英文:
How does 'type DoNotCompare [0]func()' prevent comparability in Golang
问题
根据您提供的代码，DoNotCompare 是一个空数组类型 [0]func()，它可以嵌入到结构体中以防止比较。但是，根据您的代码示例，NonCompare 结构体似乎是可比较的。
根据我理解，DoNotCompare 的作用是通过将其嵌入到结构体中，阻止该结构体进行比较操作。然而，在您的代码示例中，并没有直接使用 DoNotCompare 进行比较，而是使用了 NonCompare 结构体的方法 Compare 进行比较。
因此，根据您的代码示例，NonCompare 结构体仍然是可比较的，而 DoNotCompare 并没有起到阻止比较的作用。可能是您在代码中有其他的实现或使用方式导致了这种情况。
英文:
Per protobuf pragma
> ```
> type DoNotCompare [0]func()
> DoNotCompare can be embedded in a struct to prevent comparability
I am confused why DoNotCompare could prevent comparability, and try to understand with the following codes. Unfortunately, I am failed. The NonCompare struct seems could be comparability.
type DoNotCompare [0]func()
type NonCompare struct {
DoNotCompare
ii int8
}
type Comparer interface {
Compare(b NonCompare) int
}
func (a NonCompare) Compare(b NonCompare) int {
if a.ii > b.ii {
return 1
}
return 0
}
func main() {
var nc1 = NonCompare{ii: 3}
var nc2 = NonCompare{ii: 5}
nc1.Compare(nc2)
Is there anything am I missing?
答案1
得分: 3
预防措施适用于内置的==运算符，而不是你的Compare()方法。当然，它可以执行任何你想要的操作。
但是，如果你尝试使用==：
fmt.Println(nc1 == nc2)
你会得到一个编译时错误：
./prog.go:30:18: invalid operation: nc1 == nc2 (struct containing DoNotCompare cannot be compared)
原因是func()类型的值（或任何函数类型的值）是不可比较的，所以[0]func()类型的值也是不可比较的。而且，包含非可比较类型字段的结构体也是不可比较的。规则在规范：比较运算符中列出：
> 相等运算符==和!=适用于可比较的操作数。排序运算符<、<=、>和>=适用于有序的操作数。这些术语和比较的结果定义如下：
>
> - [...]
> - 如果结构体的所有字段都是可比较的，则结构体值是可比较的。如果对应的非空白标识符字段相等，则两个结构体值相等。
> - 如果数组元素类型的值是可比较的，则数组值是可比较的。如果对应的元素相等，则两个数组值相等。
>
> [...] 切片、映射和函数值是不可比较的。
使用大小为0的数组，这样字段的大小将为0，在使用它的结构体中不会添加任何额外的内容。请注意，这仅在将大小为0的字段放在第一位置时才能保证。详细信息请参见https://stackoverflow.com/questions/34219232/struct-has-different-size-if-the-field-order-is-different/34219916#34219916。
英文:
The prevention works on the builtin == operator, not on what your Compare() method does. Of course it can do whatever you want it to.
But if you try to use ==:
fmt.Println(nc1 == nc2)
You get a compile-time error:
./prog.go:30:18: invalid operation: nc1 == nc2 (struct containing DoNotCompare cannot be compared)
The reason is because values of func() type (or any function type) are not comparable, so [0]func() values are also not comparable. And structs having a field of a non-comparable type are also not comparable. Rules are listed at Spec: Comparison operators:
> The equality operators == and != apply to operands that are comparable. The ordering operators <, <=, >, and >= apply to operands that are ordered. These terms and the result of the comparisons are defined as follows:
>
> - [...]
> - Struct values are comparable if all their fields are comparable. Two struct values are equal if their corresponding non-blank fields are equal.
> - Array values are comparable if values of the array element type are comparable. Two array values are equal if their corresponding elements are equal.
>
> [...] Slice, map, and function values are not comparable.
An array of size 0 is used so the field's size will be 0, not adding anything extra to the struct it is used in. Please note this is only guaranteed if you place 0-size fields to the first place. For details see https://stackoverflow.com/questions/34219232/struct-has-different-size-if-the-field-order-is-different/34219916#34219916.
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