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

quoted

+# 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.