From dc425e2e64863d4e1aa0f2bb41f7bfa3cd14b558 Mon Sep 17 00:00:00 2001 From: Val Markovic Date: Wed, 4 Jul 2018 17:26:45 -0700 Subject: [PATCH] Clarifying how the alignment of the struct works The docs were not specifying how to compute the alignment of the struct, so I had to spend some time trying to figure out how that works. Found the answer [on this page](http://camlorn.net/posts/April%202017/rust-struct-field-reordering.html): > The total size of this struct is 5, but the most-aligned field is b with alignment 2, so we round up to 6 and give the struct an alignment of 2 bytes. --- src/libcore/mem.rs | 5 ++++- 1 file changed, 4 insertions(+), 1 deletion(-) diff --git a/src/libcore/mem.rs b/src/libcore/mem.rs index 84173654655eb..8fb4e0d6a02e3 100644 --- a/src/libcore/mem.rs +++ b/src/libcore/mem.rs @@ -229,6 +229,8 @@ pub fn forget(t: T) { /// 2. Round up the current size to the nearest multiple of the next field's [alignment]. /// /// Finally, round the size of the struct to the nearest multiple of its [alignment]. +/// The alignment of the struct is usually the largest alignment of all its +/// fields; this can be changed with the use of `repr(align(N))`. /// /// Unlike `C`, zero sized structs are not rounded up to one byte in size. /// @@ -283,7 +285,8 @@ pub fn forget(t: T) { /// // The size of the second field is 2, so add 2 to the size. Size is 4. /// // The alignment of the third field is 1, so add 0 to the size for padding. Size is 4. /// // The size of the third field is 1, so add 1 to the size. Size is 5. -/// // Finally, the alignment of the struct is 2, so add 1 to the size for padding. Size is 6. +/// // Finally, the alignment of the struct is 2 (because the largest alignment amongst its +/// // fields is 2), so add 1 to the size for padding. Size is 6. /// assert_eq!(6, mem::size_of::()); /// /// #[repr(C)]