What does this print on a 64-bit platform, and why?

package main

import (
    "fmt"
    "unsafe"
)

type A struct {
    a bool
    b int64
    c bool
}

type B struct {
    b int64
    a bool
    c bool
}

func main() {
    fmt.Println(unsafe.Sizeof(A{}), unsafe.Sizeof(B{}))
    fmt.Println(unsafe.Offsetof(A{}.b), unsafe.Alignof(A{}))
}

Output:

24 16
8 8

Each field is aligned to its own alignment requirement, and the struct's alignment is that of its most-aligned field (8, for int64). In A, a takes byte 0, then 7 bytes of padding so that b starts at offset 8. c is at 16, followed by 7 more bytes of tail padding, which keeps elements of []A aligned. That gives 24. In B the two bools share one 8-byte word: 8 + 1 + 1 + 6 padding = 16.

Rule of thumb: order fields from largest alignment to smallest. The fieldalignment analyzer from golang.org/x/tools reports structs that could be smaller. This matters in large slices, where saving 33% on each element adds up, and for cache-line use. Go does not reorder fields for you, because the declared layout is visible through unsafe, reflection and cgo.