1 /* 2 * Copyright (c) 2014, 2018, Oracle and/or its affiliates. All rights reserved. 3 * DO NOT ALTER OR REMOVE COPYRIGHT NOTICES OR THIS FILE HEADER. 4 * 5 * This code is free software; you can redistribute it and/or modify it 6 * under the terms of the GNU General Public License version 2 only, as 7 * published by the Free Software Foundation. 8 * 9 * This code is distributed in the hope that it will be useful, but WITHOUT 10 * ANY WARRANTY; without even the implied warranty of MERCHANTABILITY or 11 * FITNESS FOR A PARTICULAR PURPOSE. See the GNU General Public License 12 * version 2 for more details (a copy is included in the LICENSE file that 13 * accompanied this code). 14 * 15 * You should have received a copy of the GNU General Public License version 16 * 2 along with this work; if not, write to the Free Software Foundation, 17 * Inc., 51 Franklin St, Fifth Floor, Boston, MA 02110-1301 USA. 18 * 19 * Please contact Oracle, 500 Oracle Parkway, Redwood Shores, CA 94065 USA 20 * or visit www.oracle.com if you need additional information or have any 21 * questions. 22 * 23 */ 24 25 #include "precompiled.hpp" 26 #include "utilities/stringUtils.hpp" 27 28 int StringUtils::replace_no_expand(char* string, const char* from, const char* to) { 29 int replace_count = 0; 30 size_t from_len = strlen(from); 31 size_t to_len = strlen(to); 32 assert(from_len >= to_len, "must not expand input"); 33 34 for (char* dst = string; *dst && (dst = strstr(dst, from)) != NULL;) { 35 char* left_over = dst + from_len; 36 memmove(dst, to, to_len); // does not copy trailing 0 of <to> 37 dst += to_len; // skip over the replacement. 38 memmove(dst, left_over, strlen(left_over) + 1); // copies the trailing 0 of <left_over> 39 ++ replace_count; 40 } 41 42 return replace_count; 43 } 44 45 double StringUtils::similarity(const char* str1, size_t len1, const char* str2, size_t len2) { 46 size_t total = len1 + len2; 47 48 size_t hit = 0; 49 for (size_t i = 0; i < len1 - 1; i++) { 50 for (size_t j = 0; j < len2 - 1; j++) { 51 if ((str1[i] == str2[j]) && (str1[i+1] == str2[j+1])) { 52 ++hit; 53 break; 54 } 55 } 56 } 57 58 return 2.0 * (double) hit / (double) total; 59 }