从命令行输入文件

Question

I'm here to trying to take input from the file. If user run this .exe from cmd and give a filename like example.exe input.txt then it shows the file and read. But if user doesn't give the file name then it run as simply program's run.

Program is running well when I give the input from cmd during run time it run perfectly, but if I don't give the filename during running this file and run simply example.exe an exception show me the error

exception: invalid null pointer

my code is here:

// inputfile.cpp : Defines the entry point for the console application.

#include "stdafx.h"
#include<iostream>
#include<conio.h>
#include<string>
#include<fstream>

using namespace std;

void main(int argc, char* argv[])
{
    try {

        if (argc > 0)
        {
            string filename = argv[1];
            ifstream in(filename);
            in.open(filename);
            if (in.is_open())
            {
                cout << "file opened, do something with file";
            }
            else
            {
                cout << endl << "You have Entered Wrong File Name Or File Not Exist in Project's Library" << endl;

            }
        }

    }
    catch (exception e)
    {

    }

    cout << endl << "do with the simple program";

    _getch();
}

Answer 1

The logic error is in the line

  if (argc > 0)

It needs to be

  if (argc > 1)

argv[1] is NULL if the program is invoked without arguments.

argc is at least 1, the first argument being the name of the program. When the program is invoked with one argument, argc is 2 and argv[1] is the first argument.

Answer 2

When there is only one argument(always the exec file itself, eg using in the way ./exec_file or just double click the exec file), the argv[1] would throw an exception.

Here are some tips:

1. Try to learn how to debug the code(IDE like Visual Studio, Ecplise or gdb).
2. VC++ is not standard c++, which means it only runs on Windows, not cross-platform. You'd better learn to use g++ compiler(linux).