1 /* Copyright (C) 1991-1992, 1997, 1999, 2003, 2006, 2008-2010 Free Software
4 This program is free software: you can redistribute it and/or modify
5 it under the terms of the GNU General Public License as published by
6 the Free Software Foundation; either version 3 of the License, or
7 (at your option) any later version.
9 This program is distributed in the hope that it will be useful,
10 but WITHOUT ANY WARRANTY; without even the implied warranty of
11 MERCHANTABILITY or FITNESS FOR A PARTICULAR PURPOSE. See the
12 GNU General Public License for more details.
14 You should have received a copy of the GNU General Public License
15 along with this program. If not, see <http://www.gnu.org/licenses/>. */
19 /* Don't use __attribute__ __nonnull__ in this compilation unit. Otherwise gcc
20 optimizes away the nptr == NULL test below. */
21 #define _GL_ARG_NONNULL(params)
34 /* Convert NPTR to a double. If ENDPTR is not NULL, a pointer to the
35 character after the last one used in the number is put in *ENDPTR. */
37 strtod (const char *nptr, char **endptr)
39 const unsigned char *s;
40 bool negative = false;
42 /* The number so far. */
45 bool got_dot; /* Found a decimal point. */
46 bool got_digit; /* Seen any digits. */
47 bool hex = false; /* Look for hex float exponent. */
49 /* The exponent of the number. */
58 /* Use unsigned char for the ctype routines. */
59 s = (unsigned char *) nptr;
67 if (*s == '-' || *s == '+')
75 /* Check for hex float. */
76 if (*s == '0' && c_tolower (s[1]) == 'x'
77 && (c_isxdigit (s[2]) || ('.' == s[2] && c_isxdigit (s[3]))))
87 /* Make sure that multiplication by 16 will not overflow. */
88 if (num > DBL_MAX / 16)
89 /* The value of the digit doesn't matter, since we have already
90 gotten as many digits as can be represented in a `double'.
91 This doesn't necessarily mean the result will overflow.
92 The exponent may reduce it to within range.
94 We just need to record that there was another
95 digit so that we can multiply by 16 later. */
99 + (c_tolower (*s) - (c_isdigit (*s) ? '0' : 'a' - 10)));
101 /* Keep track of the number of digits after the decimal point.
102 If we just divided by 16 here, we would lose precision. */
106 else if (!got_dot && *s == '.')
107 /* Record that we have found the decimal point. */
110 /* Any other character terminates the number. */
115 /* Not a hex float. */
124 /* Make sure that multiplication by 10 will not overflow. */
125 if (num > DBL_MAX * 0.1)
126 /* The value of the digit doesn't matter, since we have already
127 gotten as many digits as can be represented in a `double'.
128 This doesn't necessarily mean the result will overflow.
129 The exponent may reduce it to within range.
131 We just need to record that there was another
132 digit so that we can multiply by 10 later. */
135 num = (num * 10.0) + (*s - '0');
137 /* Keep track of the number of digits after the decimal point.
138 If we just divided by 10 here, we would lose precision. */
142 else if (!got_dot && *s == '.')
143 /* Record that we have found the decimal point. */
146 /* Any other character terminates the number. */
153 /* Check for infinities and NaNs. */
154 if (c_tolower (*s) == 'i'
155 && c_tolower (s[1]) == 'n'
156 && c_tolower (s[2]) == 'f')
160 if (c_tolower (*s) == 'i'
161 && c_tolower (s[1]) == 'n'
162 && c_tolower (s[2]) == 'i'
163 && c_tolower (s[3]) == 't'
164 && c_tolower (s[4]) == 'y')
169 else if (c_tolower (*s) == 'n'
170 && c_tolower (s[1]) == 'a'
171 && c_tolower (s[2]) == 'n')
175 /* Since nan(<n-char-sequence>) is implementation-defined,
176 we define it by ignoring <n-char-sequence>. A nicer
177 implementation would populate the bits of the NaN
178 according to interpreting n-char-sequence as a
179 hexadecimal number, but the result is still a NaN. */
182 const unsigned char *p = s + 1;
183 while (c_isalnum (*p))
194 if (c_tolower (*s) == (hex ? 'p' : 'e') && !isspace (s[1]))
196 /* Get the exponent specified after the `e' or `E'. */
203 value = strtol ((char *) s, &end, 10);
204 if (errno == ERANGE && num)
206 /* The exponent overflowed a `long int'. It is probably a safe
207 assumption that an exponent that cannot be represented by
208 a `long int' exceeds the limits of a `double'. */
216 else if (end == (char *) s)
217 /* There was no exponent. Reset END to point to
218 the 'e' or 'E', so *ENDPTR will be set there. */
219 end = (char *) s - 1;
221 s = (unsigned char *) end;
230 /* ldexp takes care of range errors. */
231 num = ldexp (num, exponent);
235 /* Multiply NUM by 10 to the EXPONENT power,
236 checking for overflow and underflow. */
240 if (num < DBL_MIN * pow (10.0, (double) -exponent))
243 else if (exponent > 0)
245 if (num > DBL_MAX * pow (10.0, (double) -exponent))
249 num *= pow (10.0, (double) exponent);
253 *endptr = (char *) s;
254 return negative ? -num : num;
257 /* Return an overflow error. */
259 *endptr = (char *) s;
261 return negative ? -HUGE_VAL : HUGE_VAL;
264 /* Return an underflow error. */
266 *endptr = (char *) s;
268 return negative ? -0.0 : 0.0;
271 /* There was no number. */
273 *endptr = (char *) nptr;