关于接口分配

Question

Why doesn't Go think it's type mismatch error when a struct pointer be assigned to an interface?

package main

import "fmt"

type ABC interface {
    a() string
    b() int
}

type XYZ struct {
    aa string
    bb int
}

func (xyz XYZ) a() string {
    return "XYZ"
}

func (xyz XYZ) b() int {
    return 123
}

func main() {
    var xyz *XYZ
    var abc ABC = xyz // type of abc is *main.XYZ，I think that Golang can find this error here, but why not?

    fmt.Printf("%T
", abc)

    a, ret := abc.(*XYZ)
    fmt.Println(a, ret) // type of a is *main.XYZ

    fmt.Println(a.a()) // will occur a error, because the type of a(*main.XYZ) not implements the interface ABC
}

I want know why Go doesn't think this is an error at "var abc ABC = xyz"

Answer 1

will occur a error, because the type of a(*main.XYZ) not implements the interface ABC

Wrong. *main.XYZ implements ABC (else abc = xyz would fail at compile-time, try to rename the method b to c for example), but the variable a holds a nil pointer (of type *XYZ). And since the XYZ.a() method has value receiver, to call that, a pointer value of type *XYZ would have to be dereferenced. But a nil pointer points to nothing, it cannot be dereferenced, an attempt to do so results in a runtime panic, just as you experienced.

If you would initialize xyz in the beginning with a non-nil pointer, it would work:

var xyz *XYZ = new(XYZ)

Try it on the Go Playground.

Also note that if XYZ.a() and XYZ.b() would have pointer receivers, then it would also work if xyz is nil:

func (xyz *XYZ) a() string {
    return "XYZ"
}

func (xyz *XYZ) b() int {
    return 123
}

func main() {
    var xyz *XYZ
    // ...

Try it on the Go Playground. The reason for this is because if the receivers are pointers, the nil pointer does not have to be dereferenced in order to call methods with pointer receivers, so no runtime panic occurs. Of course if in the methods you would refer to the XZY.aa or XYZ.bb fields, that would be a runtime panic, but your current method implementations does not do so, so it works.