CODEWITHSUNDEEP
pointersinterfacego
Mihnea Giurgea
Mihnea Giurgea

Reputation: 447

Convert nil interface to pointer of something in Golang?

In the following code piece, trying to convert a nil interface to a pointer of something fails with the following error: interface conversion: interface is nil, not *main.Node

type Nexter interface {
    Next() Nexter
}

type Node struct {
    next Nexter
}

func (n *Node) Next() Nexter {...}

func main() {
    var p Nexter

    var n *Node
    fmt.Println(n == nil) // will print true
    n = p.(*Node) // will fail
}

Play link here: https://play.golang.org/p/2cgyfUStCI

Why does this fail exactly? It's entirely possible to do 

n = (*Node)(nil)

, so I'm wondering how can you achieve a similar effect starting from a nil interface.

Upvotes: 27

Views: 42013

Answers (1)

icza
icza

Reputation: 417512

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.

Upvotes: 54

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