c - 将结构体数组作为指针传递

Question

我想将结构体数组作为打印函数的参数传递，然后访问其成员进行打印。为什么当我不传递任何指针时会出现指针错误？

在main.c中:

struct city {
    double longitude;
    double latitute;
    char name[buf_size]; 
};
int numCitToRead = 10;
struct city cities[25];
printCities(&numCitToRead, cities);

注意:Struct 数组在文件解析函数中初始化。它总是 25 个字段长，但如果 numCitToRead 为 10，则只会填充 10 个字段

int printCities(int* t_numCitToRead, struct city t_cities[25]) {

    for (unsigned short i = 0; i < *t_numCitToRead; i++) {
        printf("\n\n\tCity %d: ", i+1);
        printf("\nname:\t\t%s", t_cities[i].name);
        printf("\nlongitude:\t%f", t_cities[i].longitude);
        printf("\nlatitude:\t%f", t_cities[i].latitute);
    }
    return 0;
}

希望有人能帮助我!

问候

Answer 1

已经修改了代码并且现在可以工作了。 让我们尝试一下:

#include <stdio.h>

struct city {
    double longitude;
    double latitute;
    char *name; 
  };

int printCities(int* t_numCitToRead, struct city t_cities[25]) {

    for (unsigned short i = 0; i < *t_numCitToRead; i++) {
    printf("\n\n\tCity %d: ", i+1);
        printf("\nname:\t\t%s", t_cities[i].name);
        printf("\nlongitude:\t%f", t_cities[i].longitude);
        printf("\nlatitude:\t%f", t_cities[i].latitute);
    }
    return 0;
}

int main() {
  int numCitToRead = 10;
  struct city cities[25];
  // create dummy data
  for(int i =1; i<=25; i++)
  {
   cities[i-1].name = "name";
   cities[i-1].longitude =  10 * i;
   cities[i-1].latitute = 10 * 20;
  } 
 printCities(&numCitToRead, cities);
 return 0;
}

谢谢!