1 /*-
2 * Copyright (c) 2001-2014 Devin Teske <[email protected]>
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/cdefs.h>
28 #include <sys/types.h>
29
30 #include <ctype.h>
31 #include <errno.h>
32 #include <stdio.h>
33 #include <stdlib.h>
34 #include <string.h>
35
36 #include "string_m.h"
37
38 /*
39 * Counts the number of occurrences of one string that appear in the source
40 * string. Return value is the total count.
41 *
42 * An example use would be if you need to know how large a block of memory
43 * needs to be for a replaceall() series.
44 */
45 unsigned int
strcount(const char * source,const char * find)46 strcount(const char *source, const char *find)
47 {
48 const char *p = source;
49 size_t flen;
50 unsigned int n = 0;
51
52 /* Both parameters are required */
53 if (source == NULL || find == NULL)
54 return (0);
55
56 /* Cache the length of find element */
57 flen = strlen(find);
58 if (strlen(source) == 0 || flen == 0)
59 return (0);
60
61 /* Loop until the end of the string */
62 while (*p != '\0') {
63 if (strncmp(p, find, flen) == 0) { /* found an instance */
64 p += flen;
65 n++;
66 } else
67 p++;
68 }
69
70 return (n);
71 }
72
73 /*
74 * Replaces all occurrences of `find' in `source' with `replace'.
75 *
76 * You should not pass a string constant as the first parameter, it needs to be
77 * a pointer to an allocated block of memory. The block of memory that source
78 * points to should be large enough to hold the result. If the length of the
79 * replacement string is greater than the length of the find string, the result
80 * will be larger than the original source string. To allocate enough space for
81 * the result, use the function strcount() declared above to determine the
82 * number of occurrences and how much larger the block size needs to be.
83 *
84 * If source is not large enough, the application will crash. The return value
85 * is the length (in bytes) of the result.
86 *
87 * When an error occurs, -1 is returned and the global variable errno is set
88 * accordingly. Returns zero on success.
89 */
90 int
replaceall(char * source,const char * find,const char * replace)91 replaceall(char *source, const char *find, const char *replace)
92 {
93 char *p;
94 char *t;
95 char *temp;
96 size_t flen;
97 size_t rlen;
98 size_t slen;
99 uint32_t n = 0;
100
101 errno = 0; /* reset global error number */
102
103 /* Check that we have non-null parameters */
104 if (source == NULL)
105 return (0);
106 if (find == NULL)
107 return (strlen(source));
108
109 /* Cache the length of the strings */
110 slen = strlen(source);
111 flen = strlen(find);
112 rlen = replace ? strlen(replace) : 0;
113
114 /* Cases where no replacements need to be made */
115 if (slen == 0 || flen == 0 || slen < flen)
116 return (slen);
117
118 /* If replace is longer than find, we'll need to create a temp copy */
119 if (rlen > flen) {
120 temp = malloc(slen + 1);
121 if (temp == NULL) /* could not allocate memory */
122 return (-1);
123 memcpy(temp, source, slen + 1);
124 } else
125 temp = source;
126
127 /* Reconstruct the string with the replacements */
128 p = source; t = temp; /* position elements */
129
130 while (*t != '\0') {
131 if (strncmp(t, find, flen) == 0) {
132 /* found an occurrence */
133 for (n = 0; replace && replace[n]; n++)
134 *p++ = replace[n];
135 t += flen;
136 } else
137 *p++ = *t++; /* copy character and increment */
138 }
139
140 /* Terminate the string */
141 *p = '\0';
142
143 /* Free the temporary allocated memory */
144 if (temp != source)
145 free(temp);
146
147 /* Return the length of the completed string */
148 return (strlen(source));
149 }
150
151 /*
152 * Expands escape sequences in a buffer pointed to by `source'. This function
153 * steps through each character, and converts escape sequences such as "\n",
154 * "\r", "\t" and others into their respective meanings.
155 *
156 * You should not pass a string constant or literal to this function or the
157 * program will likely segmentation fault when it tries to modify the data.
158 *
159 * The string length will either shorten or stay the same depending on whether
160 * any escape sequences were converted but the amount of memory allocated does
161 * not change.
162 *
163 * Interpreted sequences are:
164 *
165 * \0NNN character with octal value NNN (0 to 3 digits)
166 * \N character with octal value N (0 thru 7)
167 * \a alert (BEL)
168 * \b backslash
169 * \f form feed
170 * \n new line
171 * \r carriage return
172 * \t horizontal tab
173 * \v vertical tab
174 * \xNN byte with hexadecimal value NN (1 to 2 digits)
175 *
176 * All other sequences are unescaped (ie. '\"' and '\#').
177 */
strexpand(char * source)178 void strexpand(char *source)
179 {
180 uint8_t c;
181 char *chr;
182 char *pos;
183 char d[4];
184
185 /* Initialize position elements */
186 pos = chr = source;
187
188 /* Loop until we hit the end of the string */
189 while (*pos != '\0') {
190 if (*chr != '\\') {
191 *pos = *chr; /* copy character to current offset */
192 pos++;
193 chr++;
194 continue;
195 }
196
197 /* Replace the backslash with the correct character */
198 switch (*++chr) {
199 case 'a': *pos = '\a'; break; /* bell/alert (BEL) */
200 case 'b': *pos = '\b'; break; /* backspace */
201 case 'f': *pos = '\f'; break; /* form feed */
202 case 'n': *pos = '\n'; break; /* new line */
203 case 'r': *pos = '\r'; break; /* carriage return */
204 case 't': *pos = '\t'; break; /* horizontal tab */
205 case 'v': *pos = '\v'; break; /* vertical tab */
206 case 'x': /* hex value (1 to 2 digits)(\xNN) */
207 d[2] = '\0'; /* pre-terminate the string */
208
209 /* verify next two characters are hex */
210 d[0] = isxdigit(*(chr+1)) ? *++chr : '\0';
211 if (d[0] != '\0')
212 d[1] = isxdigit(*(chr+1)) ? *++chr : '\0';
213
214 /* convert the characters to decimal */
215 c = (uint8_t)strtoul(d, 0, 16);
216
217 /* assign the converted value */
218 *pos = (c != 0 || d[0] == '0') ? c : *++chr;
219 break;
220 case '0': /* octal value (0 to 3 digits)(\0NNN) */
221 d[3] = '\0'; /* pre-terminate the string */
222
223 /* verify next three characters are octal */
224 d[0] = (isdigit(*(chr+1)) && *(chr+1) < '8') ?
225 *++chr : '\0';
226 if (d[0] != '\0')
227 d[1] = (isdigit(*(chr+1)) && *(chr+1) < '8') ?
228 *++chr : '\0';
229 if (d[1] != '\0')
230 d[2] = (isdigit(*(chr+1)) && *(chr+1) < '8') ?
231 *++chr : '\0';
232
233 /* convert the characters to decimal */
234 c = (uint8_t)strtoul(d, 0, 8);
235
236 /* assign the converted value */
237 *pos = c;
238 break;
239 default: /* single octal (\0..7) or unknown sequence */
240 if (isdigit(*chr) && *chr < '8') {
241 d[0] = *chr;
242 d[1] = '\0';
243 *pos = (uint8_t)strtoul(d, 0, 8);
244 } else
245 *pos = *chr;
246 }
247
248 /* Increment to next offset, possible next escape sequence */
249 pos++;
250 chr++;
251 }
252 }
253
254 /*
255 * Expand only the escaped newlines in a buffer pointed to by `source'. This
256 * function steps through each character, and converts the "\n" sequence into
257 * a literal newline and the "\\n" sequence into "\n".
258 *
259 * You should not pass a string constant or literal to this function or the
260 * program will likely segmentation fault when it tries to modify the data.
261 *
262 * The string length will either shorten or stay the same depending on whether
263 * any escaped newlines were converted but the amount of memory allocated does
264 * not change.
265 */
strexpandnl(char * source)266 void strexpandnl(char *source)
267 {
268 uint8_t backslash = 0;
269 char *cp1;
270 char *cp2;
271
272 /* Replace '\n' with literal in dprompt */
273 cp1 = cp2 = source;
274 while (*cp2 != '\0') {
275 *cp1 = *cp2;
276 if (*cp2 == '\\')
277 backslash++;
278 else if (*cp2 != 'n')
279 backslash = 0;
280 else if (backslash > 0) {
281 *(--cp1) = (backslash & 1) == 1 ? '\n' : 'n';
282 backslash = 0;
283 }
284 cp1++;
285 cp2++;
286 }
287 *cp1 = *cp2;
288 }
289
290 /*
291 * Convert a string to lower case. You should not pass a string constant to
292 * this function. Only pass pointers to allocated memory with null terminated
293 * string data.
294 */
295 void
strtolower(char * source)296 strtolower(char *source)
297 {
298 char *p = source;
299
300 if (source == NULL)
301 return;
302
303 while (*p != '\0') {
304 *p = tolower(*p);
305 p++; /* would have just used `*p++' but gcc 3.x warns */
306 }
307 }
308