#include <stdbool.h>
#include <string.h>
+#include "c-ctype.h"
+
/* 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)
{
- const char *s;
+ const unsigned char *s;
bool negative = false;
/* The number so far. */
bool got_dot; /* Found a decimal point. */
bool got_digit; /* Seen any digits. */
+ bool hex = false; /* Look for hex float exponent. */
/* The exponent of the number. */
long int exponent;
s = nptr;
/* Eat whitespace. */
- while (isspace ((unsigned char) *s))
+ while (isspace (*s))
++s;
/* Get the sign. */
got_dot = false;
got_digit = false;
exponent = 0;
- for (;; ++s)
+
+ /* Check for hex float. */
+ if (*s == '0' && c_tolower (s[1]) == 'x'
+ && (c_isxdigit (s[2]) || ('.' == s[2] && c_isxdigit (s[3]))))
{
- if ('0' <= *s && *s <= '9')
+ hex = true;
+ s += 2;
+ for (;; ++s)
{
- got_digit = true;
-
- /* 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;
+ if (c_isxdigit (*s))
+ {
+ got_digit = true;
+
+ /* Make sure that multiplication by 16 will not overflow. */
+ if (num > DBL_MAX / 16)
+ /* 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 16 later. */
+ ++exponent;
+ else
+ num = ((num * 16.0)
+ + (c_tolower (*s) - (c_isdigit (*s) ? '0' : 'a' - 10)));
+
+ /* Keep track of the number of digits after the decimal point.
+ If we just divided by 16 here, we would lose precision. */
+ if (got_dot)
+ --exponent;
+ }
+ else if (!got_dot && *s == '.')
+ /* Record that we have found the decimal point. */
+ got_dot = true;
else
- num = (num * 10.0) + (*s - '0');
+ /* Any other character terminates the number. */
+ break;
+ }
+ }
+
+ /* Not a hex float. */
+ else
+ {
+ for (;; ++s)
+ {
+ if (c_isdigit (*s))
+ {
+ got_digit = true;
+
+ /* 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;
+ /* 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 = true;
+ else
+ /* Any other character terminates the number. */
+ break;
}
- else if (!got_dot && *s == '.')
- /* Record that we have found the decimal point. */
- got_dot = true;
- else
- /* Any other character terminates the number. */
- break;
}
if (!got_digit)
{
/* Check for infinities and NaNs. */
- if (tolower ((unsigned char) *s) == 'i'
- && tolower ((unsigned char) s[1]) == 'n'
- && tolower ((unsigned char) s[2]) == 'f')
+ if (c_tolower (*s) == 'i'
+ && c_tolower (s[1]) == 'n'
+ && c_tolower (s[2]) == 'f')
{
s += 3;
num = HUGE_VAL;
- if (tolower ((unsigned char) *s) == 'i'
- && tolower ((unsigned char) s[1]) == 'n'
- && tolower ((unsigned char) s[2]) == 'i'
- && tolower ((unsigned char) s[3]) == 't'
- && tolower ((unsigned char) s[4]) == 'y')
+ if (c_tolower (*s) == 'i'
+ && c_tolower (s[1]) == 'n'
+ && c_tolower (s[2]) == 'i'
+ && c_tolower (s[3]) == 't'
+ && c_tolower (s[4]) == 'y')
s += 5;
goto valid;
}
#ifdef NAN
- else if (tolower ((unsigned char) *s) == 'n'
- && tolower ((unsigned char) s[1]) == 'a'
- && tolower ((unsigned char) s[2]) == 'n')
+ else if (c_tolower (*s) == 'n'
+ && c_tolower (s[1]) == 'a'
+ && c_tolower (s[2]) == 'n')
{
s += 3;
num = NAN;
hexadecimal number, but the result is still a NaN. */
if (*s == '(')
{
- const char *p = s + 1;
- while (isalnum ((unsigned char) *p))
+ const unsigned char *p = s + 1;
+ while (c_isalnum (*p))
p++;
if (*p == ')')
s = p + 1;
goto noconv;
}
- if (tolower ((unsigned char) *s) == 'e')
+ if (c_tolower (*s) == (hex ? 'p' : 'e') && !isspace (s[1]))
{
/* Get the exponent specified after the `e' or `E'. */
int save = errno;
else
goto overflow;
}
- else if (end == s)
+ else if (end == (char *) s)
/* There was no exponent. Reset END to point to
the 'e' or 'E', so *ENDPTR will be set there. */
end = (char *) s - 1;
if (num == 0.0)
goto valid;
+ if (hex)
+ {
+ /* ldexp takes care of range errors. */
+ num = ldexp (num, exponent);
+ goto valid;
+ }
+
/* Multiply NUM by 10 to the EXPONENT power,
checking for overflow and underflow. */
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