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
}

to further joni's answer,

you can also use the environment variable PATH by scanning the envp argument for the program for: "PATH="
this means that you don't need to have the file to be opened in the same folder, without having to add any special code to have the data available, only to scan the 3rd argument to all executables "const char *argp[ ]"