How do you use int main(int argc, char* argv[])

[dlugo]

So, I have an idea of how to use int main(int argc, char* argv[]) but I don't know how to use it in the command line.

What I want to do is to copy a set of characters from the a file on my desktop, into a array, to then print it on the screen. Can anyone give me some pointers as to what I can do in the command line?

[Bazzy]

On the command line call

yourprogram yourfile

Some IDEs allow to pass command line arguments to your program without to type them every time if you are still working on it

[Skillless]

If you want to be a really cool guy you can also drag the yourfile.txt and drop it on yourprogram.exe

(Its easier if youre working with alot of files)

[dlugo]

Ok so maybe I'm confusing myself here. Lets say I created a file named test.txt with the following text:

ABCD
1234

And this is the code I run:

#include <iostream> 
using namespace std;

int main(int argc, char* argv[]) { 
   cout << "argc = " << argc << endl; 
   for(int i = 0; i < argc; i++) 
      cout << "argv[" << i << "] = " << argv[i] << endl; 
   return 0; 
}

when I type /test.out test.txt it outputs:

argc = 2
argv[0] = /test.out
argv[1] = test.txt
argv[2] = Segmentation fault (core dumped)

What do I need to do to make what's in the test.txt file from above?

[Kyon]

Well, argc holds the amount of arguments passed to the "console" (including the .exe call) and argv contains char*'s that are seen as strings (which you can refer to up to argc-1). When you check argv[1] for a file name, you can use that as the input for the creation of a filestream to that file. Check the tutorial section on file streams and you'll know what I mean.

[dlugo]

Ok so I see how it works. But... still my question is.... how can I use this to output what is inside the file?

[dlugo]

Thanks for the reply all. I found what I was looking for http://www.cprogramming.com/tutorial/lesson14.html