Instance new Type (Golang) - go

Can anyone tell me how to create a new instance of Type from a string? Reflect?
There are examples but they are for the older (pre Go 1 versions) of the language [:(]

So, if I understand your question correctly, you are asking about how you can create an object when you just have the name of the type as string. So, for example, you might have a string "MyStruct" and you want to create an object of this type.
Unfortunately, that's not easily possible because Go is a statically typed language and the linker will eliminate dead code (or inline parts of it). So, there is no guarantee, that your final executable will even contain the code of "MyStruct".
You can however, maintain a global map[string]reflect.Type manually. For example by initializing this map in the init() function of your packages which defines such discover-able types. This will also tell the compiler that you are using the types. Afterwards, you can use this map to look up the reflect.Type of the type you want to create and use reflect.New to get a pointer to a new object of that type (stored as a reflect.Value). You can extract the object into an interface with something like this:
reflect.New(yourtype).Elem().Interface()
Elem() will de-reference the pointer and Interface() will return the reflected value as an interface{}. See The Laws of Reflection for further details.
PS: There might be a better way to structure your program which doesn't even require reflection and which let the compiler catch more errors. Have you considered using a factory method for example? An other easy solution might be to maintain a map[string]func() interface{} of functions which can be invoked to create a new object with that name.

Factory with predefined constructors can be based on something like:
package main
import (
"fmt"
)
type Creator func() interface{}
type A struct {
a int
}
type B struct {
a bool
}
func NewA() interface{} {
return new(A)
}
func NewB() interface{} {
return new(B)
}
func main() {
m := map[string]Creator{}
m["A"] = NewA
m["B"] = NewB
for k, v := range m {
fmt.Printf("%v -> %v\n", k, v())
}
}

Related

When to store pointers to structs in variables instead of the struct itself

I'm currently learning Go and am following a tutorial about how to use Go with Stripe. There is this example code:
package main
import (
"fmt"
"github.com/stripe/stripe-go"
"github.com/stripe-go/customer"
)
func main() {
sc := &client.API{}
sc.Init("somekey")
c, _ := sc.Customers.Get("customerid", nil)
// ...
}
What is/could be the reason that sc stores the pointer to the struct and not the struct itself?
[To supplement the comment you received]
While in this case with the small code sample it's hard to say, in most scenarios you'll see non-trivial types passed around by pointer to enable modification. As an anti-example, consider this code which uses a variable of a struct type by value:
type S struct {
ID int
}
func (s S) UpdateID(i int) {
s.ID = i
}
func main() {
s := S{}
s.UpdateID(99)
fmt.Println(s.ID)
}
What do you think this will print? It will print 0, because methods with value receivers cannot modify the underlying type.
There's much information about this in Go - read about pointers, and about how methods should be written. This is a good reference: https://golang.org/doc/faq#methods_on_values_or_pointers, and also https://golang.org/doc/effective_go#pointers_vs_values
Back to your example: typically non-trivial types such as those representing a "client" for some services will be using pointers because method calls on such types should be able to modify the types themselves.

How to use runtime.Object to create CRD generic functions in go [duplicate]

I have an interface Model, which is implemented by struct Person.
To get a model instance, I have the following helper functions:
func newModel(c string) Model {
switch c {
case "person":
return newPerson()
}
return nil
}
func newPerson() *Person {
return &Person{}
}
The above approach allows me to return a properly typed Person instance (can easily add new models later with same approach).
When I attempted to do something similar for returning a slice of models, I get an error. Code:
func newModels(c string) []Model {
switch c {
case "person":
return newPersons()
}
return nil
}
func newPersons() *[]Person {
var models []Person
return &models
}
Go complains with: cannot use newPersons() (type []Person) as type []Model in return argument
My goal is to return a slice of whatever model type is requested (whether []Person, []FutureModel, []Terminator2000, w/e). What am I missing, and how can I properly implement such a solution?
This is very similar to a question I just answered: https://stackoverflow.com/a/12990540/727643
The short answer is that you are correct. A slice of structs is not equal to a slice of an interface the struct implements.
A []Person and a []Model have different memory layouts. This is because the types they are slices of have different memory layouts. A Model is an interface value which means that in memory it is two words in size. One word for the type information, the other for the data. A Person is a struct whose size depends on the fields it contains. In order to convert from a []Person to a []Model, you will need to loop over the array and do a type conversion for each element.
Since this conversion is an O(n) operation and would result in a new slice being created, Go refuses to do it implicitly. You can do it explicitly with the following code.
models := make([]Model, len(persons))
for i, v := range persons {
models[i] = Model(v)
}
return models
And as dskinner pointed out, you most likely want a slice of pointers and not a pointer to a slice. A pointer to a slice is not normally needed.
*[]Person // pointer to slice
[]*Person // slice of pointers
Maybe this is an issue with your return type *[]Person, where it should actually be []*Person so to reference that each index of the slice is a reference to a Person, and where a slice [] is in itself a reference to an array.
Check out the following example:
package main
import (
"fmt"
)
type Model interface {
Name() string
}
type Person struct {}
func (p *Person) Name() string {
return "Me"
}
func NewPersons() (models []*Person) {
return models
}
func main() {
var p Model
p = new(Person)
fmt.Println(p.Name())
arr := NewPersons()
arr = append(arr, new(Person))
fmt.Println(arr[0].Name())
}
As Stephen already answered the question and you're a beginner I emphasize on giving advises.
A better way of working with go's interfaces is not to have a constructor returning
the interface as you might be used to from other languages, like java, but to have
a constructor for each object independently, as they implement the interface implicitly.
Instead of
newModel(type string) Model { ... }
you should do
newPerson() *Person { ... }
newPolitician() *Politician { ... }
with Person and Politician both implementing the methods of Model.
You can still use Person or Politician everywhere where a Model
is accepted, but you can also implement other interfaces.
With your method you would be limited to Model until you do a manual conversion to
another interface type.
Suppose I have a Person which implements the method Walk() and a Model implements ShowOff(), the following would not work straight forward:
newModel("person").ShowOff()
newModel("person").Walk() // Does not compile, Model has no method Walk
However this would:
newPerson().ShowOff()
newPerson().Walk()
As others have already answered, []T is a distinct type. I'd just like to add that a simple utility can be used to convert them generically.
import "reflect"
// Convert a slice or array of a specific type to array of interface{}
func ToIntf(s interface{}) []interface{} {
v := reflect.ValueOf(s)
// There is no need to check, we want to panic if it's not slice or array
intf := make([]interface{}, v.Len())
for i := 0; i < v.Len(); i++ {
intf[i] = v.Index(i).Interface()
}
return intf
}
Now, you can use it like this:
ToIntf([]int{1,2,3})
Types T and []T are distinct types and distinct are their methods as well, even when satisfying the same interface. IOW, every type satisfying Model must implement all of the Model's methods by itself - the method receiver can be only one specific type.
Even if Go's implementation allowed this, it's unfortunately unsound: You can't assign a []Person to a variable of type []Model because a []Model has different capabilities. For example, suppose we also have Animal which implements Model:
var people []Person = ...
var models []Model = people // not allowed in real Go
models[0] = Animal{..} // ???
var person Person = people[0] // !!!
If we allow line 2, then line 3 should also work because models can perfectly well store an Animal. And line 4 should still work because people stores Persons. But then we end up with a variable of type Person holding an Animal!
Java actually allows the equivalent of line 2, and it's widely considered a mistake. (The error is caught at run time; line 3 would throw an ArrayStoreException.)

How to pass type to function argument in Go

ERROR: type CustomStruct is not an expression.
type CustomStruct struct {
}
func getTypeName(t interface{}) string {
rt := reflect.TypeOf(t).Elem()
return rt.Name()
}
getTypeName(CustomStruct)
How can I pass struct type to function without type instance?
This will work
getTypeName((*CustomStruct)(nil))
But I wonder if there is more simple version..
You can't. You can only pass a value, and CustomStruct is not a value but a type. Using a type identifier is a compile-time error.
Usually when a "type" is to be passed, you pass a reflect.Type value which describes the type. This is what you "create" inside your getTypeName(), but then the getTypeName() will have little left to do:
func getTypeName(t reflect.Type) string {
return t.Name()
}
// Calling it:
getTypeName(reflect.TypeOf(CustomStruct{}))
(Also don't forget that this returns an empty string for anonymous types such as []int.)
Another way is to pass a "typed" nil pointer value as you did, but again, you can just as well use a typed nil value to create the reflect.Type too, without creating a value of the type in question, like this:
t := reflect.TypeOf((*CustomStruct)(nil)).Elem()
fmt.Println(t.Name()) // Prints CustomStruct
Lets resurrect this!
The generics proposal for Go got approved, and that's coming, eventually. When this question was first asked, this probably made more sense as a question, but for anyone looking to implement a generics pattern now, I think I've got an alright API for it.
For now, you can't interact with abstract types, but you can interact with methods on the abstract type, and reflect allows you to examine function signatures. For a method, the 0th is the receiver.
type Example struct {int}
type Generic struct{reflect.Type}
func (p Example) Type() {}
func Reflect(generic interface{}) Generic {
real := reflect.TypeOf(generic)
if real.Kind() != reflect.Func || real.NumIn() < 1 {
panic("reflect.Type.In(n) panics if not a func and if n out of bounds")
}
return Generic{real.In(0)}
}
func (g Generic) Make() interface{} {
return reflect.Zero(g.Type).Interface()
}
func main() {
tOfp := Reflect(Example.Type)
fmt.Printf("Name of the type: %v\n", tOfp.Name())
fmt.Printf("Real (initial)value: %v\n", tOfp.Make())
}
Some quick notes:
The structure of "Example" doesn't matter, rather only that it has a method with a non-pointer receiver.
The definition of a type called "Generic" as a struct is to accomplish what I believed OP's actual intent to be.
The above definition of "Generic" is a struct instead of an interface so that it can have its own method set. Defining "Generic" as an interface, and using a methodset specific to each operand-type used with it would make tons of sense.
If you weren't aware, actual generics are coming in Go 1.18. My example above has no linter or compile protection, and will panic at runtime if used incorrectly. It does work, and will let you reason over abstract types while you wait for a native implementation.
Happy Coding!
From Go version 1.18 a new feature Generics has been introduced. In most of the case instead of passing types to function, we can use generics. Then we will also get compile time error instead of runtime error and it's more efficient than reflect also.
Example Code
func HttpGet[T](url, body) T {
var resp T
return T
}
resp := HttpGet[ResponseType]("dummy.example", nil)

How to cast interface{} back into its original struct?

I need a way to dynamically cast a struct/interface back to its original object.
I can add methods / functions inside. basically I need something like this:
MyStruct => Interface{} => MyStruct
When on the final conversion I don't know anything about the original struct besides what come inside the struct, so I can't just so:
a.(MyStruct)
You need to know at least the possible types it could be. There's a couple cases, 1. You think you might know what it is. 2. You have a list of possible types it could be, 3. Your code knows nothing about the underlying types.
If you think you know it, you can use type assertion to convert back to the original struct type.
...
package main
import (
"fmt"
)
type MyStruct struct {
Thing string
}
func (s *MyStruct) Display() {
fmt.Println(s.Thing)
}
type Thingable interface {
Display()
}
func main() {
s := &MyStruct{
Thing: "Hello",
}
// print as MyThing
s.Display()
var thinger Thingable
thinger = s
// print as thingable interface
thinger.Display()
// convert thinger back to MyStruct
s2 := thinger.(*MyStruct) // this is "type assertion", you're asserting that thinger is a pointer to MyStruct. This will panic if thinger is not a *MyStruct
s2.Display()
}
You can see this in action here: https://play.golang.org/p/rL12Lrpqsyu
Note if you want to test the type without panicking if you're wrong, do s2, ok := thinger.(*MyStruct). ok will be true if it was successful and false otherwise.
if you want to test your interface variable against a bunch of types, use a switch: (scroll to bottom)
...
package main
import (
"fmt"
"reflect"
)
type MyStruct struct {
Thing string
}
type MyStruct2 struct {
Different string
}
func (s *MyStruct) Display() {
fmt.Println(s.Thing)
}
func (s *MyStruct2) Display() {
fmt.Println(s.Different)
}
type Thingable interface {
Display()
}
func main() {
s := &MyStruct{
Thing: "Hello",
}
// print as MyThing
s.Display()
var thinger Thingable
thinger = s
// print as thingable interface
thinger.Display()
// try to identify thinger
switch t := thinger.(type) {
case *MyStruct:
fmt.Println("thinger is a *MyStruct. Thing =", t.Thing)
case *MyStruct2:
fmt.Println("thinger is a *MyStruct2. Different =", t.Different)
default:
fmt.Println("thinger is an unknown type:", reflect.TypeOf(thinger))
}
}
You can try that out here https://play.golang.org/p/7NEbwB5j6Is
If you really don't know anything about the underlying types, you'll have to expose the things you need through interface functions and call those. Chances are you can do this without knowing anything about the underlying type. If all else fails, you can use the reflect package to introspect your interface object and gather information about it. this is how the json package reads json text and returns populated structs—though this is an advanced topic and expect to sink a lot of time into it if you go this route. it’s best to hide reflection code inside a package with a clean interface(ie the package api).
No: as mentioned in this thread
Go is neither covariant nor contravariant. Types are either equal or they aren't.
You have to either take the structs apart and deal with the pieces, or use reflection.
Type assertions are only "assertions", not "coercions" of any kind.
See also this thread, which reminds us that:
A pointer is one kind of type.
A struct is another kind of type.
An integer is another kind of type.
A floating point number is another kind of type.
A boolean is another kind of type.
The principle of an interface concerns the methods attached to a type T, not what type T is.
An interface type is defined by a set of methods.
Any value that implements the methods can be assigned to an interface value of that type.
That would make the conversion from interface to concrete type quite difficult to do.

In golang, how to type assert an interface{} to a type specified by a reflect.Type?

For example, I have an interface{} named a, and I also have an reflect.Type called elemType. Now, I want to type assert a to elemType, but a.(elemType) can't be compiled successfully. How to fix it?
Sorry for my confusing expression. My meaning is that I get a type from a function, and I want to type assert an interface{} to this type, but this type is stored in a reflect.Type variable.
What I want to do is similar to the code below:
var a interface{}
//do something
func getType() reflect.Type {
var ret reflect.Type
//do something
return ret
}
targetType := getType()
result := a.(targetType)
Consider a standard type assertion in Go:
v := a.(typeName)
Here the compiler can determine the type of the variable v at compile time, and make use of that knowledge when compiling any further statements involving the variable.
With your example of using a refltect.Type variable in the assertion, it would be impossible to determine the type of v, so the code could not be compiled.
If you need to check that a particular interface variable is of a particular type at runtime, you can still do that with the reflect package. For example:
// if elemType is a normal type
if reflect.ValueOf(a).Type() == elemType {
fmt.Println("type matches")
}
// if elemType is an interface, can check if the value implements it
if reflect.ValueOf(a).Type().Implements(elemType) {
fmt.Println("value implements interface")
}
But you will need a concrete type to return back to standard variables. If you've only got a small selection of possible types, perhaps using a type switch might do what you want.

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