Why Does a Type Switch with Multiple Cases in Go Result in an 'undefined' Method Error?

Type Switch with Multiple Cases in Go

In Go, a type switch allows you to match a value against multiple types. However, when using multiple cases in a type switch, understanding variable type assignments becomes crucial.

Consider the following code snippet:

package main

import (
    "fmt"
)

type A struct {
    a int
}

func (this *A) test() {
    fmt.Println(this)
}

type B struct {
    A
}

func main() {
    var foo interface{}
    foo = A{}
    switch a := foo.(type) {
        case B, A:
            a.test()
    }
}

When executing this code, an error is raised: a.test undefined (type interface {} is interface with no methods). This error occurs because the variable a retains the type interface{}, despite the type switch.

To understand this behavior, we need to refer to the Go spec:

"The TypeSwitchGuard may include a short variable declaration. When that form is used, the variable is declared at the beginning of the implicit block in each clause. In clauses with a case listing exactly one type, the variable has that type; otherwise, the variable has the type of the expression in the TypeSwitchGuard."

In our case, since multiple types (B and A) are specified in the case clause, the variable a keeps the type of the expression in the TypeSwitchGuard, which is interface{}. This means that the compiler will not allow us to call the test() method on a because interface{} doesn't have a test() method.

To resolve this, we can use type assertions, which allow us to assert that a value has a specific type. Here's an updated version of the code that uses type assertions:

package main

import (
    "fmt"
)

type A struct {
    a int
}

func (this *A) test() {
    fmt.Println(this)
}

type B struct {
    A
}

func main() {
    var foo interface{}
    foo = &B{}
    if a, ok := foo.(tester); ok {
        fmt.Println("foo has test() method")
        a.test()
    }
}

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