现在看下强大的atof()函数，哇哈哈：
/* Convert a string to a double. */
double
atof (const char *nptr)
{
return strtod (nptr, (char **) NULL);
}
#if HAVE_CONFIG_H
# include
#endif
#include
#ifndef errno
extern int errno;
#endif
#include
#if defined (STDC_HEADERS) || (!defined (isascii) && !defined (HAVE_ISASCII))
# define IN_CTYPE_DOMAIN(c) 1
#else
# define IN_CTYPE_DOMAIN(c) isascii(c)
#endif
#define ISSPACE(c) (IN_CTYPE_DOMAIN (c) && isspace (c))
#define ISDIGIT(c) (IN_CTYPE_DOMAIN (c) && isdigit (c))
#define TOLOWER(c) (IN_CTYPE_DOMAIN (c) ? tolower(c) : (c))
#include
#include
#include
#include
/* 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) /*如果为空串,则结束转换*/
{
errno = EINVAL;
goto noconv; /*转向处理无法转换的代码*/
}
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;
}
)
