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Clarifying how the alignment of the struct works #52064

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Jul 11, 2018
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5 changes: 4 additions & 1 deletion src/libcore/mem.rs
Original file line number Diff line number Diff line change
Expand Up @@ -229,6 +229,8 @@ pub fn forget<T>(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.
///
Expand Down Expand Up @@ -283,7 +285,8 @@ pub fn forget<T>(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::<FieldStruct>());
///
/// #[repr(C)]
Expand Down