Fun with strings & chars in C

Scrag · 05-19-2004, 01:05 PM

Im programming in C.
I have an int variable called z. I have a string variable called spaces. I would like to format "spaces" with "z" number of space chars. I've tried loops and stuff but cant get it right. Any ideas?

For example, z = 8. Would like "spaces" to contain 8 spaces or ' ' chars.

Thanks!!

itsme86 · 05-19-2004, 01:20 PM

The easiest way is to use memset().

Code:

char *make_spaces(int z)
{
  static char spaces[1024]; // Or some other sufficiently large number

  memset(spaces, ' ', z);
  spaces[z] = '\0';  // Make sure you NUL-terminate the string.

  return spaces;
}

Here's an example using a loop if you prefer that:

Code:

char *make_spaces(int z)
{
  static char spaces[1024];
  int i;

  for(i = 0;i < z;++i)
    spaces[i] = ' ';
  spaces[z] = '\0';

  return spaces;
}

And another example without the need for the i counter:

Code:

char *make_spaces(int z)
{
  static char spaces[1024];

  spaces[z] = '\0';
  while(--z >= 0)
    spaces[z] = ' ';

  return spaces;
}

aluser · 05-19-2004, 02:38 PM

It would be a good idea to note that you should be careful about using these examples more than once, e.g.

Code:

char *a, *b;
a = make_spaces(3);
b = make_spaces(5);
printf("a is %d spaces long\n", strlen(a)); /* prints "a is 5 spaces long\n" */

If that is a problem, you need to either take a char* as an argument (and assume it's sufficiently long) or return a malloc()d string, leaving the caller to free it.

Scrag · 05-19-2004, 02:39 PM

Wow.....three examples

Thanks again for he help!

itsme86 · 05-19-2004, 03:03 PM

Thank you, aluser. I forgot to mention that.