CODEWITHSUNDEEP
c++windowsfileassociations
Roberto
Roberto

Reputation: 1097

Making a C++ application "Opened with..."

I'm working on a (C++) program that more or less revolves around renaming files. I would like to make it so that you can select a file, right-mouse click it and select "Open With" and then select my application. I got the context menu part figured out, I just don't know how to do the C++ part.

In other words, how to make a program (in C++) that can be opened together with a file (so by context menu or directly opening it) and process this file?

Example: In my Windows, I associate the ".roberto" extension with "C:\Program Files\MyProgram\MyProgram.exe". So if I open a ".roberto" file, a command prompt pops up, only displaying the name of the selected file.

I hope this is clear, I am not sure how to explain this. I also had some trouble with searching on this question, so please forgive me if this has been asked before. Thanks.

Upvotes: 0

Views: 193

Answers (2)

Roberto
Roberto

Reputation: 1097

I figured it out!

Using the arguments given to main was the clue. The following program prints one line if opened directly, this line is the path of the program itself, and if opened with the 'Open with...' options it also shows the selected file.

#include "stdafx.h"
#include <iostream>
using namespace std;


int main(int argc, char* argv[])
{
    cout << "Argument count: " << argc << endl << endl;

    for (int i = 0; i < argc; i++)
    {
        cout << argv[i] << endl;
    }

    cout << endl << endl << endl << endl;

    system("pause");

    return 0;
}

Upvotes: 0

Andrew Komiagin
Andrew Komiagin

Reputation: 6556

On Windows platform in MFC-based application this is done automatically by framework in InitInstance() method of your application class:

EnableShellOpen();
RegisterShellFileTypes(TRUE);

IMPORTANT: In general this functionality is framework dependent and OS speicific.

Upvotes: 1

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