c - 電流がすでに DBL_MAX*0.1 を超えているのに、なぜ strtod は数字を無視するのか

Question

ソースコード (これがどのバージョンかはわかりません。ウェブサイトからの抜粋です)。for ループの最初の部分で、「十分な数字を取得しました。残りは無視します」というコメントがあります。

なぜこれが正しいのですか？そして、なぜこの「必ずしも結果がオーバーフローすることを意味するとは限らないのですか？」

/* Convert NPTR to a double.  If ENDPTR is not NULL, a pointer to the
   character after the last one used in the number is put in *ENDPTR.  */
double
strtod (const char *nptr, char **endptr)
{
  register const char *s;
  short int sign;

  /* The number so far.  */
  double num;

  int got_dot;                  /* Found a decimal point.  */
  int got_digit;                /* Seen any digits.  */

  /* The exponent of the number.  */
  long int exponent;

  if (nptr == NULL) 
    {
      errno = EINVAL;
      goto noconv; 
    }

  s = nptr;

  /* Eat whitespace.  */
  while (ISSPACE (*s))
    ++s;

  /* Get the sign.  */
  sign = *s == '-' ? -1 : 1;
  if (*s == '-' || *s == '+')
    ++s;

  num = 0.0;
  got_dot = 0;
  got_digit = 0;
  exponent = 0;
  for (;; ++s)
    {
      if (ISDIGIT (*s))
        {
          got_digit = 1;

          /* Make sure that multiplication by 10 will not overflow.  */
          if (num > DBL_MAX * 0.1)
            /* The value of the digit doesn't matter, since we have already
               gotten as many digits as can be represented in a `double'.
               This doesn't necessarily mean the result will overflow.
               The exponent may reduce it to within range.

               We just need to record that there was another
               digit so that we can multiply by 10 later.  */
            ++exponent;
          else
            num = (num * 10.0) + (*s - '0');

          /* Keep track of the number of digits after the decimal point.
             If we just divided by 10 here, we would lose precision.  */
          if (got_dot)
            --exponent;
        }
      else if (!got_dot && *s == '.')
        /* Record that we have found the decimal point.  */
        got_dot = 1;
      else
        /* Any other character terminates the number.  */
        break;
    }

  if (!got_digit)
    goto noconv;

  if (TOLOWER (*s) == 'e')
    {
      /* Get the exponent specified after the `e' or `E'.  */
      int save = errno;
      char *end;
      long int exp;

      errno = 0;
      ++s;
      exp = strtol (s, &end, 10);
      if (errno == ERANGE)
        {
          /* The exponent overflowed a `long int'.  It is probably a safe
             assumption that an exponent that cannot be represented by
             a `long int' exceeds the limits of a `double'.  */
          if (endptr != NULL)
            *endptr = end;
          if (exp < 0)
            goto underflow;
          else
            goto overflow;
        }
      else if (end == s)
        /* There was no exponent.  Reset END to point to
           the 'e' or 'E', so *ENDPTR will be set there.  */
        end = (char *) s - 1;
      errno = save;
      s = end;
      exponent += exp;
    }

  if (endptr != NULL)
    *endptr = (char *) s;

  if (num == 0.0)
    return 0.0;

  /* Multiply NUM by 10 to the EXPONENT power,
     checking for overflow and underflow.  */

  if (exponent < 0)
    {
      if (num < DBL_MIN * pow (10.0, (double) -exponent))
        goto underflow;
    }
  else if (exponent > 0)
    {
      if (num > DBL_MAX * pow (10.0, (double) -exponent))
        goto overflow;
    }

  num *= pow (10.0, (double) exponent);

  return num * sign;

overflow:
  /* Return an overflow error.  */
  errno = ERANGE;
  return HUGE_VAL * sign;

underflow:
  /* Return an underflow error.  */
  if (endptr != NULL)
    *endptr = (char *) nptr;
  errno = ERANGE;
  return 0.0;

noconv:
  /* There was no number.  */
  if (endptr != NULL)
    *endptr = (char *) nptr;
  return 0.0;
}