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
}