[SOLVED] Create/Dump file using the relative path inside code with CMAKE project in c++

Issue

I have a project structure like this -

|Project
|--data
|  |--assets
|     |--file1.txt
|     |--file2.txt  
|--Module
|  |--src
|     |--dumpFileToAssets.cpp
|--build
|  |--out.exe

The binary is written into build/out.exe. So from my understanding the relative path is wrt the binary being generated.

Code snippet from dumpFileToAssets.cpp

Full path works -

std::string path = "/Users/Username/Project/data/assets/file1.txt"
ofstream myfile;
myfile.open (path);  
myfile << "Writing this to a file.\n";
myfile.close();

Relative path from the build directory does not work -

meaning file1.txt is not generated in the .open() call and hence not written -

std::string path = "./../data/assets/file1.txt"
ofstream myfile;
myfile.open (path);  
myfile << "Writing this to a file.\n";
myfile.close();

Question - Is my understanding that any relative path inside the code is wrt to the binary path where the binary is generated correct ?

Or can I access the Project root path inside code to achieve this ??

PS : Here are my findings -

1. When I run the binary from terminal, it works.
2. When I run an Xcode version of the project, the out.exe path is different, so it does not work, and I have to change the path wrt to the new out.exe path

Solution

Your understanding of the working directory is incorrect. There are multiple ways to run your application and each time the working directory is different

cd /Users/Username/Project/
./build/out.exe

The working directory is /Users/Username/Project/

cd build
./out.exe

The working directory is /Users/Username/Project/build

mkdir sub
cd sub
../out.exe

The working directory is /Users/Username/Project/build/sub

On Windows, the working directory may even be a different drive:

D:
cd \test
C:\Users\Username\Project\build\out.exe

The working directory is D:\test

If you want to open a file relative to the executable, then determine the path of the executable first, which is not trivial. Note that argv[0] may be a symbolic link and not the executable itself (like less is a link to more).

Answered By - Thomas Weller

Answer Checked By - Mary Flores (WPSolving Volunteer)