Struct Memory Allocation in C

Question

Does allocating a struct instance using malloc have any effect on the allocation of its members? Meaning that, is doing something like this:

typedef struct { int *i; } test; int main() { test *t = malloc(sizeof(test)); t->i = malloc(sizeof(int)); }

... is meaningless because i should be already on the heap, right ?

Or, the struct is just conceptual to help the programmer group two completely separate variables floating in memory? Meaning that: test *t = malloc(sizeof(test)) just allocates the memory for storing pointers to the members on the heap?

I'm quite baffled about this ..

t->i = malloc( sizeof(int) ); should give you a compiler warning that you are casting a pointer to an int. You do not need to allocate the memory for the int on the heap unless the struct was defined as struct { int* i; } — clcto
– clcto, Commented Oct 22, 2014 at 17:35

Mureinik · Accepted Answer · 2014-10-22 17:50:55Z

The i field of test is a primitive int, and mallocing test will take care of the memory needed by it. Assigning a result of malloc to i should produce a warning, as you're implicitly converting a pointer to an int.

EDIT:
As pointed out in the comments, the question has been edited since the above answer was posted. If your struct contains a pointer, mallocing the struct will allocate the memory to hold a pointer, but will not allocate the memory this pointer points to. For that, you'll need a second malloc call (e.g., test->i = malloc (sizeof (int))).

FYI this answer appears to be wrong due to edits to the question.

John Bollinger · Accepted Answer · 2014-10-22 17:42:36Z

Given a struct type "test", malloc(sizeof(test)) allocates space for a complete struct, including all members and any padding between. You are perhaps confused about the case where one or more of the members is a pointer:

typedef struct { char *string; char array[8]; } test;

In that case, the pointer itself is a member, and space is allocated for it, but the thing pointed to is not a member, and no space is allocated for that.

Contrast that with the case where the struct has an array member: the whole array is a member of the struct, and sufficient space is allocated in the struct for all its elements.

So if I have a struct that contains a pointer to a struct of some other kind, and I allocate it using malloc, then I'd have to use malloc again on the member struct so that it is allocated dynamically too, right ?
Yes, the sizeof the container struct (determining how much memory to allocate) includes space for the inner pointer, but not for a separate "inner" struct that could be pointed to. If you don't want to point to an independent, pre-existing inner struct (as you might want in a linked list, for example) then you need to allocate memory separately for such an inner struct.

csnate · Accepted Answer · 2014-10-22 17:36:21Z

You would only allocate the memory for i if i was a pointer to an int. See this example:

#include <stdlib.h> typedef struct { int i; } test; typedef struct { int* i; } testp; int main() { test* foo = malloc(sizeof(*foo)); foo->i = 3; testp* bar = malloc(sizeof(*bar)); bar->i = malloc(sizeof(*bar->i)); *bar->i = 3; }

When you allocate the struct, there is already space allocated for the int i.

user3767013 · Accepted Answer · 2014-10-22 17:53:28Z

When you call the first malloc, there is enough memory to hold the whole structure, including the '*i', which has enough place to hold a pointer to an integer. The field *i however remains uninitialized! so you must not use *i before you assigned it with the second malloc that reserves memory for an integer.

Collectives™ on Stack Overflow

Struct Memory Allocation in C

4 Answers 4

1 Comment

2 Comments

Comments

Comments

Linked

Hot Network Questions

Collectives™ on Stack Overflow

4 Answers 4

1 Comment

2 Comments

Comments

Comments

Linked

Related