c - 如何确定在函数参数中使用指针或普通变量？

Question

#include <stdio.h>
#include <string.h>
#include <stdlib.h>

struct contact {
    char name[30];
    int phone_number;
    char address [30];
};

int main(int argc, char **argv) {
    struct contact friend;
    strcpy(friend.name, "Jane Doe");
    friend.phone_number = 377177377;
    strcpy(friend.address, "3771 University Avenue");

    char *name;
    int number;
    char *address;

    update_contact(&friend, name, number, address);
    return 0;

}

我需要实现update_contact函数来更新联系人信息。 friend 和 number 是普通变量。 *name 和*address 是指针。但是函数调用也使用了指针friend的地址，即&friend。现在我很困惑我应该在函数参数中放什么。

我试着把指针放进去。

void update_contact (struct *c, char *name, int number, char *address) {
    c->name = name;
    c->phone_number = number;
    c->address = address;
}

然而，这会带来一堆错误，比如

error: request for member 'address' in something not a structure or union
&c->address = address;

我该如何解决这个问题？谢谢

这是固定版本。感谢 Thomas Jager 和 Some Name。

void update_contact (struct contact *c, char *name, int number, char *address) {
    strcpy(c->name, name);
    c->phone_number = number;
    strcpy(c->address, address);
}

Answer 1

正如 David C. Rankin 所建议的:

您需要初始化作为参数传递的变量。例如

char *name = "Donald Trump";
int number = 01010101010;
char *address = "White House;

---

char name[] = "Donald Trump";
int number = 01010101010;
char address[] = "White House;

---

或者你可以在调用函数时使用字面量和常量

  update_contact(&friend, "Donald Trump", 0101010010, "White House");

---

void update_contact (struct contact *c, char *name, int number, char *address) {
    strcpy(c->name, name);
    c->phone_number = number;
    strcpy(c->address, address);
}