this intead:
QString QCoreApplication::applicationDirPath() [static]
Returns the directory that contains the application executable.
For example, if you have installed Qt in the C:\Qt directory, and you run the regexp example, this function will return "C:/Qt/examples/tools/regexp".
On Mac OS X this will point to the directory actually containing the executable, which may be inside of an application bundle (if the application is bundled).
Warning: On Linux, this function will try to get the path from the /proc file system. If that fails, it assumes that argv[0] contains the absolute file name of the executable. The function also assumes that the current directory has not been changed by the application.
So, you would be writing this code:
QDesktopServices::openUrl(QString("%1/stuff/block_settings.txt")
.arg(QCoreApplication::applicationDirPath()));
I fixed the issue. Changed to:
QDesktopServices::openUrl(QUrl("file:stuff\\block_settings.txt"));
Not sure how that works because I don't see that configuration on any tutorial anywhere but w/e
I have a file saved to my desktop, when I open it with my program the working directory changes to the desktop, this means my program can not load in some files it needs as it searches for these in the working directory. Is there a way I can stop the working directory from changing like this?
There's a flag you can set to avoid the current directory from changing called OFN_NOCHANGEDIR
http://msdn.microsoft.com/en-us/library/ms646839(v=vs.85).aspx
You can just save your working directory at startup and use absolute paths. In fact, it's better to always open files with absolute paths, unless you really want to rely on the current working directory.
You would be better off determining the processes location, then using it as the key for where to find the other files? There are many ways that programs can be launched, which effect the working directory.
See: The answer here for a good description of how to get the processes location and strip out the executable filename (look in the comments)
Essentially, you use:
GetModuleFileName or GetModuleFileNameEx.
and then:
PathRemoveFileSpec to remove the file name
Opening a file doesn't change your current directory. Perhaps you using the common open file dialog? Here is an article that will explain all about how that changes your current directory.
use SetCurrentDirectory to do that.
You can locate the executable by using GetModuleFileName
TCHAR szFileName[MAX_PATH];
GetModuleFileName( NULL, szFileName, MAX_PATH )
... then compute the correct directory
SetCurrentDirectory(path);