Switch type behaving different for cases with multiple possibilities

Question

Can anyone explain why the if statements behave differently in these two cases? It does not make sense to me.

package main

import (
    "crypto/ecdsa"
    "crypto/ed25519"
    "fmt"
)

func main() {
    var publicKey interface{}
    publicKey = (ed25519.PublicKey)(nil)
    switch k := publicKey.(type) {
    case *ecdsa.PublicKey, ed25519.PublicKey:
        if k == nil {
            fmt.Println("It wont be printed")
        }
    default:
    }

    switch k := publicKey.(type) {
    case ed25519.PublicKey:
        if k == nil {
            fmt.Println("It will be printed")
        }
    default:
    }
}

score 0 · Answer 1 · answered Apr 20 '22 at 02:23

publicKey is a interface{}, which has a type and value, both type and value is empty, it will be == nil;

publicKey.(type) is a interface{}, which has a type ed25519.PublicKey and a value nil

cases with multiple possibilities will change it type:

if i run:

case ed25519.PublicKey, *ecdsa.PublicKey:

publicKey.(type)

type:  <interface{}|ed25519.PublicKey>
value: ked25519.PublicKey(nil)

publicKey.(type) == nil(because it has a type interface{})

if i run:

case ed25519.PublicKey:

publicKey.(type)

type:  <ked25519.PublicKey>
value: ked25519.PublicKey(nil)

publicKey.(type) != nil(because it only has a type ked25519.PublicKey)

Switch type behaving different for cases with multiple possibilities

1 Answers1