https://gcc.gnu.org/bugzilla/show_bug.cgi?id=118403
--- Comment #10 from Jakub Jelinek <jakub at gcc dot gnu.org> ---
(In reply to Stephen Hemminger from comment #0)
> My understanding is that both should be equivalent.
> Reference C99 Standard 6.7.8.21:
>
> If there are fewer initializers in a brace-enclosed list than there are
> elements or members of an aggregate, or fewer characters in a string literal
> used to initialize an array of known size than there are elements in the
> array, the remainder of the aggregate shall be initialized implicitly the
> same as objects that have static storage duration
A union has just one member active at a time. If you don't use designated
initializer, then {0} just initializes the first member in there. All the
remaining bits are padding.
Before C23, padding bits were guaranteed to be cleared only when initializing
static storage duration, or when doing the implicit zero initialization of
extra members.
So, say
struct A { char a; /* 7 padding bytes */ long long b; };
struct B { int c; /* 4 padding bytes */ struct A d; };
void foo ()
{
struct B e = { 0 }; // This doesn't guarantee initialization of the 4 padding
bytes,
// but does guarantee initialization of the 7 padding
bytes
struct B f = { 0, { 0 } }; // This one doesn't guarantee initialization of
any padding bytes, just the fields.
struct B g = { 0, { 0, 0 } }; // Neither this.
}
Now, C23 adds {} initializers and those are also guaranteed to zero padding
bits, so
...
struct B h = {}; // Guarantees zero initialization of all padding bits
struct B i = { 0, {} }; // Guarantees initialization of the 7 padding bytes
but not the 4.
If you use struct nic_mbx mbx = { }; then all bits including padding are zero
initialized, { .msg = { 0 } }; means that .msg field has its first member
initialized but doesn't say anything about padding bits.
You could say also { .whatever_field_is_largest = { 0 } } and initialize more
bits.
