Why does the starting directory matter when launching a OpenGL executable?

Question

I am doing a openGL porject and readering my shaders from .vert and .frag glsl files. I am using CMake with an extension for auto ninja file generation as well. This is my readfile code:

std::string readFile(const char *filePath) {
    std::string content;
    std::ifstream fileStream(filePath, std::ios::in);

    if(!fileStream.is_open()) {
        std::cerr << "Could not read file " << filePath << ". File does not exist." << std::endl;
        return "";
    }

    std::string line = "";
    while(!fileStream.eof()) {
        std::getline(fileStream, line);
        content.append(line + "\n");
    }

    fileStream.close();
    return content;
}

I spent a long time thinking I was crazy because from my build dir I was doing ./app/NAME_OF_APP.exe but the shaders were not being applied just black. It was only when I clicked on it from within the file explorer that it worked so I cd'ed into build/app/ and called ./NAME_OF_APP.exe and boom it worked. Why is this behavior happening? does the starting dir effect the relative pathing of the actual application? seems odd to me. thanks.

Answer 1

If you run a program from a command line the current directory is whatever directory is current in that command prompt. If you run a program from explorer the current directory is the directory that houses the executable.

You could use a function like GetModuleFileName (with NULL as the hModule parameter) to get the path to the executable and then chop off at the last \ and then change to that directory and then the program would not depend on the directory the program is run from.