File input without specified file name?

[Puritan2]

If the user is going to run my executable but specifies differing filenames each time it is run:

e.g., MyProg file1
e.g., MyProg file2

where file1, file2 are text files, how do I handle this? I can handle it using .open("filename") but I'm not sure how to do it otherwise...

Thanks-you.

 

[Salem]

int main ( int argc, char *argv[] ) {
    // check argc to see how many parameters were specified
    // first parameter is argv[1]
    // so perhaps
    open( argv[1] );
}

--

 

[Puritan2]

#include <cstdlib>
#include <iostream>
#include <fstream>

using namespace std;

int main(int argc, char *argv[])
{
    ifstream in;
    
    in.open(argv[1]);
}

//This worked.  Thanks!

 

[jstreich]

You want to error check to make sure the correct number of arguments are passed, else the program will exit horribly if not passed a parameter.

In general argv is a array of char pointers or a char double pointer (both are the same as far as the machine code is concerned). argv[0] is the program name, argc is the size of argv (so, no parameters should argc == 1, with program name at an index of 0; with one parameter argc == 2, with argument index at 1... and so on).