-/* Copyright (C) 1991, 1992, 1997, 1999, 2003, 2006, 2008 Free
- Software Foundation, Inc.
+/* Copyright (C) 1991-1992, 1997, 1999, 2003, 2006, 2008-2010 Free Software
+ Foundation, Inc.
This program is free software: you can redistribute it and/or modify
it under the terms of the GNU General Public License as published by
#include <config.h>
+/* Don't use __attribute__ __nonnull__ in this compilation unit. Otherwise gcc
+ optimizes away the nptr == NULL test below. */
+#define _GL_ARG_NONNULL(params)
+
#include <stdlib.h>
#include <ctype.h>
/* The number so far. */
double num;
- bool got_dot; /* Found a decimal point. */
- bool got_digit; /* Seen any digits. */
- bool hex = false; /* Look for hex float exponent. */
+ 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;
goto noconv;
}
- s = nptr;
+ /* Use unsigned char for the ctype routines. */
+ s = (unsigned char *) nptr;
/* Eat whitespace. */
while (isspace (*s))
hex = true;
s += 2;
for (;; ++s)
- {
- 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
- /* Any other character terminates the number. */
- break;
- }
+ {
+ 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
+ /* 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;
- }
- 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 (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;
+ }
+ 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 (c_tolower (*s) == 'i'
- && c_tolower (s[1]) == 'n'
- && c_tolower (s[2]) == 'f')
- {
- s += 3;
- num = HUGE_VAL;
- 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;
- }
+ && c_tolower (s[1]) == 'n'
+ && c_tolower (s[2]) == 'f')
+ {
+ s += 3;
+ num = HUGE_VAL;
+ 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 (c_tolower (*s) == 'n'
- && c_tolower (s[1]) == 'a'
- && c_tolower (s[2]) == 'n')
- {
- s += 3;
- num = NAN;
- /* Since nan(<n-char-sequence>) is implementation-defined,
- we define it by ignoring <n-char-sequence>. A nicer
- implementation would populate the bits of the NaN
- according to interpreting n-char-sequence as a
- hexadecimal number, but the result is still a NaN. */
- if (*s == '(')
- {
- const unsigned char *p = s + 1;
- while (c_isalnum (*p))
- p++;
- if (*p == ')')
- s = p + 1;
- }
- goto valid;
- }
+ && c_tolower (s[1]) == 'a'
+ && c_tolower (s[2]) == 'n')
+ {
+ s += 3;
+ num = NAN;
+ /* Since nan(<n-char-sequence>) is implementation-defined,
+ we define it by ignoring <n-char-sequence>. A nicer
+ implementation would populate the bits of the NaN
+ according to interpreting n-char-sequence as a
+ hexadecimal number, but the result is still a NaN. */
+ if (*s == '(')
+ {
+ const unsigned char *p = s + 1;
+ while (c_isalnum (*p))
+ p++;
+ if (*p == ')')
+ s = p + 1;
+ }
+ goto valid;
+ }
#endif
goto noconv;
}
/* Get the exponent specified after the `e' or `E'. */
int save = errno;
char *end;
- long int exp;
+ long int value;
errno = 0;
++s;
- exp = strtol (s, &end, 10);
+ value = strtol ((char *) s, &end, 10);
if (errno == ERANGE && num)
- {
- /* 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;
- }
+ {
+ /* 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 (value < 0)
+ goto underflow;
+ else
+ goto overflow;
+ }
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;
+ /* 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;
+ s = (unsigned char *) end;
+ exponent += value;
}
if (num == 0.0)
if (exponent < 0)
{
if (num < DBL_MIN * pow (10.0, (double) -exponent))
- goto underflow;
+ goto underflow;
}
else if (exponent > 0)
{
if (num > DBL_MAX * pow (10.0, (double) -exponent))
- goto overflow;
+ goto overflow;
}
num *= pow (10.0, (double) exponent);