C中的结构内存分配

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 ..

Answer 1

The i field of test is a primitive int , and malloc ing 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, malloc ing 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 (eg, test->i = malloc (sizeof (int)) ).

Answer 2

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 .

Answer 3

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.

Answer 4

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.