netsurf/utils/utils.c

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/*
* Copyright 2007 Rob Kendrick <rjek@netsurf-browser.org>
* Copyright 2004-2007 James Bursa <bursa@users.sourceforge.net>
* Copyright 2003 Phil Mellor <monkeyson@users.sourceforge.net>
* Copyright 2003 John M Bell <jmb202@ecs.soton.ac.uk>
* Copyright 2004 John Tytgat <joty@netsurf-browser.org>
*
* This file is part of NetSurf, http://www.netsurf-browser.org/
*
* NetSurf is free software; you can redistribute it and/or modify
* it under the terms of the GNU General Public License as published by
* the Free Software Foundation; version 2 of the License.
*
* NetSurf is distributed in the hope that it will be useful,
* but WITHOUT ANY WARRANTY; without even the implied warranty of
* MERCHANTABILITY or FITNESS FOR A PARTICULAR PURPOSE. See the
* GNU General Public License for more details.
*
* You should have received a copy of the GNU General Public License
* along with this program. If not, see <http://www.gnu.org/licenses/>.
*/
#include <assert.h>
#include <ctype.h>
#include <stdlib.h>
#include <stdio.h>
#include <string.h>
#include <strings.h>
#include <sys/stat.h>
#include <sys/types.h>
#include <sys/time.h>
#include <regex.h>
#include <time.h>
#include "utils/config.h"
#define NDEBUG
#include "utils/log.h"
#undef NDEBUG
#include "utils/messages.h"
#include "utils/utf8.h"
#include "utils/utils.h"
char * strip(char * const s)
{
size_t i;
for (i = strlen(s);
i != 0 && (s[i - 1] == ' ' || s[i - 1] == '\n' ||
s[i - 1] == '\r' || s[i - 1] == '\t');
i--)
;
s[i] = 0;
return s + strspn(s, " \t\r\n");
}
int whitespace(const char * str)
{
unsigned int i;
for (i = 0; i < strlen(str); i++)
if (!isspace(str[i]))
return 0;
return 1;
}
/**
* returns a string without its underscores
* \param replacespace true to insert a space where there was an underscore
*/
char *remove_underscores(const char *s, bool replacespace)
{
size_t i, ii, len;
char *ret;
len = strlen(s);
ret = malloc(len + 1);
if (ret == NULL)
return NULL;
for (i = 0, ii = 0; i < len; i++) {
if (s[i] != '_')
ret[ii++] = s[i];
else if (replacespace)
ret[ii++] = ' ';
}
ret[ii] = '\0';
return ret;
}
/**
* Replace consecutive whitespace with a single space.
*
* \param s source string
* \return heap allocated result, or 0 on memory exhaustion
*/
char * squash_whitespace(const char *s)
{
char *c = malloc(strlen(s) + 1);
int i = 0, j = 0;
if (!c)
return 0;
do {
if (s[i] == ' ' || s[i] == '\n' || s[i] == '\r' ||
s[i] == '\t') {
c[j++] = ' ';
while (s[i] == ' ' || s[i] == '\n' || s[i] == '\r' ||
s[i] == '\t')
i++;
}
c[j++] = s[i++];
} while (s[i - 1] != 0);
return c;
}
/**
* Converts NUL terminated UTF-8 encoded string s containing zero or more
* spaces (char 32) or TABs (char 9) to non-breaking spaces
* (0xC2 + 0xA0 in UTF-8 encoding).
*
* Caller needs to free() result. Returns NULL in case of error. No
* checking is done on validness of the UTF-8 input string.
*/
char *cnv_space2nbsp(const char *s)
{
const char *srcP;
char *d, *d0;
unsigned int numNBS;
/* Convert space & TAB into non breaking space character (0xA0) */
for (numNBS = 0, srcP = (const char *)s; *srcP != '\0'; ++srcP)
if (*srcP == ' ' || *srcP == '\t')
++numNBS;
if ((d = (char *)malloc((srcP - s) + numNBS + 1)) == NULL)
return NULL;
for (d0 = d, srcP = (const char *)s; *srcP != '\0'; ++srcP) {
if (*srcP == ' ' || *srcP == '\t') {
*d0++ = 0xC2;
*d0++ = 0xA0;
} else
*d0++ = *srcP;
}
*d0 = '\0';
return d;
}
/**
* Check if a directory exists.
*/
bool is_dir(const char *path)
{
struct stat s;
if (stat(path, &s))
return false;
return S_ISDIR(s.st_mode) ? true : false;
}
/**
* Compile a regular expression, handling errors.
*
* Parameters as for regcomp(), see man regex.
*/
void regcomp_wrapper(regex_t *preg, const char *regex, int cflags)
{
int r;
r = regcomp(preg, regex, cflags);
if (r) {
char errbuf[200];
regerror(r, preg, errbuf, sizeof errbuf);
fprintf(stderr, "Failed to compile regexp '%s'\n", regex);
die(errbuf);
}
}
/** We can have a fairly good estimate of how long the buffer needs to
* be. The unsigned long can store a value representing a maximum size
* of around 4 GB. Therefore the greatest space required is to
* represent 1023MB. Currently that would be represented as "1023MB" so 12
* including a null terminator.
* Ideally we would be able to know this value for sure, in the mean
* time the following should suffice.
**/
#define BYTESIZE_BUFFER_SIZE 20
/**
* Does a simple conversion which assumes the user speaks English. The buffer
* returned is one of three static ones so may change each time this call is
* made. Don't store the buffer for later use. It's done this way for
* convenience and to fight possible memory leaks, it is not necessarily pretty.
**/
char *human_friendly_bytesize(unsigned long bsize) {
static char buffer1[BYTESIZE_BUFFER_SIZE];
static char buffer2[BYTESIZE_BUFFER_SIZE];
static char buffer3[BYTESIZE_BUFFER_SIZE];
static char *curbuffer = buffer3;
enum {bytes, kilobytes, megabytes, gigabytes} unit = bytes;
static char units[][7] = {"Bytes", "kBytes", "MBytes", "GBytes"};
float bytesize = (float)bsize;
if (curbuffer == buffer1)
curbuffer = buffer2;
else if (curbuffer == buffer2)
curbuffer = buffer3;
else
curbuffer = buffer1;
if (bytesize > 1024) {
bytesize /= 1024;
unit = kilobytes;
}
if (bytesize > 1024) {
bytesize /= 1024;
unit = megabytes;
}
if (bytesize > 1024) {
bytesize /= 1024;
unit = gigabytes;
}
sprintf(curbuffer, "%3.2f%s", bytesize, messages_get(units[unit]));
return curbuffer;
}
/**
* Create an RFC 1123 compliant date string from a Unix timestamp
*
* \param t The timestamp to consider
* \return Pointer to buffer containing string - invalidated by next call.
*/
const char *rfc1123_date(time_t t)
{
static char ret[30];
struct tm *tm = gmtime(&t);
const char *days[] = { "Sun", "Mon", "Tue", "Wed", "Thu", "Fri", "Sat" },
*months[] = { "Jan", "Feb", "Mar", "Apr", "May", "Jun",
"Jul", "Aug", "Sep", "Oct", "Nov", "Dec" };
snprintf(ret, sizeof ret, "%s, %02d %s %d %02d:%02d:%02d GMT",
days[tm->tm_wday], tm->tm_mday, months[tm->tm_mon],
tm->tm_year + 1900, tm->tm_hour, tm->tm_min,
tm->tm_sec);
return ret;
}
/**
* Returns a number of centiseconds, that increases in real time, for the
* purposes of measuring how long something takes in wall-clock terms. It uses
* gettimeofday() for this. Should the call to gettimeofday() fail, it returns
* zero.
*
* \return number of centiseconds that increases monotonically
*/
unsigned int wallclock(void)
{
struct timeval tv;
if (gettimeofday(&tv, NULL) == -1)
return 0;
return ((tv.tv_sec * 100) + (tv.tv_usec / 10000));
}
#ifndef HAVE_STRCASESTR
/**
* Case insensitive strstr implementation
*
* \param haystack String to search in
* \param needle String to look for
* \return Pointer to start of found substring, or NULL if not found
*/
char *strcasestr(const char *haystack, const char *needle)
{
size_t needle_len = strlen(needle);
const char * last_start = haystack + (strlen(haystack) - needle_len);
while (haystack <= last_start) {
if (strncasecmp(haystack, needle, needle_len) == 0)
return (char *)haystack;
haystack++;
}
return NULL;
}
#endif
#ifndef HAVE_STRNDUP
/**
* Duplicate up to n characters of a string.
*/
char *strndup(const char *s, size_t n)
{
size_t len;
char *s2;
for (len = 0; len != n && s[len]; len++)
continue;
s2 = malloc(len + 1);
if (!s2)
return 0;
memcpy(s2, s, len);
s2[len] = 0;
return s2;
}
#endif
#ifndef HAVE_STRCHRNUL
/**
* Find the first occurrence of C in S or the final NUL byte.
*
* \note This implementation came from glibc 2.2.5
*/
char *strchrnul (const char *s, int c_in)
{
const unsigned char *char_ptr;
const unsigned long int *longword_ptr;
unsigned long int longword, magic_bits, charmask;
unsigned char c;
c = (unsigned char) c_in;
/* Handle the first few characters by reading one character at a time.
Do this until CHAR_PTR is aligned on a longword boundary. */
for (char_ptr = (const unsigned char *)s; ((unsigned long int) char_ptr
& (sizeof (longword) - 1)) != 0;
++char_ptr)
if (*char_ptr == c || *char_ptr == '\0')
return (void *) char_ptr;
/* All these elucidatory comments refer to 4-byte longwords,
but the theory applies equally well to 8-byte longwords. */
longword_ptr = (unsigned long int *) char_ptr;
/* Bits 31, 24, 16, and 8 of this number are zero. Call these bits
the "holes." Note that there is a hole just to the left of
each byte, with an extra at the end:
bits: 01111110 11111110 11111110 11111111
bytes: AAAAAAAA BBBBBBBB CCCCCCCC DDDDDDDD
The 1-bits make sure that carries propagate to the next 0-bit.
The 0-bits provide holes for carries to fall into. */
switch (sizeof (longword))
{
case 4: magic_bits = 0x7efefeffL; break;
case 8: magic_bits = ((0x7efefefeL << 16) << 16) | 0xfefefeffL; break;
default:
abort ();
}
/* Set up a longword, each of whose bytes is C. */
charmask = c | (c << 8);
charmask |= charmask << 16;
if (sizeof (longword) > 4)
/* Do the shift in two steps to avoid a warning if long has 32 bits. */
charmask |= (charmask << 16) << 16;
if (sizeof (longword) > 8)
abort ();
/* Instead of the traditional loop which tests each character,
we will test a longword at a time. The tricky part is testing
if *any of the four* bytes in the longword in question are zero. */
for (;;)
{
/* We tentatively exit the loop if adding MAGIC_BITS to
LONGWORD fails to change any of the hole bits of LONGWORD.
1) Is this safe? Will it catch all the zero bytes?
Suppose there is a byte with all zeros. Any carry bits
propagating from its left will fall into the hole at its
least significant bit and stop. Since there will be no
carry from its most significant bit, the LSB of the
byte to the left will be unchanged, and the zero will be
detected.
2) Is this worthwhile? Will it ignore everything except
zero bytes? Suppose every byte of LONGWORD has a bit set
somewhere. There will be a carry into bit 8. If bit 8
is set, this will carry into bit 16. If bit 8 is clear,
one of bits 9-15 must be set, so there will be a carry
into bit 16. Similarly, there will be a carry into bit
24. If one of bits 24-30 is set, there will be a carry
into bit 31, so all of the hole bits will be changed.
The one misfire occurs when bits 24-30 are clear and bit
31 is set; in this case, the hole at bit 31 is not
changed. If we had access to the processor carry flag,
we could close this loophole by putting the fourth hole
at bit 32!
So it ignores everything except 128's, when they're aligned
properly.
3) But wait! Aren't we looking for C as well as zero?
Good point. So what we do is XOR LONGWORD with a longword,
each of whose bytes is C. This turns each byte that is C
into a zero. */
longword = *longword_ptr++;
/* Add MAGIC_BITS to LONGWORD. */
if ((((longword + magic_bits)
/* Set those bits that were unchanged by the addition. */
^ ~longword)
/* Look at only the hole bits. If any of the hole bits
are unchanged, most likely one of the bytes was a
zero. */
& ~magic_bits) != 0 ||
/* That caught zeroes. Now test for C. */
((((longword ^ charmask) + magic_bits) ^ ~(longword ^ charmask))
& ~magic_bits) != 0)
{
/* Which of the bytes was C or zero?
If none of them were, it was a misfire; continue the search. */
const unsigned char *cp = (const unsigned char *) (longword_ptr - 1);
if (*cp == c || *cp == '\0')
return (char *) cp;
if (*++cp == c || *cp == '\0')
return (char *) cp;
if (*++cp == c || *cp == '\0')
return (char *) cp;
if (*++cp == c || *cp == '\0')
return (char *) cp;
if (sizeof (longword) > 4)
{
if (*++cp == c || *cp == '\0')
return (char *) cp;
if (*++cp == c || *cp == '\0')
return (char *) cp;
if (*++cp == c || *cp == '\0')
return (char *) cp;
if (*++cp == c || *cp == '\0')
return (char *) cp;
}
}
}
/* This should never happen. */
return NULL;
}
#endif