Regular-Expression
regmatch_t 數組中未使用的結構元素不是 -1
regcomp 和 regexec的手冊頁指出，“任何未使用的結構元素都將包含值 -1。”
然而，在我最後一場比賽之後檢查值的邏輯中
rm_so，情況似乎並非如此。它似乎等於0：#include <stdlib.h> #include <stdio.h> #include <string.h> #include <sys/types.h> #include <regex.h> #define MATCH_CNT 4 void print_matches(regmatch_t matches[]) { printf("Regexec successful.\n"); int i = 0; for(; i < MATCH_CNT; i++) { //########### HERE'S THE RUB #################### if(matches[i].rm_so != -1) { printf("Match %i; Beginning:\t%i\n", i, matches[i].rm_so); printf("Match %i; End:\t\t%i\n", i, matches[i].rm_eo); } } } int main(int argc, char *argv[]) { if(argc != 3) fprintf(stderr, "retester <pattern> <string>\n"); char *pat = argv[1]; char *str = argv[2]; regex_t compreg; memset(&compreg, 0, sizeof(regex_t)); regmatch_t matches[MATCH_CNT]; memset(&matches, 0, MATCH_CNT*sizeof(regmatch_t)); int matchcnt = -1; char *errbuff; printf("Trying to match with extended regex...\n"); int compret = -2; //set REG_EXTENDED flag if((compret = regcomp(&compreg, pat, REG_EXTENDED)) == 0) { printf("Compiling successful.\n"); int execret = -2; if((execret = regexec(&compreg, str, matchcnt, matches, 0)) == 0) { print_matches(matches); } else if(execret != 0) { printf("Regexec failed.\n"); size_t errbuffsz = regerror(execret, &compreg, 0, 0); errbuff = malloc(errbuffsz); memset(errbuff, '\0', errbuffsz); regerror(execret, &compreg, errbuff, errbuffsz); fprintf(stderr, "Regexec error: %s\n", errbuff); } } else { printf("Compiling failed.\n"); size_t errbuffsz = regerror(compret, &compreg, 0, 0); errbuff = malloc(errbuffsz); memset(errbuff, '\0', errbuffsz); regerror(compret, &compreg, errbuff, errbuffsz); fprintf(stderr, "Regexec error: %s", errbuff); } free(errbuff); errbuff = NULL; regfree(&compreg); return 0; }這是上面的輸出。我對模式的理解是它只有一個匹配組。根據文件，這意味著我應該看到兩個填充
regmatch_t元素：第一個包含整個字元串，第二個包含匹配的組。[chb]$ gcc -g -o foo foo.c [chb]$ ./foo "^ROOM\s{1}NAME:\s{1}([[:alpha:]]{6})" "ROOM NAME: BriCol" Trying to match with extended regex... Compiling successful. Regexec successful. Match 0; Beginning: 0 Match 0; End: 17 Match 1; Beginning: 11 Match 1; End: 17 Match 2; Beginning: 0 Match 2; End: 0 Match 3; Beginning: 0 Match 3; End: 0我什至註釋掉了
memset將數組清零的內容，認為regexec命令內部的邏輯在某種程度上受到了它的影響。
如果你想
regexec用 -1 填充未使用的數組元素，你需要首先告訴它有多少。在此行中替換matchcnt為：MATCH_CNTif((execret = regexec(&compreg, str, matchcnt, matches, 0)) == 0)所以它變成了
if((execret = regexec(&compreg, str, MATCH_CNT, matches, 0)) == 0)
int（並在之前添加缺少的compret = -2;內容）並且您的程序將按預期工作：$ ./549805 "^ROOM\s{1}NAME:\s{1}([[:alpha:]]{6})" "ROOM NAME: BriCol" Trying to match with extended regex... Compiling successful. Regexec successful. Match 0; Beginning: 0 Match 0; End: 17 Match 1; Beginning: 11 Match 1; End: 17 $您還應該初始化
errbuff為NULL，以避免free()使用未定義的值呼叫（何時errbuff從未初始化）。當你在做的時候，檢查malloc()的返回值——分配可能會失敗。