当golang是本机类型时，为什么golang禁止分配给相同的基础类型？

Question

Consider this code:

package main
import "fmt"

type specialString string

func printString(s string) {
    fmt.Println(s)
}

// unlike, say, C++, this is not legal GO, because it redeclares printString
//func printString(s specialString) {    
//  fmt.Println("Special: " + s)
//}

func main() {
    ss := specialString("cheese")
    // ... so then why shouldn't this be allowed?
    printString(ss)
}

My question is: why is the language defined so that the call to printString(ss) in main() is not allowed? (I'm not looking for answers that point to the Golang rules on assignment; I have already read them, and I see that both specialString and string have the same 'underlying type' and both types are 'named' -- if you consider the generic type 'string' to be named, which Golang apparently does -- and so they are not assignable under the rules.)

But why are the rules like that? What problem is solved by treating the built-in types as 'named' types, and preventing you from passing named types to all the standard library functions that accepting the same underlying built-in type? Does anybody know what the language designers had in mind here?

From my point of view, it seems to create a lot of pointless type conversion in the code, and discourages the use of strong typing where it actually would make sense..

Answer 1

I believe the initial authors' logic here is that named type is named for a reason - it represents something different, not just underlying type.

I guess I've read it somewhere in golang-nuts, but can't remember exact discussion.

Consider the following example:

type Email string

You named it Email, because you need to represent e-mail entity, and 'string' is just simplified representation of it, sufficient for the very start. But later, you may want to change Email to something more complex, like:

type Email struct {
    Address string
    Name    string
    Surname string
}

And that will break all your code that work with Email implicitly assuming it's a string.