using command line to open file

Question

I am new to command line, i have a problem. i have searched to solve the problem but none of them work . I want to use command line to open file but i cant open it and i don't know why. I checked if i created my file yet.I did change the working directory, but it still cant find my file.

My code

#include<iostream>
#include<string>
using namespace std;
void main(int i ,char *a[])
{
    if (i != 1)
    {
        cout << "Wrong!!!";
    }
    fstream fp;
    fp.open(a[1]);
    if (!fp.is_open())
    {
        cout << "Cant open file";
    }

}

What did i do wrong??

Stick with the convention: argc and argv. You do konw the exe name counts, so that is arg 1. You say anything other than 1 arg is wrong but then open the 2nd arg anyway... — John3136
– John3136, Commented Apr 6, 2017 at 5:15
@radbrawler i use command line to run it first i g to disk D: then D: cd>New folder\ConsoleAp8\Debug then ConsoleAp8.exe test1.txt that's it. — Van
– Van, Commented Apr 6, 2017 at 5:17
@John3136 i thought the exe name count as argv[0] ?, argc just count the statement ,so the file name is first statement ?? — Van
– Van, Commented Apr 6, 2017 at 5:19
@Van. Right, but an array with 1 member (argc == 1) only has position 1. You say if (argc != 1) { print wrong } that implies you expect no args (just the exe name) — John3136
– John3136, Commented Apr 6, 2017 at 5:27

radbrawler · Accepted Answer · 2017-04-06 15:35:52Z

1

Well I changed the program a bit and it worked. Mainly it was argument length

#include<iostream>
#include<string>
#include<fstream>
using namespace std;
int main(int i ,char *a[])
{
    if (i != 2)
    {
        cout << "Wrong!!! "<<i;
    }
    fstream fp;
    fp.open(a[1]);
    if (!fp.is_open())
    {
        cout << "Cant open file";
    }
    return 0;
}

That worked for me.

edit- I ran it on ubuntu. I didn't had windows.

edited Apr 6, 2017 at 15:35

answered Apr 6, 2017 at 6:18

radbrawler

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