From Newsgroup: comp.lang.c

Kaz Kylheku <643-408-1753@kylheku.com> writes:

Jonas Lund of https://whizzter.woorlic.org/ mentioned this
trick in a HackerNews comment:

Given:

struct S {
// ...
T A[];
};

Don't do this:

malloc(offsetof(S, A) + n * sizeof (T));

But rather this:

malloc(offsetof(S, A[n]));

It's easy to forget that the second argument of offsetof is a
designator, not simply a member name.

Clever but inadvisable. Even if the offsetof() expression compiles,
in most cases it either gives a value that is too small or is
undefined behavior.
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From Newsgroup: comp.lang.c

pozz <pozzugno@gmail.com> writes:

Il 08/10/2025 08:35, Kaz Kylheku ha scritto:

Jonas Lund of https://whizzter.woorlic.org/ mentioned this
trick in a HackerNews comment:

Given:

struct S {
// ...
T A[];
};

Don't do this:

malloc(offsetof(S, A) + n * sizeof (T));

But rather this:

malloc(offsetof(S, A[n]));

It's easy to forget that the second argument of offsetof is a
designator, not simply a member name.

struct S {
unsigned int size;
unsigned char mode;
unsigned char array[];
}

In a 32-bits integer machine, sizeof(struct S) is 8, because there are
3 bytes of padding after mode and array is considered empty.

Now I want to store 9 bytes in array[]. I could use:

malloc(sizeof(struct S) + 9 * sizeof(unsigned char))=malloc(17)

or I could use:

malloc(offsetof(struct S, array[9]))=malloc(14)

Is the second better (and correct) than the first?

And another question. Suppose I need an array of struct S. All
elements have 7-bytes array[] member. How to allocate this array and
access each element?

I think I can't use the first malloc (17), neither the second
(14). Both aren't a multiple of alignment length.

Right. Getting the (minimum) size exactly right is tricky enough
to merit writing a macro to compute the desired value.
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From Newsgroup: comp.lang.c

Keith Thompson <Keith.S.Thompson+u@gmail.com> writes:

BGB <cr88192@gmail.com> writes:

On 10/8/2025 1:35 AM, Kaz Kylheku wrote:

Jonas Lund of https://whizzter.woorlic.org/ mentioned this
trick in a HackerNews comment:
Given:
struct S {
// ...
T A[];
};
Don't do this:
malloc(offsetof(S, A) + n * sizeof (T));
But rather this:
malloc(offsetof(S, A[n]));
It's easy to forget that the second argument of offsetof is a
designator, not simply a member name.

This is assuming offsetof and can deal with general expressions (vs
just field names). IIRC, it is only required to work with field names
(and with plain structs).

I just read that part of the standard, and it's not clear whether
the second argument to offsetof() has to be a member name or whether
it can be something more elaborate.

Quoting the N3096 draft of C23, 7.21:

offsetof(type, member-designator)

which expands to an integer constant expression that has type
`size_t`, the value of which is the offset in bytes, to the
subobject (designated by *member-designator*), from the beginning
of any object of type *type*. The type and member designator
shall be such that given

static type t;

then the expression &(t. *member-designator*) evaluates to
an address constant. If the specified *type* defines a new
type or if the specified member is a bit-field, the behavior
is undefined.

The requirements imply that the type can be a struct or a union.

The term "member designator" is not used elsewhere in the standard.
If the term to be taken literally, then it has to designate a
*member*, not an element of a member. But the term "subobject",
along with the address constant requirement, could imply that it
could be an arbitrary sequence of members and array elements.

Clearly the italicized token member-designator is just a placeholder
with no semantics implied, and the controlling text here is the word "subobject", which applies recursively.

Note also that the phrase "from the beginning of any object of type
*type*" precludes the use of 'offsetof(S, A[n])', if n is too large,
since A[n] will not be a subobject of an arbitrary object of type
*type*.
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