How to print out dash or dot using fprintf/printf?

As of now I'm using below line to print with out dot's

fprintf( stdout, "%-40s[%d]", tag, data);

I'm expecting the output would be something like following,

Number of cards..................................[500]
Fixed prize amount [in whole dollars]............[10]
Is this a high winner prize?.....................[yes]

How to print out dash or dot using fprintf/printf?

From stackoverflow

• You are going to have to output the string with the dot or dash padding yourself.

  Something like (forgive my C, it's rusty):

  printAmount(char *txt, int amt) {
      printf("%s",txt);
      for(int xa=strlen(txt); xa<40; xa++) { putc('.'); }
      printf("[%d]",amt);
      printf("\n");
      }
  

• You can't do it in one statement. You can use sprintf, then substitute dots for spaces yourself, or do something like

  int chars_so_far;
  char padder[40+1]= '..........'; //assume this is 40 dots.
  printf("%.40s%n",tag,&chars_so_far);
  printf("%s[%d]",padder+chars_so_far,data);
  

  ---

  Edit: Simplified my example above based on @Ates' padder concept. This way doesn't require any 'leaps of faith', about whether the tag string is too big or too small - it always starts the data in column 41.

  Ates Goral : I think it can be done in one statement.

  Christoph : @Ates: it could, but imo not without two calls to `strlen()` if you want to properly check boundaries - therefore, you need to cache this value -> two statements!

• I'd suggest writing a function that pads a string with X characters and use that to generate the first argument to your printf string. Something like:

  char s[40];
  pad_str(tag, s, 40, '.');
  fprintf( stdout, "%-40s[%d]", s, data);
  

  Note that the third line of your sample data would need this format:

  "%-40s[%s]"
  

• A faster approach:

  If the maximum amount of padding that you'll ever need is known in advance (which is normally the case when you're formatting a fixed-width table like the one you have), you can use a static "padder" string and just grab a chunk out of it. This will be faster than calling printf or cout in a loop.

  static const char padder[] = "......................"; // Many chars
  
  size_t title_len = strlen(title);
  size_t pad_amount = sizeof(padder) - 1 - title_len;
  
  printf(title); // Output title
  
  if (pad_amount > 0) {
      printf(padder + title_len); // Chop!
  }
  
  printf("[%d]", data);
  

  You could even do it in one statement, with some leap of faith:

  printf("%s%s[%d]", title, padder + strlen(title), data);
  

  Christoph : your last suggestion breaks if `strlen(title) > strlen(padder)`

  Ates Goral : @Christoph, hence the proclaimed "leap of faith" :)

  Christoph : @Ates: interesting concept, this 'faithful programming' of yours ;)

• Another solution using a tiny helper function

  static inline size_t min(size_t a, size_t b)
  {
      return a < b ? a : b;
  }
  

  Then, you can do the following:

  char padder[] = "........................................";
  int len = min(strlen(tag), sizeof(padder) - 1);
  printf("%.*s%s[%d]", len, tag, padder + len, data);
  

  This is essentially what Ates posted, but I actually figured this out on my own ;)

  Jonathan Leffler : Change the size_t to int; you must pass an int to the '*' in the printf() family of functions.

  Ates Goral : Or make the helper more helpful: printf("%s%s[%d]", title, getPadding(title), data);

  Christoph : @Jonathan: thanks, corrected it...

• You can easily do this with iostreams instead of printf

  cout << setw(40) << setfill('.') << left << tag[i] << '[' << data[i] << ']' << endl;
  

  Or if you really need to use fprintf (say, you are passed a FILE* to write to)

  strstream str;
  str << setw(40) << setfill('.') << left << tag[i] << '[' << data[i] << ']' << endl;
  printf(%s", str.str());
  

  Nick Presta : +1 for being easy to read and understand and not breaking under certain (albeit somewhat uncommon situations).

• I think there's a better way.

  #include <string.h>
  #include <stdio.h>
  
  #define MIN(A,B) ((A)<(B)?(A):(B))
  char *dashpad(char *buf, int len, const char *instr) {
      memset(buf, '-', len);
      buf[len] = 0;
      int inlen = strlen(instr);
      memcpy(buf, instr, MIN(len, inlen));
      return buf;
  }
  main() {
      char buf[40];
      printf("%s\n", dashpad(buf, 40, "Hello world, dash padded "));
  }
  

  Ates Goral : First filling the entire buffer with the pad character, and then overwriting them seems a bit wasteful though...

  Software Monkey : Also, you have a buffer overrun (don't you love C) - s/be: buf[len-1]=0;

  Software Monkey : And you have a null-erasure bug in you min(), which should also be len-1.

  Thi : is it possible to call a function inside printf?

  Software Monkey : @Thi: Yes this is legitimate - the function is invoked and it's return passed as a parameter to printf... this is normal coding in any procedural, OOP or functional language.