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 */ 44unsigned int 45strcount(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 */ 89int 90replaceall(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 */ 177void 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 */ 265void 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 */ 294void 295strtolower(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