Is it guaranteed that `align_of::<uX>() == align_of::<iX>()`?

[joshlf]

The reference page on type layout says:

---

The size of most primitives is given in this table.

Type	size_of::<Type>()
bool	1
u8 / i8	1
u16 / i16	2
u32 / i32	4
u64 / i64	8
u128 / i128	16
usize / isize	See below
f32	4
f64	8
char	4

usize and isize have a size big enough to contain every address on the target platform. For example, on a 32 bit target, this is 4 bytes, and on a 64 bit target, this is 8 bytes.

The alignment of primitives is platform-specific. In most cases, their alignment is equal to their size, but it may be less. In particular, i128 and u128 are often aligned to 4 or 8 bytes even though their size is 16, and on many 32-bit platforms, i64, u64, and f64 are only aligned to 4 bytes, not 8.

---

Is it guaranteed that the alignments of u16 and i16 are the same, that the alignments of u32 and i32 are the same, etc? It doesn't appear to be guaranteed by the text.

(Zerocopy doesn't need this – I'm just curious.)

cc @jswrenn