为什么在Go类型开关中声明了一个单独的变量？

Question

I am having trouble understanding why type switches are written with an additional variable defined in the switch statement. The below code seems to be the sanctioned way of doing things:

func test_func(i interface{}) {
    switch v := i.(type) {
    case int:
        fmt.Printf("%T
", v)
    case float64:
        fmt.Printf("%T
", v)
    case int:
        fmt.Printf("I don't know about type %T!
", v)
    }
}

func main() {
    test_func(float64(34))
    test_func(int(34))
    test_func("hello world")
}

As expected, this returns:

float64
int
I don't know about type string!

However, I can change test_func slightly so that v is not defined in the switch statement, and instead we use i inside our case statements:

func test_func(i interface{}) {
    switch i.(type) {
    case int:
        fmt.Printf("%T
", i)
    case float64:
        fmt.Printf("%T
", i)
    case int:
        fmt.Printf("I don't know about type %T!
", i)
    }
}

func main() {
    test_func(float64(34))
    test_func(int(34))
    test_func("hello world")
}

And the output is not changed. It seems the two forms are interchangeable. Why would I go to the trouble of defining v when I could just use i? The latter case is simpler since there is one less variable to keep track of; maybe it is even more performant.

Answer 1

They are not interchangeable; you're just passing i to a function that can take it regardless of its type (fmt.Printf's arguments after the format string are of type interface{}). i is still its original type, because the type of a variable cannot change.

If you actually wanted to do something with it based on its type, you would need the first form, so that v would be of the type in the case statement. Whether or not you assign the typed value to a variable, the original variable i retains its original type.

This is covered well in the Tour of Go: Type switches