printf printing twice?

I have a short main that interacts with the user, grabbing commands to call on files passed as arguments to the program. The program queries the user for a command, gets the command, and then processes. If the user types 'n' the program goes to the next file. On the second file, and I'm assuming the following files, the query printf prints twice. Anyone have an idea why that happened?

2nd EDIT: Upon further investigation i've found that it is carrying out the command related the the user-entered char, and then for some reason it loops again and calls execute_option with a blank, perhaps the \n char, which calls the default case. Any idea how to avoid this? I've used a for(;; ) with a break or exit before with no trouble, but this while loop structure seems to be posing an issue.

for example with "a.out wut huh" (n is next file, q is quit program).

Code:

---

for the file ``wut'': please enter a command command:
n
for the file ``huh'': please enter a command command:
for the file ``huh'': please enter a command command:
q

---

Code below: I've chopped out includes. They are all fine. EDIT: I realize the bottom of main got chopped off, ignore that. The loops are the important parts.

Code:

---

int execute_choice(char c, char *filename)
{
  switch(c)
  {
    case('c'): list_file(filename); return 1;
    case('d'): copy_file(filename); return 1;
    case('r'): rename_file(filename); return 1;
    case('u'): remove_file(filename); return 1;
    case('t'): truncate_file(filename); return 1;
    case('a'): append_file(filename); return 1;
    case('l'): tail_file(filename); return 1;
    case('m'): change_permissions(filename); return 1;
    case('x'): change_access(filename); return 1;
    case('n'): return 0;
    case('q'): exit(0);
    default: return 1;
  };
}

int main(int argc, char *argv[])
{
  if(!(argc>1))
  {
    fprintf(stderr, "Usage: wash <file1> <file2>...\n");
    exit(-1);
  }
  int go_next = 1;
  for(argv++; *argv != NULL; )
  {
    char command;
    while(go_next == 1)
    {
      printf("for the file ``%s'': please enter a command command:\n", *argv);
      command = getchar();
      go_next = execute_choice(command, *argv);
    }
    argv++;
    go_next = 1;
  }

---

Thanks folks.

Welcome to the most common problem with scanf() and getchar(). They leave the newline char, still in the input buffer.

Add this:
(void) getchar();

right after the getchar that you now have, and that will end the problem. That will remove one '\n' char from the input stream.

Ah. Is there anyway to avoid that? Any other function that avoids the return on the buff?

Code:

---

char buff[BUFSIZ];
if ( fgets( buff, sizeof buff, stdin ) != NULL ) {
  // parse buff to see what was input
}

---

If you use fgets for ALL input, then you always know what state the input stream is in, and you can have a meaningful go at parsing the buffer.

scanf(), which attempts input and conversion all at the same time, is a PITA to get right when you try to implement full error checking and recovery.

The key is, if you have a consistent approach which always works, then you can just do it, and start worring about something specific to your actual problem. As opposed to worrying about which new scanf trap you're going to fall into.

Oh and you don't need that (void) in front of getchar().
That's only for a certain anchient, outdated, dead, and irrelevant compiler, that a select few quirky people still drag along with them, whilst it kicks and screams "please let me die in peace!".

Truth be known, scanf() really stinks for user input. Every book uses it, so the teachers do as well, and it's fine for formatted input, but not good for user's input.

fgets(), when you're ready, is definitely better for user input.

And Malcolm, my Turbo C screams "please, don't let me die on the scrap heap!". :D :D