How to pass parameter in stdin

[gundalav]

Hi all,

Currently this program of mine hard code
the input data (i.e. mydata.txt) inside it.

How can I modify it so that I can pass parameter for input file
when executing the program in command line?

#include <iostream>
#include <fstream>
#include <string>
using namespace std;

int main () {
    string line;
    ifstream myfile ("mydata.txt");
    if (myfile.is_open())
    {
        while (! myfile.eof() )
        {
            getline (myfile,line);
            cout << line << endl;
        }
        myfile.close();
    }

    else cout << "Unable to open file";

    return 0;
}

I am thinking like this command line

$ ./mycode  mydata2.txt

[Zhuge]

I think you make main() take two parameters; int argc and char** argv.

int main(int argc, char** argv) {

argv is an array of C-style strings of the command line parameters, with element 0 being the name used to call the executable. argv is the number of parameters.

[gundalav]

Hi zhude,


Thanks, but then how can I replace "mydata.txt"
in this line?

ifstream myfile ("mydata.txt");

I tried

ifstream myfile (argv);

but fail.

[chris]

int main(int argc, char** argv)

In this code, argc is how many parameters are passed to your program, argv contains them. You treat argv as an array:

argv[0] - should contain the command used to execute the program (ie mycode.exe)
argv[1] - will be your first parameter (so hopeflly "mydata2.txt")

Then you need to do something like:

if(argc==2) // if there is a filename passed to it (only 1 extra parameter)
{
// all code goes here
ifstream myfile (argv[1]); // this is opening the file in the parameter
}
else // if 1 extra parameter was not passed
{
cout << "No file specified"; // tell the user they did not enter a file
}

Hope that helps.

[gundalav]

thanks chris.