Consider the following code snippet.

printf("%g", strtof("0.0", NULL))
printf("%lg", strtod("0.0", NULL))
printf("%llg", strtold("0.0", NULL))

When executed, expecting to see zeroes, the output is: 

2.216671e-231
2.216671e-231
2.216671e-231

This seems to be due to strtod_l.c setting
 int_no is 0
 lead_zero = -1 
and later 
 dig_no -= lead_zero
This increases the number of digits considered for fractional digits since
 dig_no - int_no
is passed to str_to_mpn.  This causes the null string terminator to be
considered as a digit in the fractional digits.

------- Additional Comment #1 From Jakub Jelinek 2008-04-29 18:48 -------

It would be helpful if you could say on which target that is, because I
definitely can't reproduce it, neither on x86_64, nor on ppc32.

#define _GNU_SOURCE
#include <stdlib.h>
#include <stdio.h>

int
main (void)
{
  printf("%g\n", strtof("0.0", NULL));
  printf("%g\n", strtod("0.0", NULL));
  printf("%Lg\n", strtold("0.0", NULL));
  printf("%g\n", strtof("0", NULL));
  printf("%g\n", strtod("0", NULL));
  printf("%Lg\n", strtold("0", NULL));
  printf("%g\n", strtof("0.000", NULL));
  printf("%g\n", strtod("0.000", NULL));
  printf("%Lg\n", strtold("0.000", NULL));
  printf("%g\n", strtof("0.000e0", NULL));
  printf("%g\n", strtod("0.000e0", NULL));
  printf("%Lg\n", strtold("0.000e0", NULL));
  return 0;
}

prints just 0s.

------- Additional Comment #2 From Pete Eberlein 2008-04-29 19:29 -------

Bah, close this due to user error.  I neglected to include stdlib.h.

Aside: I discovered the problem reading the null terminator while working on the
dfp version and assumed it affected the bfp version as well.  It looks like
str_to_mpn isn't affected by the null so it just works.

------- Additional Comment #3 From Jakub Jelinek 2008-04-29 19:43 -------

Invalid.