英文:
Why type assertion of an interface equal to nil in type switch non-nil branch?
问题
给定的代码如下：
type Int struct {
v int
}
func typeAssert(val any) {
switch v := val.(type) {
case nil:
fmt.Println("val type is nil")
case *Int:
fmt.Println("val type is *Int")
if v == nil {
fmt.Println("val after type assertion is nil?")
}
}
}
func main() {
var i1 *Int
typeAssert(i1)
}
输出结果为：
val type is *Int
val after type assertion is nil?
我感到困惑的是，既然在switch v := val.(type)中匹配到了*Int，为什么v == nil会为true？如果v == nil为true，那么case nil应该也能匹配，但实际上并没有匹配到。
英文:
Given the codes
type Int struct {
v int
}
func typeAssert(val any) {
switch v := val.(type) {
case nil:
fmt.Println("val type is nil")
case *Int:
fmt.Println("val type is *Int")
if v == nil {
fmt.Println("val after type assertion is nil?")
}
}
}
func main() {
var i1 *Int
typeAssert(i1)
}
Output:
val type is *Int
val after type assertion is nil?
What confuses me is that since the *Int is matched in switch v := val.(type), why could the v == nil be true? If the v == nil be true, the case nil could be matched, actually, it does not.
答案1
得分: 1
因为有两种带有接口值的nil：
- 一个
nil接口，会触发类型切换的第一个分支。就像var x interface{} = nil一样-不仅值是nil，类型也是nil。 - 一个
nil值，比如var x interface = (nil)(*Int)。这会触发第二个分支，因为它有一个类型，类型与switch中的类型匹配，但值是nil。在代码中，v是一个值为nil的*Int。
英文:
Because there are two kinds of nil with an interface value:
- A
nilinterface, which would hit the first branch of your type switch. This is likevar x interface{} = nil- not only is the valuenil, the type is alsonil. - A
nilvalue, likevar x interface = (nil)(*Int). This hits the second branch, because it has a type, the type matches the type in theswitch, but the value isnil. In the code,vis a*Intwhose value isnil.
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