Structs, pointers and memory allocation

Question

//The struct
typedef struct {
    int nr;
    char *nume, **mesaj;
} utilizator;

void citire_mesaje(utilizator *user, int n)
{
    user->nr = malloc (n * sizeof(int));
    scanf("%d", &user[0].nr);
    printf("user[0].nr = %d\n", user[0].nr);
}
int main()
{
    int n;
    utilizator *user = malloc (sizeof(utilizator)); 
    citire_mesaje(user, n);

    return 0;
}

What am I doing wrong? I used user[0].nr just to test it more easily. I can get it work if I only use one element of type struct (utilizator user;), but I can't figure it out if I use a pointer. I get:

 warning: assignment makes integer from pointer without a cast [enabled by default]
 user->nr = malloc (n * sizeof(int));
           ^

Any suggestions?

you do not need user->nr = malloc (n * sizeof(int)) because it is already secured area. On the contrary nr update by scanf (is making an area that can not be released) memory leak. — BLUEPIXY
– BLUEPIXY, Commented Dec 7, 2013 at 9:22

Domi · Accepted Answer · 2013-12-07 09:34:50Z

1

It seems, nr is an array of int and not just int. Fix it's declaration:

typedef struct {
    int * nr;
    char *nume, **mesaj;
} utilizator;

If you only want one int, don't call malloc. It will be allocated as part of your utilizator object (funny word btw).

edited Dec 7, 2013 at 9:34

answered Dec 7, 2013 at 9:14

Domi

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3 Comments

Radu Gabitza Over a year ago

I want it to be just an int. I dont have to allocate memory for it?

Domi Over a year ago

This line user->nr = malloc (n * sizeof(int)); suggests that you want n ints, and not just one. If you just want one int, don't call malloc.

Abhineet Over a year ago

You don't have to do malloc for nr. You can use it directly by assigning a value.

Abhineet · Accepted Answer · 2013-12-07 09:52:56Z

First, you do not need to user->nr = malloc (n * sizeof(int)); because nr is just an int and it have its own memory space of length sizeof(int) bytes.

Secondly and most important,

You are not including the header file #include <stdlib.h>. Beacuse of this, the malloc function is implicitly declared to return int. Since you are treating the result as a pointer, there is a type-mismatch warning there.

All you have to do is, include the file stdlib.h.

void* will be safely promoted to any other pointer type in this case.

So, your code should be something like this::

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

typedef struct {
    int nr;
    char *nume, **mesaj;
} utilizator;

void citire_mesaje(utilizator *user, int n)
{
    scanf("%d", &user[0].nr);
    printf("user[0].nr = %d\n", user[0].nr);
    return ;

}
int main()
{
    int n = 0 ;
    utilizator *user = malloc (sizeof(utilizator)); 
    citire_mesaje(user, n);
    free( user ) ;

    return 0;
}

rullof · Accepted Answer · 2013-12-07 17:51:35Z

You are allocating the memory for your variables the wrong way. You should use the casting (change the type of the value returned by malloc to the type of the variable):

// In the citire_mesaje function
user->nr = (int) malloc (n * sizeof(int));
// In the main function
utilizator *user = (utilizator *) malloc (sizeof(utilizator));

Your final code should look like this:

//The struct
typedef struct {
    int nr;
    char *nume, **mesaj;
} utilizator;

void citire_mesaje(utilizator *user, int n)
{
    user->nr = (int) malloc (n * sizeof(int));
    printf("Enter a value for user[0].nr: ");
    scanf("%d", &user[0].nr);
    printf("user[0].nr = %d\n", user[0].nr);
}
int main()
{
    int n; // make sure to inisialize the n variable
    utilizator *user = (utilizator *) malloc (sizeof(utilizator));
    citire_mesaje(user, n);

    return 0;
}

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