将结构指针作为函数参数传递（Seg错误未知）

Question

Header file #1 "city.h"

typedef struct City{

    double* lat;
    double* lon;
    double* pop;


    char* airport;
    char* name;


}City; 

Header file #2 "vector.h"

typedef struct Vector{


  City* cityArray[26];  // 26 pointers to struct City


 }Vector;

C file

#include "vector.h"
#include "city.h"
#include <stdlib.h>


void init(Vector *ptr) {

ptr->cityArray[0]->name = "hi"; // Error Seg Fault!

}

Hi, Your suggestion did work, but for some reason I'm getting Seg faults now even though the I'm 100% sure the code didn't change. Could you see whats wrong?

Answer 1

Try this -

  ptr->cityArray[0]->name = "hi";       // ptr is pointer to struct vector

As cityArray is a member variable of struct Vector , access it using a struct variable or pointer .

Can't do this cityArray[0]->name = "hi"; because cityArray is not any independent array of pointers.

EDIT

As you get segmentation fault , you need to allocate memory to struct pointer ptr as well as to char * present in struct city .

In function do it like this -

ptr=malloc(sizeof(Vector));
ptr->cityArray[0]->name=malloc(3);     // size 3 to store "hi" you can give desired size.

But remember to free allocated memory.

Answer 2

In your case, vector.h appears before city.h . You need to put city.h before vector.h to make the definition of city visible to vector .

That said, cityArray itself is not an independent variable , it is called a member variable. You need to have a variable of the structure type to make use of cityArray . Something like

Vector Vec;
Vec.cityArray[0]->name ....

and so on. Also make a note about the type of the variable and the initializer used. They both should match.