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.