strstr()函式工作原理與自寫實現
阿新 • • 發佈:2019-02-09
#include <stdio.h> #include <string.h> void putOneByOneP(char *s); char *strstrM1 (char str[], char mod[]); char *strstrM2 (char str[], char mod[]); int main(void) { char s[] = "What's your name?"; char s2[] = "name"; putOneByOneP(s);//列印每個字元地址,以便檢查strstr()函式是否正確工作 char *p = strstrM2(s, s2);//strstr()函式返回索引,以下校驗的結果 if(p != NULL) printf("%s找到了%s!返回地址%p,儲存的是 %c\n", s, s2, p, *p); else printf("%s未找到%s!\n",s, s2); printf("%s", (p != NULL) ? "真的找到了!\n":"真的沒找到!\n");//較為簡明的檢驗 return 0; } void putOneByOneP(char *s) { char *p =s;//最後要用s初始值,所以用p拷貝一份 while (*s != '\0') { printf("[%c] %p\n", *s, s); ++s; } } char *strstrM1 (char str[], char mod[]) { char *s = str, *m = mod;//沒必要的城市化,但能明確意圖 for (;*str != '\0'; ++str) { for (s = str, m = mod; *m != '\0' && *s == *m; ++s,++m) ; if (*m == '\0') return (char *)str; } return NULL; } char *strstrM2 (char str[], char mod[]) { int i, j, k; for (i = 0; str[i]!= '\0'; ++i) { for (j = 0,k = i; mod[j] != '\0' && str[k] == mod[j]; ++j, ++k) ; if (mod[j] == '\0') return &str[i]; } return NULL; }