Question

I posted a similar code earlier, but I think this is a different issue now. I just can’t figure out why my run code won’t go past “infile open”. (“-e” prints out, “-d” doesn’t) I’m trying to open my file, use a command line option to determine if I will print out a certain multiple of characters.

For example, a.out -d 2 < sample.txt will print out every second letter.

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

   if (infile.good())
      printf("infile open \n");

   int c;    
   int number = 0;
   int count = 0; 


   string str1 = argv[1];
   string str2 = "-d";
   string str3 = "-e";


   if (str1.compare(str2) == 0)
   { 
      printf("entered -d\n");
      c = infile.get();       

         while(!infile.eof()) {

             number = atoi(argv[2]);  

              if (count == number)
            {
              cout.put(c);
                      count = 0;
                }
                  else
                      count++;

             c = infile.get();         

}//end while 

}//end if

           if (str1.compare(str3) == 0)       
                printf("entered -e\n");


}//end main

Answer 1

infile is never opened:

ifstream infile; // Does not open a file as no file name is supplied.

Either use cin or pass "sample.txt" as another command-line argument and open it:

ifstream inFile(argv[3]);
if (inFile.is_open())
{
}

Other points:

• Use std::cout instead of mixing printf() and std::cout.
• atoi() returns 0 if the argument is invalid or if the argument is a valid 0. See strtol() for an alternative.
• There is no reason to convert argv[2] on every iteration of the while. Just do it once prior the while.
• Always check argc before accessing the elements of argv, to avoid invalid memory access.
• std::string instances can be compared using operator==.