Why is Fallthrough Not Allowed in Go's Type Switch?

Mary-Kate Olsen
Release: 2024-11-12 19:54:02
Original
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Why is Fallthrough Not Allowed in Go's Type Switch?

Fallthrough in Type Switch: Why Not Permitted?

Question:

In Go, attempting to use the fallthrough statement in a type switch raises an error. Why is fallthrough not allowed in this context?

Answer:

In a type switch, the variable being switched on changes type depending on the specific case that is executed. For example, the variable i in the following code has a type that depends on the case that is invoked:

var x interface{}
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!")
}
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If fallthrough were allowed, how would it be expected to behave? In the case of bool, i is typed as bool. However, in the case of string, i becomes typed as string.

Allowing fallthrough would require either magical type morphing (impossible) or variable shadowing (with no meaningful value). Consider the following example:

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 is the type here? Should be "string", but what if it falls through from bool?
}
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The only possible solution would be to have fallthrough implicitly cast subsequent cases to interface{}, but this would be confusing and poorly defined.

If the desired type-checking behavior for switch-case expressions is essential, it can be implemented with 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
}
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