int main(int argc, char * argv[])
{
    //argc has count of no of arguments on command line.
    // argv[] contains array of strings seperated by space.
    // so argv[1] will give u, second argument, with first
    // argument being ur ./a.out(in unix)/test.exe(in win)
    // so this way u can access CL args.

    // ur code.
}

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

argc is nothing but Argument Count
argv is nothing Argument Vector (or) Argument Value which 
is array of Strings.

The first argument to the Command line is always the 
Program Name. 

argc contains no. of Arguments that it passed including the 
Program Name.So Leave the 1st Argument,

int i;
for(i=1;i<=argc;i++)
cout<<argv[i];

By using -->
   int main(int argc, char * argv[])

In this statement the variable argc holds the at least one 
argument of type int , the name of the file it self first 
argument and *argv[] is array of char which takes the name 
of file.......
      If any correction plz E-mail me ...
                   Bye