[C] Pointer member in structure confusion

[yoonkwun]

Hi, I have a typedef'd struct with a pointer member:

typedef struct
{
    int *i;
} TEMP;

I want to set the VALUE of this member:

TEMP temp;
*(temp.i) = 42;

I would think that temp.i would be a pointer, then I would dereference it with *, then set the value. The code compiles but the program crashes at that point. Can someone explain what I'm doing wrong?

Thanks.

[R0mai]

i points to a garbage address in the memory. You're going to need to allocate memory for i.

[yoonkwun]

Ah right, silly mistake. Thanks.

[yoonkwun]

Another Q. If I wanted to use malloc, would I have to malloc each member in the struct, or can I just malloc the pointer to the struct? The latter seems to crash on me:

TEMP *temp;
temp = malloc(sizeof(*temp));

*(temp->i) = 42;
printf("%d", *(temp->i));

But, after experimenting I found out that it won't crash if I just malloc the member and not the struct:

TEMP *temp;
temp->i = malloc(sizeof(*temp->i));

*(temp->i) = 42;
printf("%d", *(temp->i));

It would be a pain if I had to malloc a struct with a lot of pointers, so can somebody tell me if I'm doing something wrong or if there's an easier way?

[firedraco]

You must malloc each member separately.

If you want to do it for all objects of your type, and you are using C++, then you can just use a constructor to do it for you.

[R0mai]

If you have to use C, you can still imitate C++:

typedef struct
{
    int *i; 
    /*Other members*/
} TEMP;

void construct_TEMP(TEMP *t) { 
    t->i = malloc(sizeof(int));
    /*Init other members*/
}

void destruct_TEMP(TEMP *t) { 
    t->i = free(t->i);
    /*free other members*/
}

int main() {
    TEMP t;
    construct_TEMP(&t);
    /*use t*/
    destruct_TEMP(&t);
    return 0;
}

EDIT: fixed code, thx yoonkwun and Disch

[yoonkwun]

Thanks, I have just two more questions. If the structure contained non-pointer members, do I have to malloc them in the same way I would malloc a pointer? Also, in your code R0mai, I noticed in line 8:

t->i = malloc(sizeof(t->i));

It would return a pointer to a memory block of 4 bytes, since that's the size of a pointer right? Isn't it supposed to be:

t->i = malloc(sizeof(*t->i));

so that it returns a memory block of how big t->i is, and not how big a pointer is?

[Disch]

t->i = malloc(sizeof(t->i));

This code isn't right.

t->i is a pointer, so you're allocating the size of a pointer, which may not be the size of an int.

The right way to do it:

t->i = malloc(sizeof(int) * however_many_ints_you_want_to_allocate);

[R0mai]

If the structure contained non-pointer members, do I have to malloc them in the same way I would malloc a pointer?

No, you don't have to allocate those.

[jsmith]

I would advise two things:

1) use new instead of malloc;
2) declare the struct as:

struct TEMP {
   int* i;
};

The way you've declared it is a bad C-ism that you should avoid in C++.

[Disch]

Based on the code and thread title I assumed he was coding C.