Question

I’m wondering why fallthrough isn’t allowed in a type switch statement in golang.

According to the specification: “The “fallthrough” statement is not permitted in a type switch.”, which doesn’t explain much about WHY it isn’t allowed.

The code attached is to simulate a possible scenario were a fallthrough in a type switch statement might have been useful.

Notice! This code doesn’t work, it will produce the error: “cannot fallthrough in type switch”. I’m just wondering what possible reasons might have been for not allowing the fallthrough statement in a type switch.

//A type switch question
package main

import "fmt"

//Why isn't fallthrough in type switch allowed?
func main() {
    //Empty interface
    var x interface{}

    x = //A int, float64, bool or string value

    switch i := x.(type) {
    case int:
        fmt.Println(i + 1)
    case float64:
        fmt.Println(i + 2.0)
    case bool:
        fallthrough
    case string:
        fmt.Printf("%v", i)
    default:
        fmt.Println("Unknown type. Sorry!")
    }
}

Answer 1

How would you expect fallthrough to work? In this type switch, the i variable has a type that depends on the particular case that’s invoked. So in the case bool the i variable is typed as bool. But in case string it’s typed as string. So either you’re asking for i to magically morph its type, which isn’t possible, or you’re asking for it to be shadowed by a new variable i string, which will have no value because its value comes from x which is not, in fact, a string.

---

Here’s an example to try and illustrate the problem:

switch i := x.(type) {
case int:
    // i is an int
    fmt.Printf("%T\n", i); // prints "int"
case bool:
    // i is a bool
    fmt.Printf("%T\n", i); // prints "bool"
    fallthrough
case string:
    fmt.Printf("%T\n", i);
    // What does that type? It should type "string", but if
    // the type was bool and we hit the fallthrough, what would it do then?
}

The only possible solution would be to make the fallthrough cause the subsequent case expression to leave i as an interface{}, but that would be a confusing and bad definition.

If you really need this behavior you can already accomplish this with the existing functionality:

switch i := x.(type) {
case bool, string:
    if b, ok := i.(bool); ok {
        // b is a bool
    }
    // i is an interface{} that contains either a bool or a string
}