C: Struct pointer-array

Question

I'm trying to create and initialise a pointer to an array of pointers to structs. This array will be passed around to many parts of my program.

This is my code:

file.h

#ifndef FILE_H
#define FILE_H

typedef struct Object Object;

void init(Object*** objs);

#endif

file.c

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

struct Object
{
    int a, b;
};

void init(Object*** objs)
{
    *objs = malloc(5 * sizeof(Object*));

    for(int i = 0; i < 5; i++)
    {
        *objs[i] = malloc(sizeof(struct Object));
        *objs[i] -> a = i;     // arbitrary member access
        *objs[i] -> b = i * 2; // arbitrary member access
    }

}

main.c

#include "file.c"

int main(int argc, char** argv)
{
    Object** prog_objs;

    init(&prog_objs);
    // should now have a pointer to an array to pass around

    for(int i = 0; i < 5; i++)
    {
        printf("Obj: %d, %d\n", prog_objs[i] -> a, prog_objs[i] -> b);
    }

    return 0;
}

I'm not entirely sure why this ins't working. I'm positive my main() is correct and the issue is somewhere in the init() function. I've tried a variety of different ways to initialise the array elements to be struct pointers, but I keep getting compile errors or segfaults.

Any advice on where my issues are is much appreciated! Thank you

melpomene · Accepted Answer · 2018-12-08 03:22:28Z

1

*objs[i] is a precedence error. It parses as *(objs[i]).

It should be (*objs)[i] instead.

answered Dec 8, 2018 at 3:22

melpomene

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1 Comment

user9506231 Over a year ago

This has fixed it! Thanks very much

Ruks · Accepted Answer · 2018-12-08 03:29:08Z

1

*objs[i] does not mean what you think it means, it means:

*(objs[i])

Dereferencing the element at position i... But what you need is:

(*obj)[i]

Which means dereferencing obj, then accessing the dereferenced member...

As to why this happens, this is due to the [] operator having higher precedence than * operator... So the [] operator evaluates first and then the * operator does...

Another way you can solve your problem is by:

obj[0][i]

answered Dec 8, 2018 at 3:29

Ruks

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Comments

H.S. · Accepted Answer · 2018-12-08 03:47:05Z

This question is already answered by @melpomene as the problem is occurring due to the operator precedence.

Since the init() function is doing allocation as well as initialization work, to avoid such kind of problems, you can modify the init() to return Object** type instead of taking Object*** objs as parameter. You can do:

Object** init(void)
{
    Object** objs;
    objs = malloc(5 * sizeof(Object*));
    if (objs == NULL) {
       fprintf(stderr, "Failed to allocate memory");
       exit(EXIT_FAILURE); //You can  return NULL from here and handle it in calling function
    }

    for(int i = 0; i < 5; i++)
    {
        objs[i] = malloc(sizeof(struct Object));
        if (objs[i] == NULL) {
           fprintf (stderr, "Failed to allocate memory");
           exit(EXIT_FAILURE);
        }
        objs[i] -> a = i;     // arbitrary member access
        objs[i] -> b = i * 2; // arbitrary member access
    }
    return objs;
}

In the calling function, you can do:

Object** prog_objs = init();

Also, follow good programming practice. Always check malloc return and ensure to free the allocated memory once you are done with it.

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