I'm trying to figure out how go handles type conversion between structs. Everything I have read tells me that types with the same underlying type are considered compatible and type conversion happens implicitly. If that is the case, why does the code below not work? Both Foo and Bar implement the FooI interface, and they both add an x property of type string. Yet, when I pass a struct of type Bar to AcceptBarOrFoo that expects a struct of type Foo, I get a type mismatch compile error.
Go Playground
package main
import (
"play.ground/bar"
"play.ground/foo"
)
func main() {
AcceptBarOrFoo(bar.Bar{})
}
func AcceptBarOrFoo(foo.Foo) interface{} {
return nil
}
// -- iface/iface.go --
package iface
type FooI interface {
a() string
b(int) int
}
// -- foo/foo.go --
package foo
import (
"play.ground/iface"
)
type Foo struct {
iface.FooI
x string
}
// -- bar/bar.go --
package bar
import (
"play.ground/iface"
)
type Bar struct {
iface.FooI
x string
}
Foo's x is different from Bar's x because non-exported identifiers are never equal across package boundaries. Fix by exporting the fields in foo.Foo and bar.Bar:
type Foo struct {
iface.FooI
X string // <-- start with capital letter to export
}
To use a foo.Foo or bar.Bar as an argument value, a foo.Foo and bar.Bar must be assignable to the argument's type. It does not work to use foo.Foo as the argument type because named types are not assignable to each other. However, named types are assignable to unnamed types when the two types share the same underlying type. Declare the argument as an unnamed type:
func AcceptBarOrFoo(struct {
iface.FooI
X string
}) interface{} {
return nil
}
With these changes, the following code compiles:
AcceptBarOrFoo(bar.Bar{})
AcceptBarOrFoo(foo.Foo{})
Run the example on the Go playground
Another option is to use a conversion to a common type. In the following code, foo.Foo is the common type and bar.Bar is converted to a foo.Foo.
func Accept(foo.Foo) interface{} {
return nil
}
...
Accept(foo.Foo{})
Accept(foo.Foo(bar.Bar{}))
Run the example on the Go playground.
Note: foo.Foo and bar.Bar must have the same fields for the above to work (field names are exported, fields in same order, fields have same types).
Some notes about Go:
There are conversions from one concrete type to another.
Go is famous for having no implicit conversions in expressions, but there are implicit conversions in some assignment scenarios.
You cannot convert one concrete type to another concrete type, ever. They are not the same. There is no way to define this type of automatic casting in Go. At best, you could define a function that accepts a Bar and builds and returns a new Foo with its fields set to the same values as the input Bar.
Everything I have read tells me that if the underlying types are the same, the higher order types are considered compatible and type conversion happens implicitly
It's unclear what your source for this is, but nothing would have ever stated or implied this, it's simply not true. Go does no implicit conversion, of anything. That's a big, loudly advertised feature of Go. Given type Foo struct { a int } and type Bar struct { a int }, you can never assign an object of type Bar to a variable of type Foo.
You can convert from either concrete type to an interface type, when the type satisfies the interface. Your AcceptBarOrFoo method should accept an interface type (which both Foo and Bar satisfy), not a concrete type. Given that interfaces only define methods (not members), and given neither Foo or Bar have any methods, your interface would be the empty interface, interface{}. The value passed in would serve no purpose, except for you to later convert it back to a concrete type to access its members, but that's not really what interfaces are for.
Related
This question already has an answer here:
Go generics: invalid composite literal type T
(1 answer)
Closed last month.
The following code results in the error "invalid composite literal type T".
package main
import "fmt"
func main() {
fmt.Println(createThing[foo]())
}
type thing interface {
foo | bar
}
type foo struct {
id int
a string
}
type bar struct {
id int
b int
}
func createThing[T thing, P *T]() P {
return &T{}
}
If I only include foo in interface thing, or remove a string and b int so foo and bar are exactly the same, the code will run without error. However, doesn't this defeat the purpose of generics? Why can't I instantiate a generic type like this, especially when I'm not even accessing any fields?
Possibly related to https://github.com/golang/go/issues/48522
Most generic types are not valid types for composite literals. This isn't a problem though, as there are other ways to create values of generic types.
To create a pointer to a new zero value:
func createThing[T thing]() *T {
return new(T)
}
Or to create a non-pointer zero value:
func createThing[T thing]() T {
var value T
return value
}
As for why the error occurs in this way, here's the explanation from the spec, revised to address your specific question.
For composite literals:
The LiteralType's core type T must be a struct, array, slice, or map type
What is the core type?
An interface T has a core type if [...] there is a single type U which is the underlying type of all types in the type set of T
No other interfaces have a core type.
What is the underlying type?
Each type T has an underlying type: If T is one of the predeclared boolean, numeric, or string types, or a type literal, the corresponding underlying type is T itself. Otherwise, T's underlying type is the underlying type of the type to which T refers in its declaration.
A "type literal" can refer to a literal struct type, like struct{int id}. So, when foo and bar both have an underlying type of struct{int id}, then thing has a core type of struct{int id}, and so composite literals are possible. When foo and bar don't have the same underlying type, then thing has no core type, and composite literals are not possible, hence your error.
The formal definition may seem complicated, but the result and practical takeaway is simple: generic code is only capable of expressing common behaviour across the possible types. Besides in the special case where all underlying types are the same, literal values are not a common behaviour.
Say I have a generic struct called foo and I create two objects from it. I can determine the concrete type of each using reflect.TypeOf(), like so:
package main
import (
"fmt"
"reflect"
)
type foo[T any] struct {
data T
}
func main() {
a := foo[string]{"cheese"}
b := foo[int]{42}
fmt.Println(reflect.TypeOf(a))
fmt.Println(reflect.TypeOf(b))
}
// main.foo[string]
// main.foo[int]
What I am interested in is determining just the generic type of these objects (i.e., foo) and not the concrete type (i.e., foo[string] and foo[int]). Is this possible or do I need to manually extract the generic type from these strings (e.g., with regex)?
Edit
Regex might look something like this:
func GetGenericType(x any) string {
// Get type as a string
s := reflect.TypeOf(x).String()
// Regex to run
r := regexp.MustCompile(`\.(.*)\[`)
// Return capture
return r.FindStringSubmatch(s)[1]
}
fmt.Println(GetGenericType(a))
fmt.Println(GetGenericType(b))
// foo
// foo
I've also seen this question but this doesn't answer this question because it gives the concrete type (i.e., main.foo[string]) rather than the generic type (i.e., foo).
Reflection doesn't see the name of the "base" generic type, because at run time that base type doesn't exist.
The relevant passage from the Go spec is Instantiations:
Instantiating a type results in a new non-generic named type; instantiating a function produces a new non-generic function.
So when you write:
b := foo[int]{42}
name := reflect.TypeOf(b).Name()
the name of that type is precisely foo[int].
It's worth noting that the identifier foo without the type parameter list is relevant at compile time, because it prevents you from redeclaring it in the same package. Type definitions:
A type definition creates a new, distinct type with the same
underlying type and operations as the given type and binds an
identifier, the type name, to it.
TypeDef = identifier [ TypeParameters ] Type .
But instantiations, as defined above, result in a new named type which is different than foo; and at run time when you can use reflection, you deal with instantiations only.
In conclusion, I think your solution with regex is acceptable, until some helper function is added to the stdlib (if ever). Reposting it here for clarity:
func GetGenericType(x any) string {
// Get type as a string
s := reflect.TypeOf(x).String()
// Regex to run
r := regexp.MustCompile(`\.(.*)\[`)
// Return capture
return r.FindStringSubmatch(s)[1]
}
Just keep in mind the difference between Type.String() and Type.Name(): any type can have a string representation, but only named types have a name. (Obviously, right?). So for example if you wrote:
b := &foo[int]{42}
then the type of b is *foo[int], which is an anonymous composite type, and Name() returns an empty string.
In Go, because of it's static type language we can see lot of type conversion for example,
var x float64
var y =32
x=float64(32)
Even though Go supports type casting, as far as I know it's very rarely used in. Can someone explain why and usecases that type casting can be used. Following is a one example that use type casting.
type Common struct {
Gender int
From string
To string
}
type Foo struct {
Id string
Name string
Extra Common
}
type Bar struct {
Id string
Name string
Extra Common
}
foo:=Foo{
Id:"123",
Name:"damitha",
Extra: struct {
Gender int
From string
To string
}{Gender:1 , From:"xx", To:"yy" },
}
bar:=*(*Bar)(unsafe.Pointer(&foo))
fmt.Printf("%+v\n",bar)
Even this mapping also possible to achieve in more safer ways.
The unsafe shenanigans in the question are similar to type casting. These shenanigans are not needed. Use a conversion:
bar := *(*Bar)(&foo)
or more simply:
bar := Bar(foo)
The relevant quote from the specification is:
A non-constant value x can be converted to type T in any of these cases:
...
ignoring struct tags (see below), x's type and T have identical underlying types.
ignoring struct tags (see below), x's type and T are pointer types that are not defined types, and their pointer base types have identical underlying types.
The first point in the quote applies to Bar(foo) because Foo and Bar have identical underlying types.
The second point in the quote applies to the conversion (*Bar)(&foo) because:
*Foo and *Bar are pointer types
*Foo and *Bar are not defined types.
The base types Foo and Bar have the same underlying type.
Go supports neither type casting nor implicit type conversion while lot of static support languages like Java C/C++ supporting both. Go only supports for type conversion.
I was wondering why you can't do:
type Foo struct { A int }
type Bar Foo
foos := []Foo{Foo{1}, Foo{2}}
bars := []Bar(foos)
//cannot convert foos (type []Foo) to type []Bar
and I found out that this would require the runtime to perform a loop over the slice to convert each of the elements, which would be non-idiomatic Go. This makes sense.
However, could this not be solved by the compiler just aliasing Bar as Foo, so internally they're the same and they use the same type header underneath? I'm guessing the answer is no though I'm curious as to why.
This:
[]Bar(foos)
is a type conversion. Conversions have specific rules according to the spec:
A non-constant value x can be converted to type T in any of these cases:
x is assignable to T.
x's type and T have identical underlying types.
x's type and T are unnamed pointer types and their pointer base types have identical underlying types.
x's type and T are both integer or floating point types.
x's type and T are both complex types.
x is an integer or a slice of bytes or runes and T is a string type.
x is a string and T is a slice of bytes or runes.
None applies here. Why?
Because the underlying type of []Foo is not the same as the underlying type of []Bar. And a value of type []Foo is not assignable to a variable of type []Bar, see Assignability rules here.
The underlying type of Foo is the same as the underlying type of Bar, but the same does not apply to slices where the element type is Foo and Bar.
So the following works:
type Foo struct{ A int }
type Foos []Foo
type Bars Foos
func main() {
foos := []Foo{Foo{1}, Foo{2}}
bars := Bars(foos)
fmt.Println(bars)
}
Output (try it on the Go Playground):
[{1} {2}]
Note that since the actual memory representation of Foo and Bar is the same (because the underlying type of Bar is Foo), in this case using the package unsafe you can "view" a value of []Foo as a value of []Bar:
type Foo struct{ A int }
type Bar Foo
func main() {
foos := []Foo{Foo{1}, Foo{2}}
bars := *(*[]Bar)(unsafe.Pointer(&foos))
fmt.Println(bars)
fmt.Printf("%T", bars)
}
This: *(*[]Bar)(unsafe.Pointer(&foos)) means that take the address of foos, convert it to unsafe.Pointer (according to spec all pointers can be converted to unsafe.Pointer), then this Pointer is converted to *[]Bar (again according to the spec Pointer can be converted to any other pointer type), and then this pointer is dereferenced (* operator), so the result is a value of type []Bar as can be seen in the output.
Output (try it on the Go Playground):
[{1} {2}]
[]main.Bar
Notes:
Quoting the package doc of unsafe:
Package unsafe contains operations that step around the type safety of Go programs.
Packages that import unsafe may be non-portable and are not protected by the Go 1 compatibility guidelines.
What does this mean? It means that you shouldn't revert to using package usafe every time it makes your life easier. You should only use it in exceptional cases, when not using it would make your program really slow and complicated.
In your program this is not the case as I proposed a working example with just a little refactoring (Foos and Bars being slices).
unsafe steps around the type safety of Go. What does this mean? If you would change the type of foos (e.g. drastically like foos := "trap!"), your program would still compile and run, but most likely runtime panic would occur. Using usafe you lose type checks of the compiler.
While if you use my other proposal (Foos and Bars), such changes/typos are detected at compile time.
As mentioned in "Why can I type alias functions and use them without casting?"
In Go, there is no such thing as a type alias.
The type keyword introduces new named types. They are not aliases
If you compare two named types, the names must match in order for them to be interchangeable
This is what the spec mentions:
A type declaration binds an identifier, the type name, to a new type that has the same underlying type as an existing type, and operations defined for the existing type are also defined for the new type.
The new type is different from the existing type.
When I have a function that returns an interface type, the returned value doesn't work as I would expect. That is, it acts strictly as the defined interface, and to access the methods and values not defined in the interface, I have to do a type assertion. Why?
Consider the following sample code:
package main
import (
"fmt"
)
type Frobnicator interface {
Frobnicate()
}
type Foo struct {
Value string
}
func (f *Foo) Frobnicate() {
fmt.Printf("The value is %s\n", f.Value)
}
func fooFactory () Frobnicator {
return &Foo{"chicken"}
}
func main() {
foo := fooFactory( )
foo.Frobnicate()
// foo.Value undefined (type Frobnicator has no field or method Value)
// fmt.Printf("foo value = %s\n", foo.Value)
bar := foo.(*Foo)
fmt.Printf("bar value = %s\n", bar.Value)
}
Is there a better, easier, more idiomatic way to get at foo.Value? Or is a type assertion really the best way?
Not sure what to answer here. Maybe there is a misconception what interface types are. Interface types are absolutely normal types. And you can do with a interface value what the interface says: Invoke the interface methods. For a struct type you may access fields and invoke interface methods as defined by the struct type. So everything is plain and simple: A type allows what it allows, no matter whether interface or struct.
It now happens that a value of interface type may contain some struct value (say). Up to now this is hidden. Type asserting reveals the struct value (and there is no more interface). You may hide an other struct value in the interface (given it implements the right methods) this might not have a Value field. This makes it clear that you cannot access the Value field without a type assertion, because it might not be there.
If you need access to inner value of interface implementation (Value), you must either expose it via interface itself or do a type assertion. That's because nothing in Frobnicator suggests whether it's Foo or some other implementing struct.
It is not different than many other languages. In Java you will have to cast also under similar circumstances.