Re: [PATCH] checkpatch: prefer = {} initializations to = {0}

Al Viro
2021-08-14T14:57:14Z

On Sat, Aug 14, 2021 at 03:59:22PM +0200, Christophe JAILLET wrote:

> > +# prefer = {}; to = {0};
> > + if ($line =~ /= \{ *0 *\}/) {
> > + WARN("ZERO_INITIALIZER",
> > + "= {} is preferred over = {0}\n" . $herecurr);

Sigh... "is preferred over" by whom? Use the active voice, would you?

> [1] and [2] state that {} and {0} don't have the same effect. So if correct,
> this is not only a matter of style.
>
> When testing with gcc 10.3.0, I arrived at the conclusion that both {} and
> {0} HAVE the same behavior (i.e the whole structure and included structures
> are completely zeroed) and I don't have a C standard to check what the rules
> are.
> gcc online doc didn't help me either.

http://www.open-std.org/jtc1/sc22/wg14/www/docs/n1256.pdf, but empty
initializer-list is gccism anyway.

Section 6.7.8 is the one to look through there.

> Can someone provide some rational or compiler output that confirms that {}
> and {0} are not the same?

Easily: compare
int x[] = {0};
and
int x[] = {};

For more obscure example,
int x = {0};
is valid, if pointless, but
int x = {};
will be rejected even by gcc.

Incidentally, do *NOT* assume that initializer will do anything with padding
in a structure, no matter how you spell it. Neither {} nor {0} nor explicit
initializer for each member of struct do anything to the padding. memset()
does, but anything short of that leaves the padding contents unspecified.