Freeing subordinates?

Question

typedef struct A
{
    int* whatever;  
} A;


int main(void)
{
    A* foo = (A)malloc(sizeof(A));
    foo->whatever = (int)malloc(sizeof(int));
    free(A); // leak? (foo->whatever)
    return 0;
}

Do I have to free each component of a struct / composite data type, or can I just free the struct?

Compile with all warning levels set to maximum. That should have told you assigning an int to a int* is invalid. — Zan Lynx, Nov 10 '14 at 21:03
You might like `talloc`. See http://stackoverflow.com/a/3065381/13422 — Zan Lynx, Nov 10 '14 at 21:05

score 3 · Accepted Answer · answered Nov 10 '14 at 21:01

3

Anything that is malloc'd needs to be freed

answered Nov 10 '14 at 21:01

Lawrence Aiello

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score 1 · Answer 2 · answered Nov 10 '14 at 21:01

1

For each malloc in a program, there must be a free. That's the rule, plain and simple.

Since you malloc twice, free twice.

answered Nov 10 '14 at 21:01

nneonneo

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And do it in the correct order. Last memory malloced should be first memory freed. – Zan Lynx Nov 10 '14 at 21:02
@ZanLynx: That's less important. Objects can outlive other objects. – nneonneo Nov 10 '14 at 21:02
Not when they are nested. If the container object is free'd first then it is invalid to use its pointer to get at the pointer inside. – Zan Lynx Nov 10 '14 at 21:04
@ZanLynx: Yes, but that's not a _general_ rule about `malloc` and `free`. You could just as easily copy the contained object's pointer somewhere and use that to free it later. – nneonneo Nov 10 '14 at 21:07
It's important in this case...if you free the struct first, the pointer inside is no longer valid and can't be freed. – Lee Daniel Crocker Nov 10 '14 at 21:07

Freeing subordinates?

2 Answers2