Impossible to use malloc and typedef

Question

It seems that it is impossible to use typedef and incorporate a malloc function in my program: More specifically:

groups.h

#include "headers.h"
#include "info.h"
#include "sub_list.h"

typedef struct group_entity *group;

group new_group(int id);

void free_group(group group);

groups.c

#include "groups.h"

struct group_entity {
    int gId;
    info gFirst;
    info gLast;
    subinfo gSub;
};

group new_group(int id) {
    group new_group = malloc(sizeof(group));
    if (!new_group) {
        return NULL;
    }
    new_group->gId = id;
    new_group->gFirst = NULL;
    new_group->gLast = NULL;
    new_group->gSub = NULL;
    return new_group;
}


void free_group(group group) {
    free(group);
}

header.h

#pragma once
#include <stdlib.h>
#include <stdio.h>
#include <malloc.h>

#define MG 5 /* Length of Groups Array. */

main.c

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

#include "headers.h"
#include "groups.h"
#include "info.h"
#include "sub_list.h"

int main() {
    group new_object = new_group(5);
    return 0;
}

Whenever I run this piece of code, I get halted with a breakpoint

exe_common.inl - ln 297.

    if (!has_cctor)
        _cexit();

I've tried using a free_group(group group) function in my main.c - but it seems that I hit a breakpoint. By using the VS 2019 debugger, I tried to find out which line causes the error. It seems that the breakpoint is caused AFTER my main is executed. Quite curious behavior, since main won't even return 0.

`void free_group(group group);` While it might work, I would call this a terrible idea to use identical names for types and variables. Already typedef'ing a pointer type is really bad idea. It confuses the reader by hiding important information. — Gerhardh, Nov 23 '22 at 07:07
`malloc(sizeof(group));` This only allocates size of a pointer, not size of the struct. That is why hiding pointers is not recommended. — Gerhardh, Nov 23 '22 at 07:09
See [Is it a good idea to typedef pointers?](https://stackoverflow.com/q/750178/15168) — TL;DR — The answer's usually "No" with a possible exception for pointers to functions. — Jonathan Leffler, Nov 23 '22 at 13:52

score 5 · Answer 1 · answered Nov 23 '22 at 07:08

Typedef pointers are confusing. sizeof(group) is the size of the pointer not the size of the data. Consider using a typedef to the data, and adding * everywhere you are using a pointer, to indicate, clearly, that you are working with pointers.

typedef struct group_entity group;
group *new_group(int id);
void free_group(group *group);

group *new_group(int id) {
    group *new_group = malloc(sizeof(group));

Alternatively, you can do sizeof(*new_group) or sizeof(struct group_entity).

probably the best idea is not to use types in malloc at all. Better use the pointer variable. `group new_group = malloc(sizeof *new_group);` — tstanisl, Nov 23 '22 at 11:55

Gabriel GRONDIN - GGLinnk · Answer 2 · 2022-11-25T05:14:30.807

In addition to what KamilCuk said here, I would say that you can use a define directive in your groups.h file if your intent was to avoid to specify the * everytime.

typedef struct group_entity group;
#define *group GROUP_PTR
GROUP_PTR new_group(int id);

Then in your groups.c replace with the following:

GROUP_PTR new_group(int id) {
  GROUP_PTR new_group = malloc(sizeof(group));
  [...]
  return new_group;
}

It should work like a charm.

Impossible to use malloc and typedef

2 Answers2