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 <sys/types.h>
28
29 #include <ctype.h>
30 #include <errno.h>
31 #include <stdio.h>
32 #include <stdlib.h>
33 #include <string.h>
34
35 #include "string_m.h"
36
37 /*
38 * Counts the number of occurrences of one string that appear in the source
39 * string. Return value is the total count.
40 *
41 * An example use would be if you need to know how large a block of memory
42 * needs to be for a replaceall() series.
43 */
44 unsigned int
strcount(const char * source,const char * find)45 strcount(const char *source, const char *find)
46 {
47 const char *p = source;
48 size_t flen;
49 unsigned int n = 0;
50
51 /* Both parameters are required */
52 if (source == NULL || find == NULL)
53 return (0);
54
55 /* Cache the length of find element */
56 flen = strlen(find);
57 if (strlen(source) == 0 || flen == 0)
58 return (0);
59
60 /* Loop until the end of the string */
61 while (*p != '\0') {
62 if (strncmp(p, find, flen) == 0) { /* found an instance */
63 p += flen;
64 n++;
65 } else
66 p++;
67 }
68
69 return (n);
70 }
71
72 /*
73 * Replaces all occurrences of `find' in `source' with `replace'.
74 *
75 * You should not pass a string constant as the first parameter, it needs to be
76 * a pointer to an allocated block of memory. The block of memory that source
77 * points to should be large enough to hold the result. If the length of the
78 * replacement string is greater than the length of the find string, the result
79 * will be larger than the original source string. To allocate enough space for
80 * the result, use the function strcount() declared above to determine the
81 * number of occurrences and how much larger the block size needs to be.
82 *
83 * If source is not large enough, the application will crash. The return value
84 * is the length (in bytes) of the result.
85 *
86 * When an error occurs, -1 is returned and the global variable errno is set
87 * accordingly. Returns zero on success.
88 */
89 int
replaceall(char * source,const char * find,const char * replace)90 replaceall(char *source, const char *find, const char *replace)
91 {
92 char *p;
93 char *t;
94 char *temp;
95 size_t flen;
96 size_t rlen;
97 size_t slen;
98 uint32_t n = 0;
99
100 errno = 0; /* reset global error number */
101
102 /* Check that we have non-null parameters */
103 if (source == NULL)
104 return (0);
105 if (find == NULL)
106 return (strlen(source));
107
108 /* Cache the length of the strings */
109 slen = strlen(source);
110 flen = strlen(find);
111 rlen = replace ? strlen(replace) : 0;
112
113 /* Cases where no replacements need to be made */
114 if (slen == 0 || flen == 0 || slen < flen)
115 return (slen);
116
117 /* If replace is longer than find, we'll need to create a temp copy */
118 if (rlen > flen) {
119 temp = malloc(slen + 1);
120 if (temp == NULL) /* could not allocate memory */
121 return (-1);
122 memcpy(temp, source, slen + 1);
123 } else
124 temp = source;
125
126 /* Reconstruct the string with the replacements */
127 p = source; t = temp; /* position elements */
128
129 while (*t != '\0') {
130 if (strncmp(t, find, flen) == 0) {
131 /* found an occurrence */
132 for (n = 0; replace && replace[n]; n++)
133 *p++ = replace[n];
134 t += flen;
135 } else
136 *p++ = *t++; /* copy character and increment */
137 }
138
139 /* Terminate the string */
140 *p = '\0';
141
142 /* Free the temporary allocated memory */
143 if (temp != source)
144 free(temp);
145
146 /* Return the length of the completed string */
147 return (strlen(source));
148 }
149
150 /*
151 * Expands escape sequences in a buffer pointed to by `source'. This function
152 * steps through each character, and converts escape sequences such as "\n",
153 * "\r", "\t" and others into their respective meanings.
154 *
155 * You should not pass a string constant or literal to this function or the
156 * program will likely segmentation fault when it tries to modify the data.
157 *
158 * The string length will either shorten or stay the same depending on whether
159 * any escape sequences were converted but the amount of memory allocated does
160 * not change.
161 *
162 * Interpreted sequences are:
163 *
164 * \0NNN character with octal value NNN (0 to 3 digits)
165 * \N character with octal value N (0 thru 7)
166 * \a alert (BEL)
167 * \b backslash
168 * \f form feed
169 * \n new line
170 * \r carriage return
171 * \t horizontal tab
172 * \v vertical tab
173 * \xNN byte with hexadecimal value NN (1 to 2 digits)
174 *
175 * All other sequences are unescaped (ie. '\"' and '\#').
176 */
strexpand(char * source)177 void strexpand(char *source)
178 {
179 uint8_t c;
180 char *chr;
181 char *pos;
182 char d[4];
183
184 /* Initialize position elements */
185 pos = chr = source;
186
187 /* Loop until we hit the end of the string */
188 while (*pos != '\0') {
189 if (*chr != '\\') {
190 *pos = *chr; /* copy character to current offset */
191 pos++;
192 chr++;
193 continue;
194 }
195
196 /* Replace the backslash with the correct character */
197 switch (*++chr) {
198 case 'a': *pos = '\a'; break; /* bell/alert (BEL) */
199 case 'b': *pos = '\b'; break; /* backspace */
200 case 'f': *pos = '\f'; break; /* form feed */
201 case 'n': *pos = '\n'; break; /* new line */
202 case 'r': *pos = '\r'; break; /* carriage return */
203 case 't': *pos = '\t'; break; /* horizontal tab */
204 case 'v': *pos = '\v'; break; /* vertical tab */
205 case 'x': /* hex value (1 to 2 digits)(\xNN) */
206 d[2] = '\0'; /* pre-terminate the string */
207
208 /* verify next two characters are hex */
209 d[0] = isxdigit(*(chr+1)) ? *++chr : '\0';
210 if (d[0] != '\0')
211 d[1] = isxdigit(*(chr+1)) ? *++chr : '\0';
212
213 /* convert the characters to decimal */
214 c = (uint8_t)strtoul(d, 0, 16);
215
216 /* assign the converted value */
217 *pos = (c != 0 || d[0] == '0') ? c : *++chr;
218 break;
219 case '0': /* octal value (0 to 3 digits)(\0NNN) */
220 d[3] = '\0'; /* pre-terminate the string */
221
222 /* verify next three characters are octal */
223 d[0] = (isdigit(*(chr+1)) && *(chr+1) < '8') ?
224 *++chr : '\0';
225 if (d[0] != '\0')
226 d[1] = (isdigit(*(chr+1)) && *(chr+1) < '8') ?
227 *++chr : '\0';
228 if (d[1] != '\0')
229 d[2] = (isdigit(*(chr+1)) && *(chr+1) < '8') ?
230 *++chr : '\0';
231
232 /* convert the characters to decimal */
233 c = (uint8_t)strtoul(d, 0, 8);
234
235 /* assign the converted value */
236 *pos = c;
237 break;
238 default: /* single octal (\0..7) or unknown sequence */
239 if (isdigit(*chr) && *chr < '8') {
240 d[0] = *chr;
241 d[1] = '\0';
242 *pos = (uint8_t)strtoul(d, 0, 8);
243 } else
244 *pos = *chr;
245 }
246
247 /* Increment to next offset, possible next escape sequence */
248 pos++;
249 chr++;
250 }
251 }
252
253 /*
254 * Expand only the escaped newlines in a buffer pointed to by `source'. This
255 * function steps through each character, and converts the "\n" sequence into
256 * a literal newline and the "\\n" sequence into "\n".
257 *
258 * You should not pass a string constant or literal to this function or the
259 * program will likely segmentation fault when it tries to modify the data.
260 *
261 * The string length will either shorten or stay the same depending on whether
262 * any escaped newlines were converted but the amount of memory allocated does
263 * not change.
264 */
strexpandnl(char * source)265 void strexpandnl(char *source)
266 {
267 uint8_t backslash = 0;
268 char *cp1;
269 char *cp2;
270
271 /* Replace '\n' with literal in dprompt */
272 cp1 = cp2 = source;
273 while (*cp2 != '\0') {
274 *cp1 = *cp2;
275 if (*cp2 == '\\')
276 backslash++;
277 else if (*cp2 != 'n')
278 backslash = 0;
279 else if (backslash > 0) {
280 *(--cp1) = (backslash & 1) == 1 ? '\n' : 'n';
281 backslash = 0;
282 }
283 cp1++;
284 cp2++;
285 }
286 *cp1 = *cp2;
287 }
288
289 /*
290 * Convert a string to lower case. You should not pass a string constant to
291 * this function. Only pass pointers to allocated memory with null terminated
292 * string data.
293 */
294 void
strtolower(char * source)295 strtolower(char *source)
296 {
297 char *p = source;
298
299 if (source == NULL)
300 return;
301
302 while (*p != '\0') {
303 *p = tolower(*p);
304 p++; /* would have just used `*p++' but gcc 3.x warns */
305 }
306 }
307