#include

#include

//strlen()函数 int sub_strlen(const char *p) { int control = 0; do { if (p[control] != '\0') { control++; } } while (p[control] == '\0');

return control;

}

//strcmp()函数 int sub_strcmp(const char p1, const char p2) { int sum = 0; int control; do { sum += (p1[control] - p2[control]); control++; } while (p1[control] != '\0' && p2[control] != '\0');

return sum;

}

//strcpy函数（如何返回一个地址？？？） void sub_strcpy(char restrict p1,const char restrict p2) { int control = 0; do { p1[control] = p2[control]; } while (p2[control++] != '\0');

}

//strcat()函数(无返回地址) void sub_strcat(char p1,const char p2) { int count = 0; while (p1[count] != '\0') { count++; }

int control=0;
do
{
    p1[count] = p2[control];
    count++;
} while (p2[control++] != '\0');

}

//strchr()函数（无返回地址） void sub_strchr(const char *p1, int c) { int control = 0; while (p1[control] != '\0') { control++; }

int result = 1;
int count = 0;
while ((int)p1[count] != c) {
    if (count == control) {
        result = 0;
        break;
    } else {
        count++;
    }
}

switch (result) {
    case 1:    
        for (;count <=control;count++) {
            printf("%c",p1[count]);
        }
        break;
    default:
        printf("NULL");
        break;
}

} //strrchr()函数如上相似

int main(void) { char shuzu[1000] ="Hello hihihfkabj"; char shuzu2[] = "nimade"; sub_strchr(shuzu,'e'); printf("%s\n",strchr(shuzu,'e')); return 0; }