Thread: More Unix Woes

I am writing a program that counts the number of words in a text file.
How would I go about useing a command line argument as the text file name that will be opened? Right now i am opening a file in the same directory as the program.(test.txt file)

for example: wordcount somefile.txt

would open "somefile" and count the words in it.

Code:

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

using namespace std;

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

int wordcount = 0;
int letters = 0;
double average = 0;
char ch;
	ifstream infile;

	infile.open("test.txt");
	if (infile.fail( )) 
	{
	cout << "Input file opening failed.\n";
	system("pause");
	exit(1);
	}
}

Instead of "test.txt" you use argv[1]. But of course you should first check that there is a command line argument, so

Code:

if (argc > 1)
{
	infile.open(argv[1]);
}
else
{
	tell the user s/he should give a file name
}