1 /*-
2 * Copyright (c) 2001-2014 Devin Teske <dteske@FreeBSD.org>
3 * All rights reserved.
4 *
5 * Redistribution and use in source and binary forms, with or without
6 * modification, are permitted provided that the following conditions
7 * are met:
8 * 1. Redistributions of source code must retain the above copyright
9 * notice, this list of conditions and the following disclaimer.
10 * 2. Redistributions in binary form must reproduce the above copyright
11 * notice, this list of conditions and the following disclaimer in the
12 * documentation and/or other materials provided with the distribution.
13 *
14 * THIS SOFTWARE IS PROVIDED BY THE AUTHOR AND CONTRIBUTORS ``AS IS'' AND
15 * ANY EXPRESS OR IMPLIED WARRANTIES, INCLUDING, BUT NOT LIMITED TO, THE
16 * IMPLIED WARRANTIES OF MERCHANTABILITY AND FITNESS FOR A PARTICULAR PURPOSE
17 * ARE DISCLAIMED. IN NO EVENT SHALL THE AUTHOR OR CONTRIBUTORS BE LIABLE
18 * FOR ANY DIRECT, INDIRECT, INCIDENTAL, SPECIAL, EXEMPLARY, OR CONSEQUENTIAL
19 * DAMAGES (INCLUDING, BUT NOT LIMITED TO, PROCUREMENT OF SUBSTITUTE GOODS
20 * OR SERVICES; LOSS OF USE, DATA, OR PROFITS; OR BUSINESS INTERRUPTION)
21 * HOWEVER CAUSED AND ON ANY THEORY OF LIABILITY, WHETHER IN CONTRACT, STRICT
22 * LIABILITY, OR TORT (INCLUDING NEGLIGENCE OR OTHERWISE) ARISING IN ANY WAY
23 * OUT OF THE USE OF THIS SOFTWARE, EVEN IF ADVISED OF THE POSSIBILITY OF
24 * SUCH DAMAGE.
25 */
26
27 #include <ctype.h>
28 #include <errno.h>
29 #include <stdint.h>
30 #include <stdlib.h>
31 #include <string.h>
32
33 #include "string_m.h"
34
35 /*
36 * Counts the number of occurrences of one string that appear in the source
37 * string. Return value is the total count.
38 *
39 * An example use would be if you need to know how large a block of memory
40 * needs to be for a replaceall() series.
41 */
42 unsigned int
strcount(const char * source,const char * find)43 strcount(const char *source, const char *find)
44 {
45 const char *p = source;
46 size_t flen;
47 unsigned int n = 0;
48
49 /* Both parameters are required */
50 if (source == NULL || find == NULL)
51 return (0);
52
53 /* Cache the length of find element */
54 flen = strlen(find);
55 if (strlen(source) == 0 || flen == 0)
56 return (0);
57
58 /* Loop until the end of the string */
59 while (*p != '\0') {
60 if (strncmp(p, find, flen) == 0) { /* found an instance */
61 p += flen;
62 n++;
63 } else
64 p++;
65 }
66
67 return (n);
68 }
69
70 /*
71 * Replaces all occurrences of `find' in `source' with `replace'.
72 *
73 * You should not pass a string constant as the first parameter, it needs to be
74 * a pointer to an allocated block of memory. The block of memory that source
75 * points to should be large enough to hold the result. If the length of the
76 * replacement string is greater than the length of the find string, the result
77 * will be larger than the original source string. To allocate enough space for
78 * the result, use the function strcount() declared above to determine the
79 * number of occurrences and how much larger the block size needs to be.
80 *
81 * If source is not large enough, the application will crash. The return value
82 * is the length (in bytes) of the result.
83 *
84 * When an error occurs, -1 is returned and the global variable errno is set
85 * accordingly. Returns zero on success.
86 */
87 int
replaceall(char * source,const char * find,const char * replace)88 replaceall(char *source, const char *find, const char *replace)
89 {
90 char *p;
91 char *t;
92 char *temp;
93 size_t flen;
94 size_t rlen;
95 size_t slen;
96 uint32_t n = 0;
97
98 errno = 0; /* reset global error number */
99
100 /* Check that we have non-null parameters */
101 if (source == NULL)
102 return (0);
103 if (find == NULL)
104 return (strlen(source));
105
106 /* Cache the length of the strings */
107 slen = strlen(source);
108 flen = strlen(find);
109 rlen = replace ? strlen(replace) : 0;
110
111 /* Cases where no replacements need to be made */
112 if (slen == 0 || flen == 0 || slen < flen)
113 return (slen);
114
115 /* If replace is longer than find, we'll need to create a temp copy */
116 if (rlen > flen) {
117 temp = malloc(slen + 1);
118 if (temp == NULL) /* could not allocate memory */
119 return (-1);
120 memcpy(temp, source, slen + 1);
121 } else
122 temp = source;
123
124 /* Reconstruct the string with the replacements */
125 p = source; t = temp; /* position elements */
126
127 while (*t != '\0') {
128 if (strncmp(t, find, flen) == 0) {
129 /* found an occurrence */
130 for (n = 0; replace && replace[n]; n++)
131 *p++ = replace[n];
132 t += flen;
133 } else
134 *p++ = *t++; /* copy character and increment */
135 }
136
137 /* Terminate the string */
138 *p = '\0';
139
140 /* Free the temporary allocated memory */
141 if (temp != source)
142 free(temp);
143
144 /* Return the length of the completed string */
145 return (strlen(source));
146 }
147
148 /*
149 * Expands escape sequences in a buffer pointed to by `source'. This function
150 * steps through each character, and converts escape sequences such as "\n",
151 * "\r", "\t" and others into their respective meanings.
152 *
153 * You should not pass a string constant or literal to this function or the
154 * program will likely segmentation fault when it tries to modify the data.
155 *
156 * The string length will either shorten or stay the same depending on whether
157 * any escape sequences were converted but the amount of memory allocated does
158 * not change.
159 *
160 * Interpreted sequences are:
161 *
162 * \0NNN character with octal value NNN (0 to 3 digits)
163 * \N character with octal value N (0 thru 7)
164 * \a alert (BEL)
165 * \b backslash
166 * \f form feed
167 * \n new line
168 * \r carriage return
169 * \t horizontal tab
170 * \v vertical tab
171 * \xNN byte with hexadecimal value NN (1 to 2 digits)
172 *
173 * All other sequences are unescaped (ie. '\"' and '\#').
174 */
strexpand(char * source)175 void strexpand(char *source)
176 {
177 uint8_t c;
178 char *chr;
179 char *pos;
180 char d[4];
181
182 /* Initialize position elements */
183 pos = chr = source;
184
185 /* Loop until we hit the end of the string */
186 while (*pos != '\0') {
187 if (*chr != '\\') {
188 *pos = *chr; /* copy character to current offset */
189 pos++;
190 chr++;
191 continue;
192 }
193
194 /* Replace the backslash with the correct character */
195 switch (*++chr) {
196 case 'a': *pos = '\a'; break; /* bell/alert (BEL) */
197 case 'b': *pos = '\b'; break; /* backspace */
198 case 'f': *pos = '\f'; break; /* form feed */
199 case 'n': *pos = '\n'; break; /* new line */
200 case 'r': *pos = '\r'; break; /* carriage return */
201 case 't': *pos = '\t'; break; /* horizontal tab */
202 case 'v': *pos = '\v'; break; /* vertical tab */
203 case 'x': /* hex value (1 to 2 digits)(\xNN) */
204 d[2] = '\0'; /* pre-terminate the string */
205
206 /* verify next two characters are hex */
207 d[0] = isxdigit(*(chr+1)) ? *++chr : '\0';
208 if (d[0] != '\0')
209 d[1] = isxdigit(*(chr+1)) ? *++chr : '\0';
210
211 /* convert the characters to decimal */
212 c = (uint8_t)strtoul(d, 0, 16);
213
214 /* assign the converted value */
215 *pos = (c != 0 || d[0] == '0') ? c : *++chr;
216 break;
217 case '0': /* octal value (0 to 3 digits)(\0NNN) */
218 d[3] = '\0'; /* pre-terminate the string */
219
220 /* verify next three characters are octal */
221 d[0] = (isdigit(*(chr+1)) && *(chr+1) < '8') ?
222 *++chr : '\0';
223 if (d[0] != '\0')
224 d[1] = (isdigit(*(chr+1)) && *(chr+1) < '8') ?
225 *++chr : '\0';
226 if (d[1] != '\0')
227 d[2] = (isdigit(*(chr+1)) && *(chr+1) < '8') ?
228 *++chr : '\0';
229
230 /* convert the characters to decimal */
231 c = (uint8_t)strtoul(d, 0, 8);
232
233 /* assign the converted value */
234 *pos = c;
235 break;
236 default: /* single octal (\0..7) or unknown sequence */
237 if (isdigit(*chr) && *chr < '8') {
238 d[0] = *chr;
239 d[1] = '\0';
240 *pos = (uint8_t)strtoul(d, 0, 8);
241 } else
242 *pos = *chr;
243 }
244
245 /* Increment to next offset, possible next escape sequence */
246 pos++;
247 chr++;
248 }
249 }
250
251 /*
252 * Expand only the escaped newlines in a buffer pointed to by `source'. This
253 * function steps through each character, and converts the "\n" sequence into
254 * a literal newline and the "\\n" sequence into "\n".
255 *
256 * You should not pass a string constant or literal to this function or the
257 * program will likely segmentation fault when it tries to modify the data.
258 *
259 * The string length will either shorten or stay the same depending on whether
260 * any escaped newlines were converted but the amount of memory allocated does
261 * not change.
262 */
strexpandnl(char * source)263 void strexpandnl(char *source)
264 {
265 uint8_t backslash = 0;
266 char *cp1;
267 char *cp2;
268
269 /* Replace '\n' with literal in dprompt */
270 cp1 = cp2 = source;
271 while (*cp2 != '\0') {
272 *cp1 = *cp2;
273 if (*cp2 == '\\')
274 backslash++;
275 else if (*cp2 != 'n')
276 backslash = 0;
277 else if (backslash > 0) {
278 *(--cp1) = (backslash & 1) == 1 ? '\n' : 'n';
279 backslash = 0;
280 }
281 cp1++;
282 cp2++;
283 }
284 *cp1 = *cp2;
285 }
286
287 /*
288 * Convert a string to lower case. You should not pass a string constant to
289 * this function. Only pass pointers to allocated memory with null terminated
290 * string data.
291 */
292 void
strtolower(char * source)293 strtolower(char *source)
294 {
295 char *p = source;
296
297 if (source == NULL)
298 return;
299
300 while (*p != '\0') {
301 *p = tolower(*p);
302 p++; /* would have just used `*p++' but gcc 3.x warns */
303 }
304 }
305