Command Line problems

[jiahwa008]

Hello C Board,
I'm a first time caller I reading thru C++ Without Fear-Overland. Okay, here's the example exercise I'm having a problem with:

//the project name is readtxt2 and the text file is name output.txt 
#include <stdafx.h>
#include <iostream>
#include <fstream>
#include <string.h>
using namespace std;

int main(int argc, char *argv[]) {
    int c;   // input character
    int i;   // loop counter
    char filename[81];
    char input_line[81];

    if (argc > 1)
        strncpy(filename, argv[1], 80);
    else {
        cout << "Enter a file name and press ENTER: ";
        cin.getline(filename, 80);
    }

    ifstream file_in(filename);

    if (! file_in) {
        cout << "File " << filename << " could not be opened.";
        return -1;
    }

    while (1) {
        for (i = 1; i <= 24 && ! file_in.eof(); i++) {
            file_in.getline(input_line, 80);
            cout << input_line;
        }
        if (file_in.eof())
            break;
        cout << endl << "More? (Press 'Q' and ENTER to quit.)";
        cin.getline(input_line, 80);
        c = input_line[0];
        if (c == 'Q' || c == 'q')
            break;
    } 
    return 0;
}

Okay, that is the sample answer, i didn't change it.

I am told I can enter in the command line this:
readtxt2 output.txt
and it should open output.txt and display for me, but when I run it and enter:
readtxt2 output.txt
I am told file not cannot open, now when I just enter:
output
it works fine. why is that?
btw I'm using microsoft visual studio 2010 on windows vista.

I know this is a beginner question, so please bear with me.
I can use all the help I can get, since I am trying to learn by homeschool, thanks.

[jimblumberg]

I would change strncpy() to strcpy(), the padding that may happen with strncpy() may cause problems. And in your if statement print out the value of argv[1] to insure the program is receiving what you entered.


Jim

[Elysia]

I would change

    char filename[81];
    char input_line[81];

to

    std::string filename;
    std::string input_line;

strncpy(filename, argv[1], 80);

to

filename = argv[1];

        cin.getline(filename, 80);

to

        std::getline(std::cin, filename);

ifstream file_in(filename);

to

ifstream file_in(filename.c_str());

(Requirement.)

I'll leave the rest as an exercise.
After you made the changes, test again, and see if it works better this time.

[jiahwa008]

Thank for replying guys.  I've used both styles and they both work fine.  I may not have expressed my question clearly.  I am wondering why the book asks me to enter 2 commands to open output.txt.  (it works fine when entering just filename).
I am instructed to enter "readtxt2" "output.txt" into the command-line when promted.
When I enter just "output"  the programs runs fine, but when I enter 
"readtxt2" "output" the file is not found.

I've saved a txt file named output in the project folder readtxt2.  I'm reading C++ without fear by overland, I am using MS Visual Studio 2010, on a MS Windows Vista OS.

So I guess what I'm asking is why will it not recognize the "readtxt2" or even the ".txt"?  Is that just the book lingo and I am not supposed to use it?  (even thought it clearly states I enter "readtxt2 output.txt" into the command-line.)