访问动态分配的结构指针数组的成员

Question

I need to have a global dynamic array of pointers, in which I will store my structs, beacuse later I will need to iterate through this array to list all the stored information, I also need to be able to read the name , age and job variables from the console, and store them in a person_t in the iterator array.

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

typedef struct Person
{
    char name[30];
    int age;
    char job[30];
} person_t;

person_t **iterator;
int capacity = 10;
int size = 0;

int main()
{
    int i;
    *iterator = (person_t *)malloc(capacity * sizeof(person_t));
    for (i = 0; i < capacity; ++i)
    {
        person_t p;
        p.age = i;
        *iterator[i] = p;
    }
    return 0;
}

I get no errors/warnings compiling this code (gcc -ansi -pedantic -Wall -Wextra), but when I try to run it, I get a Segmentation fault immediately.

Answer 1

When you do this:

*iterator = (person_t *)malloc(capacity * sizeof(person_t));

You're deferencing iterator , however as a file-scope pointer variable it's initialized to NULL. Attempting to dereference a NULL pointer invokes undefined behavior .

I suspect what you really want is an array of structs, not an array of pointers to structs. That being the case, define iterator as:

person_t *iterator;

Then you allocate memory for it like this:

iterator = malloc(capacity * sizeof(person_t));

Then assign to array elements like this:

iterator[i] = p;

Answer 2

Your stated purpose is to create a "global dynamic array of pointers, in which I will store my structs" . The following modification of your code (see comments) will do this:

person_t p[10] = {0};

int main()
{
    int i;
    // with declaration: person_t **iterator = NULL;, 
    //following is all that is needed to create an array of pointers:
    iterator = malloc(capacity * sizeof(person_t *));//no need to cast return of malloc
   
        
        for (i = 0; i < capacity; ++i)
        {
            //person_t p;//moved to scope that will exist outside of main()
            p[i].age = i;
            iterator[i] = &p[i];//assign the address of the object to the pointer
                             //iterator[i] is the ith pointer in a collection of 
                             //pointers to be assigned to point to 
                             //instances of struct person_t 
        }
        //Once all fields are populated (to-do), the following will display the results:
        for (i = 0; i < capacity; ++i)
        { 
           printf("%d) Name: %s Age: %d  Job: %s\n",  i, iterator[i]->name,iterator[i]->age,iterator[i]->job);
        }

    return 0;
}

Answer 3

you are not allocating memory correctly

First you need to allocate memory for a pointer which can store capacity number of address ie done through iterator = malloc(capacity * sizeof(person_t*)); and then you need to allocate memory for holding each structure element ie iterator[i] = malloc(sizeof(person_t));

all the malloc 'ed memory should be free 'd once we are done with it.

Also, have not done the error check for malloc 's, that is left as an exercise for you.

int main()
{
    int i;
    // test data
    char *names[] = {"ABC", "DEF"};
    char *jobs[] = {"Accountant", "Security"};
    int ages[] = {50, 60};
    
    // first allocate memory for iterator , which can hold pointers to store iterator poniters
    iterator = malloc(capacity * sizeof(person_t*)); 
    
    for (i = 0; i < capacity; ++i)
    {
        // now allocate memory for individual iterator
        iterator[i] = malloc(sizeof(person_t)); 
        strcpy(iterator[i]->name,names[i]);
        iterator[i]->age = ages[i];
        strcpy(iterator[i]->job, jobs[i]);
    }
    
    for (i = 0; i < capacity; ++i)
    {
        printf("name = %s ", iterator[i]->name);
        printf("Age = %d ", iterator[i]->age);
        printf("Job = %s\n", iterator[i]->job);
    }
    return 0;
}