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strstr()函式工作原理與自寫實現

#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;
}