malloc an array of struct pointers, need help please

Question

I am trying to use malloc to allocate heap memory for an array of struct pointers, but I cannot get it to work. Below is my code, but when I compile with gcc, I got errors like this "error: invalid type argument of ‘->’ "

The array I want to set up an array of mystruct_pointer, which should point to the actual __mystruct_t, and I think I can use "->" on its member field. Where is wrong with my code? I think it should work. thanks

typedef struct
{
    int id;
    bool status;
} __mystruct_t;

typedef __mystruct_t* mystruct_pointer;

mystruct_pointer struct_ptr_array;

void my_init(int number)
{
    struct_ptr_array = (mystruct_pointer) malloc(sizeof(__mystruct_t) * number);

    int i;
    for (i = 0; i < number; i++)  /* initialize the array of struct pointers */
    {
         struct_ptr_array[i]->id = i;
         struct_ptr_array[i]->status = false;
     }
}

yinqiwen · Accepted Answer · 2014-02-25 07:15:39Z

4

Replace '->' by '.'. Since 'struct_ptr_array[i]' already dereference the pointer.

answered Feb 25, 2014 at 7:15

yinqiwen

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Comments

Lundin · Accepted Answer · 2014-02-25 07:31:12Z

2

Problems like these come from doing obscure things. You typedef-hide a pointer and instantly you have made the program unreadable to yourself. And that's the only thing the typedef achieved. So never hide pointers with typedefs, it is very bad practice.

Other issues:

Avoid double underscore because that's reserved for compiler identifiers.
Don't cast the result of malloc, because doing so is completely pointless in C and also potentially dangerous on old C compilers.
Handle the case where malloc fails because there's no heap space available.
Never use variables with global scope, it is very bad practice that leads to spaghetti code. And there is never a reason to do so in C.

The code should be fixed as follows:

typedef struct
{
    int id;
    bool status;
} mystruct_t;



static mystruct_t* struct_ptr_array;

void my_init(int number)
{
    struct_ptr_array = malloc(sizeof(mystruct_t) * number);

    if(struct_ptr_array == NULL)
    {
        handle_error();
        return ;
    }

    for (int i = 0; i < number; i++)  /* initialize the array of struct pointers */
    {
        struct_ptr_array[i].id = i;
        struct_ptr_array[i].status = false;
    }
}

answered Feb 25, 2014 at 7:31

Lundin

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3 Comments

Subhajit Over a year ago

Can you please explain why do you stick to "mystruct_t*" instead of taking "mystruct_t**" for array of pointers?

Lundin Over a year ago

@Subhajit There is no array of pointers here.... and there is no need for one either.

Subhajit Over a year ago

understood,struct_ptr_array holds the address of the array of structures and struct_ptr_array[i] denotes one single structure instance,correct me if I am wrong

user707650 · Accepted Answer · 2014-02-25 07:15:10Z

1

struct_ptr_array is a pointer, but you index it, so you get an actual __mystruct_t, not the pointer. Thus, simply use:

     struct_ptr_array[i].id = i;
     struct_ptr_array[i].status = false;

answered Feb 25, 2014 at 7:15

user707650

Comments

quetzalcoatl · Accepted Answer · 2014-02-25 10:16:16Z

1

Replace two lines by

struct_ptr_array[i].id = i;
struct_ptr_array[i].status = false;

edited Feb 25, 2014 at 10:16

quetzalcoatl

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answered Feb 25, 2014 at 9:01

Amit

915 bronze badges

1 Comment

quetzalcoatl Over a year ago

Would you mind paying a bit more attention to formatting? btw. also, don't get me wrong, but your answer seems like a copy from already-existing Evert's answer. I don't accuse you, but if you are doing it, that's not a good way to 'score'.

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