如果字段顺序不同，结构的大小也将不同

Question

package main

import (
    "fmt"
    "unsafe"
)

type A struct {
    a bool
    b int64
    c int
}

type B struct {
    b int64
    a bool
    c int
}

type C struct {
}

func main() {
    // output 24
    fmt.Println(unsafe.Sizeof(A{}))

    // output 16
    fmt.Println(unsafe.Sizeof(B{}))

    // output 0
    fmt.Println(unsafe.Sizeof(C{}))
}

1. Struct A and B have the same fields, but if specified in different order they result in different size. why?

2. Size of struct C is zero. How much memory is allocated by the system for a := C{}?

Thanks.

Answer 1

1. Struct sizes

TL;DR; (Summary): Different implicit padding will be used if you reorder the fields, and the implicit padding counts towards the size of the struct.

Note that the result depends on the target architecture; results you posted applies when GOARCH=386, but when GOARCH=amd64, sizes of both A{} and B{} will be 24 bytes.

Address of fields of a struct must be aligned, and the address of fields of type int64 must be a multiple of 8 bytes. Spec: Package unsafe:

Computer architectures may require memory addresses to be aligned; that is, for addresses of a variable to be a multiple of a factor, the variable's type's alignment. The function Alignof takes an expression denoting a variable of any type and returns the alignment of the (type of the) variable in bytes.

Align of int64 is 8 bytes:

fmt.Println(unsafe.Alignof((int64(0)))) // Prints 8

So in case of A since first field is bool, there is a 7-byte implicit padding after A.a so that A.b which is of type int64 can start on an address that is a multiple of 8. This (that 7-byte padding is needed exactly) is guaranteed as the struct itself is aligned to an address which is a multiple of 8, because that is the largest size of all of its fields. See: Spec: Size alignment guarantees:

For a variable x of struct type: unsafe.Alignof(x) is the largest of all the values unsafe.Alignof(x.f) for each field f of x, but at least 1.

In case of B (and if GOARCH=386 which is your case) there will only be a 3-byte implicit padding after the B.a field of type bool because this field is followed by a field of type int (which has size of 4 bytes) and not int64.

Align of int is 4 bytes if GOARCH=386, and 8 bytes if GOARCH=amd64:

fmt.Println(unsafe.Alignof((int(0))))   // Prints 4 if GOARCH=386, and 8 if GOARCH=amd64

Use unsafe.Offsetof() to find out the offsets of fields:

// output 24
a := A{}
fmt.Println(unsafe.Sizeof(a),
    unsafe.Offsetof(a.a), unsafe.Offsetof(a.b), unsafe.Offsetof(a.c))

// output 16
b := B{}
fmt.Println(unsafe.Sizeof(b),
    unsafe.Offsetof(b.b), unsafe.Offsetof(b.a), unsafe.Offsetof(b.c))

// output 0
fmt.Println(unsafe.Sizeof(C{}))

var i int
fmt.Println(unsafe.Sizeof(i))

Output if GOARCH=386 (try it on the Go Playground):

24 0 8 16
16 0 8 12
0
4

Output if GOARCH=amd64:

24 0 8 16
24 0 8 16
0
8

2. Zero size values

Spec: Size alignment guarantees:

A struct or array type has size zero if it contains no fields (or elements, respectively) that have a size greater than zero. Two distinct zero-size variables may have the same address in memory.

So the spec just gives a hint to use the same memory address but it's not a requirement. But current implementations follow it. That is, no memory will be allocated for values of types having a size of zero, this includes the empty struct struct{} and arrays of zero length, e.g. [0]int, or arrays whose elements has a size of zero (and with arbitrary length).

See this example:

a := struct{}{}
b := struct{}{}
c := [0]int{}
d := [3]struct{}{}

fmt.Printf("%p %p %p %p %p", &a, &b, &c, &d, &d[2])

Output (try it on the Go Playground): All the addresses are the same.

0x21cd7c 0x21cd7c 0x21cd7c 0x21cd7c 0x21cd7c

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For an interesting and related topic, read: Dave Cheney: Padding is hard