sizeof and incomplete types

[aegis]

sizeof could not possibly evaluate to some size for a type, unless
two things for a type occur:

a) the type is complete and sizeof will evaluate to appropriate size
b) type is not complete and any sizeof use with incomplete type
results in undefined behavior.

Is 'b' so? If so, where in the standard does it say this?

I only bring this up because of the following:

Breakpoint 1, main () at malloc.c:9
9 foo = malloc(sizeof foo);
(gdb) n
11 return 0;
(gdb) p foo
$1 = 0x80495d8
(gdb) p sizeof foo
$2 = 4
(gdb) list malloc.c:1,15
1 #include <stdlib.h>
2
3 typedef struct foo *FOO;
4
5 int main(void)
6 {
7 FOO foo;
8
9 foo = malloc(sizeof foo);
10
11 return 0;
12 }
(gdb)
--
aegis

[Keith Thompson]

"aegis" <ae***@mad.scie ntist.com> writes:

sizeof could not possibly evaluate to some size for a type, unless
two things for a type occur:

a) the type is complete and sizeof will evaluate to appropriate size
b) type is not complete and any sizeof use with incomplete type
results in undefined behavior.

Is 'b' so? If so, where in the standard does it say this?

I only bring this up because of the following:

Breakpoint 1, main () at malloc.c:9
9 foo = malloc(sizeof foo);
(gdb) n
11 return 0;
(gdb) p foo
$1 = 0x80495d8
(gdb) p sizeof foo
$2 = 4
(gdb) list malloc.c:1,15
1 #include <stdlib.h>
2
3 typedef struct foo *FOO;
4
5 int main(void)
6 {
7 FOO foo;
8
9 foo = malloc(sizeof foo);
10
11 return 0;
12 }
(gdb)

foo is a pointer object. struct foo is an incomplete type, but
struct foo* (aliased as FOO) isn't.

Your use of the identifier foo both as a struct tag and as the name of
a pointer object to that struct is potentially misleading.

--
Keith Thompson (The_Other_Keit h) ks***@mib.org <http://www.ghoti.net/~kst>
San Diego Supercomputer Center <*> <http://users.sdsc.edu/~kst>
We must do something. This is something. Therefore, we must do this.

[Eric Sosman]

aegis wrote:

sizeof could not possibly evaluate to some size for a type, unless
two things for a type occur:

a) the type is complete and sizeof will evaluate to appropriate size
b) type is not complete and any sizeof use with incomplete type
results in undefined behavior.

Is 'b' so? If so, where in the standard does it say this?
By 6.5.3.4/1 it is a constraint violation to apply `sizeof'
to an incomplete type. By 5.1.1.3/1 the constraint violation
must elicit a diagnostic message.
I only bring this up because of the following:

Breakpoint 1, main () at malloc.c:9
9 foo = malloc(sizeof foo);
(gdb) n
11 return 0;
(gdb) p foo
$1 = 0x80495d8
(gdb) p sizeof foo
$2 = 4
(gdb) list malloc.c:1,15
1 #include <stdlib.h>
2
3 typedef struct foo *FOO;
4
5 int main(void)
6 {
7 FOO foo;
8
9 foo = malloc(sizeof foo);

`foo' has the type `pointer to struct foo'. While
`struct foo' is an incomplete type, `pointer to struct foo'
is not: it is a "complete" type whose size is known, even
though the size of what it points to is not known. Hence,
`sizeof' can be applied to it.

However, the allocation is very likely wrong. If the
intent is to allocate memory for `foo' to point at, this
works only if an actual `struct foo' turns out to be no
larger than a pointer. If you had used the recommended style

foo = malloc(sizeof *foo);

.... you would have received a diagnostic because `*foo' has
an incomplete type.

--
Eric Sosman
es*****@acm-dot-org.invalid