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/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 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 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 */ 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 */ 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 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