Is there a way to make portable file creation?

Question

I'm trying to make a C++ function that creates a file called log.txt in the same directory as the binary and outputs data.

When I use ofstream it creates the file where the source code is.

If I move the binary to a different directory it doesn't create any file.

Code:

ofstream logFile("log.txt");
void print(string msg)
{
    if (USING_DEBUG == true) {
        cout << "Log: " << msg;
    }
    logFile << "Log: " << msg;
}

Answer 1

You can parse argv[0] get the directory of the executable relative to the current working directory at startup. This should be reasonable for what you want. Then append "log.txt" and attempt to open the file at that path.

So if argv[0] passed in simply foo.exe or foo without a path qualifier, then you'd open a log.txt file without an explicit path. If argv[0] is something like ../project/bin, then you'd open a log file at ../project/log.txt. If it's something like c:\users\selbie\foo.exe, then you'd open c:\users\jselbie\log.txt

Some code below that makes an attempt at this.

bool openLogFile(const std::string& argv0, std::ofstream& logFile)
{
#if WIN32 || _WIN32
    const char delimiter = '\\';
#else
    const char delimiter = '/';
#endif

    std::string directory;
    std::string logFilePath;

    // get the directory of argv[0], if any
    size_t index = argv0.find_last_of(delimiter);
    if (index != std::string::npos)
    {
        directory = argv0.substr(0, index+1); // +1 to include the delimiter
    }

    logFilePath = directory + "log.txt";

    std::cout << "opening log file at: " << logFilePath << " ";

    logFile.open(logFilePath.c_str());
    bool success = logFile.is_open();

    std::cout << (success ? "succeeded" : "failed") << "\n";

    if (success)
    {
        logFile << "First log file line\n";
    }

    return success;
}

int main(int argc, const char** argv)
{
    std::ofstream logFile;
    openLogFile(argv[0], logFile);
    return 0;
}