NewObject(..)函数可以在泛型的支持下实现，不需要在类型约束中列出所有支持的类型。

func NewObject[T any](id string) T {
 var vehicle any

 switch id {

 case "truck":
  vehicle = Truck{
   Foo: "foo",
  }
 case "car":
  vehicle = Car{
   Bar: "bar",
  }
 }

 if val, ok := vehicle.(T); ok {
  return val
 }

 var otherVehicle T
 fmt.Printf("Not implemented. Returning with default values for \"%v\"\n", id)
 return otherVehicle
}

您可以看到完整的示例here。

我有个奇怪的例子
此示例将返回construct of struct
希望这能激励你

package main

import (
    "fmt"
)

type testA struct {
    a uint
}
type testB struct {
    b string
}

var mp map[string]func(...interface{}) interface{} = map[string]func(...interface{}) interface{}{
    "struct1": func(para ...interface{}) interface{} {
        return &testA{
            a: para[0].(uint),
        }
    },
    "struct2": func(para ...interface{}) interface{} {
        return &testB{
            b: para[0].(string),
        }
    },
}

func NewSpecial(type_check_str string) func(...interface{}) interface{} {
    if val, ok := mp[type_check_str]; ok {
        return val
    }
    return nil
}

func main() {

    test1 := NewSpecial("struct1")(uint(5))
    test2 := NewSpecial("struct2")("It's Ok?")
    fmt.Println(test1)
    fmt.Println(test2)
}