c - Would this union work if char had stricter alignment requirements than int? -

recently came across following snippet, attempt ensure bytes of i (nad no more) accessible individual elements of c:

union {   int i;   char c[sizeof(int)]; }; 

now seems idea, wonder if standard allows case alignment requirements char more restrictive int.

in other words, possible have four-byte int required aligned on four-byte boundary one-byte char (it 1 byte, definition, see below) required aligned on sixteen-byte boundary?

and stuff use of union above?

two things note.

1. i'm talking standard allows here, not sane implementor/architecture provide.

2. i'm using term "byte" in iso c sense, it's width of char, not 8 bits.

no type can ever have stricter alignment requirements size (because of how arrays work), , sizeof(char) 1.

in case it's not obvious:

• sizeof(t [n]) sizeof(t)*n.
• sizeof in units of char; types represented fixed number of bytes (char), number being size. see 6.2.6 (representation of types) details.
• given t a[2];, (char *)&a[1] - (char *)&a[0] equal sizeof a[0].
• therefore alignment requirement t no greater sizeof(t) (in fact divides sizeof(t))