Why type assertion of an interface equal to nil in type switch non-nil branch?

Question

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.

Interface is a type and a value pair. The type is `*Int`, on which the type assert switch works and the value is `nil` — Inian, Sep 26 '22 at 13:42

score 1 · Accepted Answer · answered Sep 26 '22 at 13:44

Because there are two kinds of nil with an interface value:

A nil interface, which would hit the first branch of your type switch. This is like var x interface{} = nil - not only is the value nil, the type is also nil.
A nil value, like var x interface = (nil)(*Int). This hits the second branch, because it has a type, the type matches the type in the switch, but the value is nil. In the code, v is a *Int whose value is nil.

Why type assertion of an interface equal to nil in type switch non-nil branch?

1 Answers1