Why does argc returns 2 when only 1 argument was passed?

Question

I wrote this simple code to understand how the argument system works. I dragged a textfile to the .exe file, and get 2 as output instead of 1 as i expected. Why 2? Is Arg 1 the .exe itself? How can i find out the filenames of the arguments?

#include<iostream>
using namespace std;

int main(int argc, char* argv[])
{
    int i = 0;
    const int max = 100000;

    if (argc > 0)
    {
        cout<< argc <<endl;
    }

    cin.get(); cin.get();

  return 0;
}

And an additional question. Where can i inform on how to access each argument and use the informations. My goal is to open all files passed as arguments to the .exe.

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This is not a duplicate question, i asked why 2 is returned when you pass 1 argument. The question in the Link is another...

Answer 1

argv[0] is normally the name of the program being run, and is counted in argc. If argc >= 2, the first of your text-file names should be in argv[1]. You can simply loop over them...

for (size_t i = 1; i < argc; ++i)
    if (std::ifstream infile(argv[i]))
        ...use infile...
    else
        std::cerr << "ERROR - unable to open " << argv[i] << '\n';

For more complex requirements, you may want to use getopt() et al if your system provides it, or the boost library equivalents.