Is this hack valid according to standard? - c

Question

This is just like struct hack.
Is it valid according to standard C?
// error check omitted!
typedef struct foo {
void *data;
char *comment;
size_t num_foo;
}foo;
foo *new_Foo(size_t num, blah blah)
{
foo *f;
f = malloc(num + sizeof(foo) + MAX_COMMENT_SIZE );
f->data = f + 1; // is this OK?
f->comment = f + 1 + num;
f->num_foo = num;
...
return f;
}

Answer 1

Yes, it's completely valid. And I would strongly encourage doing this when it allows you to avoid unnecessary additional allocations (and the error handling and memory fragmentation they entail). Others may have different opinions.
By the way, if your data isn't void * but something you can access directly, it's even easier (and more efficient because it saves space and avoids the extra indirection) to declare your structure as:
struct foo {
size_t num_foo;
type data[];
};
and allocate space for the amount of data you need. The [] syntax is only valid in C99, so for C89-compatibility you should use [1] instead, but this may waste a few bytes.

Answer 2

The line you question is valid - as others have said.
Interestingly, the next line, which you did not query, is syntactically valid but is not giving you the answer you want (except in the case where num == 0).
typedef struct foo
{
void *data;
char *comment;
size_t num_foo;
} foo;
foo *new_Foo(size_t num, blah blah)
{
foo *f;
f = malloc(num + sizeof(foo) + MAX_COMMENT_SIZE );
f->data = f + 1; // This is OK
f->comment = f + 1 + num; // This is !!BAD!!
f->num_foo = num;
...
return f;
}
The value of f + 1 is a foo * (implicitly coerced into a void * by the assignment).
The value of f + 1 + num is also a foo *; it points to the num+1th foo.
What you probably had in mind was:
foo->comment = (char *)f->data + num;
Or:
foo->comment = (char *)(f + 1) + num;
Note that while GCC will allow you to add num to a void pointer, and it will treat it as if sizeof(void) == 1, the C Standard does not give you that permission.

Answer 3

That is an old game, though the usual form is like
struct foo {
size_t size
char data[1]
}
and then allocate the space as big as you want and use array as if it had the desired size.
It is valid, but I would encourage you to find another way if possible: there are lots of chance to screw this up.

Answer 4

Yes, the general idea of the hack is valid, but at least as I read it, you haven't implemented it quite correctly. This much you've done right:
f = malloc(num + sizeof(foo) + MAX_COMMENT_SIZE );
f->data = f + 1; // is this OK?
But this is wrong:
f->comment = f + 1 + num;
Since f is foo *, the f+1+num is computed in terms of sizeof(foo) -- i.e., it's equivalent to saying f[1+num] -- it (attempts to) index to the 1+numth foo in an array. I'm pretty sure that's not what you want. When you allocate the data, you're passing sizeof(foo)+num+MAX_COMMENT_SIZE, so what you're allocating space for is num chars, and what you (presumably) want is to point f->comment to a spot in memory that's num chars after f->data, which would be more like this:
f->comment = (char *)f + sizeof(foo) + num;
Casting f to a char * forces the math to be done in terms of chars instead of foos.
OTOH, since you're always allocating MAX_COMMENT_SIZE for comment, I'd probably simplify things (quite) a bit, and use something like this:
typedef struct foo {
char comment[MAX_COMMENT_SIZE];
size_t num_foo;
char data[1];
}foo;
And then allocate it like:
foo *f = malloc(sizeof(foo) + num-1);
f->num_foo = num;
and it'll work without any pointer manipulation at all. If you have a C99 compiler, you can modify this slightly:
typedef struct foo {
char comment[MAX_COMMENT_SIZE];
size_t num_foo;
char data[];
}foo;
and allocate:
foo *f = malloc(sizeof(foo) + num);
f->num_foo = num;
This has the additional advantage that the standard actually blesses it, though in this case the advantage is pretty minor (I believe the version with data[1] will work with every C89/90 compiler in existence).

Answer 5

Another possible problem might be alignment.
If you simply malloc your f->data, then you can safely e.g. convert your void* to double* and use it to read/write a double (provided that num is sufficiently large). However, in your example you can no longer do that, as f->data might not be properly aligned. For example, to store a double in f->data, you will need to use something like memcpy instead of a simple typecast.

Answer 6

I'd rather use some function to allocate the data dynamically and free it correctly instead.
Using this trick only saves you the trouble of initializing the data structure, and can lead to very bad problems (see Jerry's comment).
I'd do something like this:
typedef struct foo {
void *data;
char *comment;
size_t num_foo;
}foo;
foo *alloc_foo( void * data, size_t data_size, const char *comment)
{
foo *elem = calloc(1,sizeof(foo));
void *elem_data = calloc(data_size, sizeof(char));
char *elem_comment = calloc(strlen(comment)+1, sizeof(char));
elem->data = elem_data;
elem->comment = elem_comment;
memcpy(elem_data, data, data_size);
memcpy(elem_comment, comment, strlen(comment)+1);
elem->num_foo = data_size + strlen(comment) + 1;
}
void free_foo(foo *f)
{
if(f->data)
free(f->data);
if(f->comment)
free(f->comment);
free(f);
}
Note that I did no check on data validity, and my alloc can be optimized (replacing strlen() calls by a stored lenght value).
It seems to me that this behavior is more secure... at the price of a disseminated data chunk maybe.