How to get the current directory and the file name

**megareix** · 01-07-2009

Hi, I'm looking for a code that gets the directory where the file is being executed (for example: C:\some\directory) and the name of the file that is being executed.(for example: someprogram.exe)

Thanks in advance.

**C_ntua** · 01-07-2009

In windows there is GetCurrentDirectory
In linux maybe this

For the program name you can do int main(int argc, char** argv) and read argv[0]. Which is a char array that contains the name of the executable. But NOT always. But you can try it. I believe it is true at least for windows and some linux OS.

**Codeplug** · 01-07-2009

GetModuleFileName() will return both.

gg

**cpjust** · 01-07-2009

Originally Posted by C_ntua

In windows there is GetCurrentDirectory
In linux maybe this

For the program name you can do int main(int argc, char** argv) and read argv[0]. Which is a char array that contains the name of the executable. But NOT always. But you can try it. I believe it is true at least for windows and some linux OS.

Windows has getcwd() also, but Microsoft likes to annoy people by putting a '_' infront of POSIX function names.

**Elkvis** · 01-07-2009

Originally Posted by C_ntua

In windows there is GetCurrentDirectory
In linux maybe this

For the program name you can do int main(int argc, char** argv) and read argv[0]. Which is a char array that contains the name of the executable. But NOT always. But you can try it. I believe it is true at least for windows and some linux OS.

another option on linux (this example assumes C++, but it can easily be done in C as well):

Code:

#include <string>
#include <cstdlib>

int main(void)
{
  std::string mycwd(getenv("PWD"));
  std::cout << "Current Directory: " << mycwd << std::endl;
  return 0;
}

**matsp** · 01-08-2009

Originally Posted by Elkvis

another option on linux (this example assumes C++, but it can easily be done in C as well):

Code:

#include <string>
#include <cstdlib>

int main(void)
{
  std::string mycwd(getenv("PWD"));
  std::cout << "Current Directory: " << mycwd << std::endl;
  return 0;
}

That assumes that the environment variable PWD is actually in sync with the current directory.

Note also that an application can be started in a different directory than the application actually lives in, so if you want to know what directory the application actually lives in, it's can be difficult. In a well-behaved system, argv[0] contains the application name and location, but there's nothing preventing someone from writing code like this (in Linux/Unix):

Code:

   execl("/somepath/someapp", "rubbish", "arg1", NULL);

argv[0] will then be "rubbish", but the application is "someapp" and loaded from "/somepath".

--
Mats

**megareix** · 01-08-2009

Thanks for all of your answers. One thing I forgot to mention is that I'm using C, not C++.
The code in fact was very simple, and is made, as you said, using GetModuleFileName. Here is the code:

Code:

#include <stdio.h>

#include <stdlib.h>

#include <windows.h>



int main()

{

    char path[100];



    GetModuleFileName(NULL, path, sizeof(path));

    printf("%s\n", path);

    system("pause");

    return 0;

}

Thanks for all of your help =D

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