This is because a variable of static type Nexter
(which is just an interface) may hold values of many different dynamic types.
Yes, since *Node
implements Nexter
, your p
variable may hold a value of type *Node
, but it may hold other types as well which implement Nexter
; or it may hold nothing at all (nil
value). And Type assertion cannot be used here because quoting from the spec:
x.(T)
asserts that x
is not nil
and that the value stored in x
is of type T
.
But x
in your case is nil
. And if the type assertion is false, a run-time panic occurs.
If you change your program to initialize your p
variable with:
var p Nexter = (*Node)(nil)
Your program will run and type assertion succeeds. This is because an interface value actually holds a pair in the form of: (value, dynamic type)
, and in this case your p
will not be nil
, but will hold a pair of (nil, *Node)
; for details see The Laws of Reflection #The representation of an interface.
If you also want to handle nil
values of interface types, you may check it explicitly like this:
if p != nil {
n = p.(*Node) // will not fail IF p really contains a value of type *Node
}
Or better: use the special "comma-ok" form:
// This will never fail:
if n, ok := p.(*Node); ok {
fmt.Printf("n=%#v\n", n)
}
Using the "comma-ok" form:
The value of ok
is true
if the assertion holds. Otherwise it is false
and the value of n
is the zero value for type T
. No run-time panic occurs in this case.