I would like to get rid of the variable temp in the following code:
type myinterface interface {
f1()
}
type a struct {
val int
}
type b struct {
mi *myinterface
}
func (a) f1() {
}
func demo() {
a1 := a{3}
var temp myinterface = a1
b1 := b{&temp}
fmt.Println(b1)
But if I try to write
b1 := b{&myinterface(a1)}
I get the message
cannot take the address of myinterface(a1) ( undefined )
what is the correct way to do this?
Update:
I did not a pointer to an interface, since an interface can hold a struct or a pointer to a struct, as also detailed in this question:
"<type> is pointer to interface, not interface" confusion
Let me know if this is what you are looking for:
https://play.golang.org/p/ZGRyIqN7bPR
Full code:
package main
import (
"fmt"
)
type myinterface interface {
f1()
}
type a struct {
val int
}
type b struct {
mi myinterface
}
func (a) f1() {}
func main() {
fmt.Println("Hello, playground")
a1 := &a{3}
b1 := b{a1}
fmt.Println(b1)
}
You almost never need a pointer to an interface, since interfaces are just pointers themselves.
So just change the struct b to:
type b struct {
mi myinterface
}
myinterface(a1) is a type conversion, it converts a1 to type myinteface.
Type conversion expressions are not addressable, so you cannot take the address of it. What is addressable is listed explicitly in the Spec: Address operators:
The operand must be addressable, that is, either a variable, pointer indirection, or slice indexing operation; or a field selector of an addressable struct operand; or an array indexing operation of an addressable array. As an exception to the addressability requirement, x [in the expression of &x] may also be a (possibly parenthesized) composite literal.
This related answer lists several options how to obtain the address of such expressions: How do I do a literal *int64 in Go?
For example if you use a composite literal to create a slice of type []myinterface and put a1 in it, you can take the address of its first element (which will be of type *myinterface):
b1 := b{&[]myinterface{a1}[0]}
And it will work (try it on the Go Playground):
a1 := a{3}
b1 := b{&[]myinterface{a1}[0]}
fmt.Println(b1)
But know that using pointers to interfaces is very rarely needed, so is a field of type *myinterface really what you want in the first place?
Interface values can be nil, and also nil values (e.g. nil pointers) can also be wrapped in interfaces, so most likely you don't need a pointer to interface. We would have to know your "wider" goal to tell if this is what you really need.
Related
package main
import (
"fmt"
)
type outer struct {
in *int
}
func main() {
i := 4
o := outer{&i}
fmt.Printf("%+v", o)
}
I'd like to see {in:4} at the end of this, not {in:0x......}, i.e. pretty print the data structure.
I'd like to accomplish this in a similar manner to the code posted (e.g. with a fmt shortcut similar to %+v or an analogous solution).
This is for autogenerated code from a required field of a thrift struct.
What's the best way to go about this?
When you use &i it does not dereference i. Rather it references i, which means that it copies the address of i into o. See the documentation for the Address operators.
From what I gather, you should be able to use *o to dereference the pointer; in other words, go from the address back to the original variable.
For an operand x of pointer type *T, the pointer indirection *x denotes the variable of type T pointed to by x. If x is nil, an attempt to evaluate *x will cause a run-time panic.
There are already several Q&As on this "X does not implement Y (... method has a pointer receiver)" thing, but to me, they seems to be talking about different things, and not applying to my specific case.
So, instead of making the question very specific, I'm making it broad and abstract -- Seems like there are several different cases that can make this error happen, can someone summary it up please?
I.e., how to avoid the problem, and if it occurs, what are the possibilities? Thx.
This compile-time error arises when you try to assign or pass (or convert) a concrete type to an interface type; and the type itself does not implement the interface, only a pointer to the type.
Short summary: An assignment to a variable of interface type is valid if the value being assigned implements the interface it is assigned to. It implements it if its method set is a superset of the interface. The method set of pointer types includes methods with both pointer and non-pointer receiver. The method set of non-pointer types only includes methods with non-pointer receiver.
Let's see an example:
type Stringer interface {
String() string
}
type MyType struct {
value string
}
func (m *MyType) String() string { return m.value }
The Stringer interface type has one method only: String(). Any value that is stored in an interface value Stringer must have this method. We also created a MyType, and we created a method MyType.String() with pointer receiver. This means the String() method is in the method set of the *MyType type, but not in that of MyType.
When we attempt to assign a value of MyType to a variable of type Stringer, we get the error in question:
m := MyType{value: "something"}
var s Stringer
s = m // cannot use m (type MyType) as type Stringer in assignment:
// MyType does not implement Stringer (String method has pointer receiver)
But everything is ok if we try to assign a value of type *MyType to Stringer:
s = &m
fmt.Println(s)
And we get the expected outcome (try it on the Go Playground):
something
So the requirements to get this compile-time error:
A value of non-pointer concrete type being assigned (or passed or converted)
An interface type being assigned to (or passed to, or converted to)
The concrete type has the required method of the interface, but with a pointer receiver
Possibilities to resolve the issue:
A pointer to the value must be used, whose method set will include the method with the pointer receiver
Or the receiver type must be changed to non-pointer, so the method set of the non-pointer concrete type will also contain the method (and thus satisfy the interface). This may or may not be viable, as if the method has to modify the value, a non-pointer receiver is not an option.
Structs and embedding
When using structs and embedding, often it's not "you" that implement an interface (provide a method implementation), but a type you embed in your struct. Like in this example:
type MyType2 struct {
MyType
}
m := MyType{value: "something"}
m2 := MyType2{MyType: m}
var s Stringer
s = m2 // Compile-time error again
Again, compile-time error, because the method set of MyType2 does not contain the String() method of the embedded MyType, only the method set of *MyType2, so the following works (try it on the Go Playground):
var s Stringer
s = &m2
We can also make it work, if we embed *MyType and using only a non-pointer MyType2 (try it on the Go Playground):
type MyType2 struct {
*MyType
}
m := MyType{value: "something"}
m2 := MyType2{MyType: &m}
var s Stringer
s = m2
Also, whatever we embed (either MyType or *MyType), if we use a pointer *MyType2, it will always work (try it on the Go Playground):
type MyType2 struct {
*MyType
}
m := MyType{value: "something"}
m2 := MyType2{MyType: &m}
var s Stringer
s = &m2
Relevant section from the spec (from section Struct types):
Given a struct type S and a type named T, promoted methods are included in the method set of the struct as follows:
If S contains an anonymous field T, the method sets of S and *S both include promoted methods with receiver T. The method set of *S also includes promoted methods with receiver *T.
If S contains an anonymous field *T, the method sets of S and *S both include promoted methods with receiver T or *T.
So in other words: if we embed a non-pointer type, the method set of the non-pointer embedder only gets the methods with non-pointer receivers (from the embedded type).
If we embed a pointer type, the method set of the non-pointer embedder gets methods with both pointer and non-pointer receivers (from the embedded type).
If we use a pointer value to the embedder, regardless of whether the embedded type is pointer or not, the method set of the pointer to the embedder always gets methods with both the pointer and non-pointer receivers (from the embedded type).
Note:
There is a very similar case, namely when you have an interface value which wraps a value of MyType, and you try to type assert another interface value from it, Stringer. In this case the assertion will not hold for the reasons described above, but we get a slightly different runtime-error:
m := MyType{value: "something"}
var i interface{} = m
fmt.Println(i.(Stringer))
Runtime panic (try it on the Go Playground):
panic: interface conversion: main.MyType is not main.Stringer:
missing method String
Attempting to convert instead of type assert, we get the compile-time error we're talking about:
m := MyType{value: "something"}
fmt.Println(Stringer(m))
To keep it short and simple, let say you have a Loader interface and a WebLoader that implements this interface.
package main
import "fmt"
// Loader defines a content loader
type Loader interface {
load(src string) string
}
// WebLoader is a web content loader
type WebLoader struct{}
// load loads the content of a page
func (w *WebLoader) load(src string) string {
return fmt.Sprintf("I loaded this page %s", src)
}
func main() {
webLoader := WebLoader{}
loadContent(webLoader)
}
func loadContent(loader Loader) {
loader.load("google.com")
}
The above code will give you this compile time error
./main.go:20:13: cannot use webLoader (type WebLoader) as type Loader
in argument to loadContent:
WebLoader does not implement Loader (Load method has pointer receiver)
To fix it you only need to change webLoader := WebLoader{} to following:
webLoader := &WebLoader{}
Why this will fix the issue? Because you defined this function func (w *WebLoader) Load to accept a pointer receiver. For more explanation please read #icza and #karora answers
Another case when I have seen this kind of thing happening is if I want to create an interface where some methods will modify an internal value and others will not.
type GetterSetter interface {
GetVal() int
SetVal(x int) int
}
Something that then implements this interface could be like:
type MyTypeA struct {
a int
}
func (m MyTypeA) GetVal() int {
return a
}
func (m *MyTypeA) SetVal(newVal int) int {
int oldVal = m.a
m.a = newVal
return oldVal
}
So the implementing type will likely have some methods which are pointer receivers and some which are not and since I have quite a variety of these various things that are GetterSetters I'd like to check in my tests that they are all doing the expected.
If I were to do something like this:
myTypeInstance := MyType{ 7 }
... maybe some code doing other stuff ...
var f interface{} = myTypeInstance
_, ok := f.(GetterSetter)
if !ok {
t.Fail()
}
Then I won't get the aforementioned "X does not implement Y (Z method has pointer receiver)" error (since it is a compile-time error) but I will have a bad day chasing down exactly why my test is failing...
Instead I have to make sure I do the type check using a pointer, such as:
var f interface{} = new(&MyTypeA)
...
Or:
myTypeInstance := MyType{ 7 }
var f interface{} = &myTypeInstance
...
Then all is happy with the tests!
But wait! In my code, perhaps I have methods which accept a GetterSetter somewhere:
func SomeStuff(g GetterSetter, x int) int {
if x > 10 {
return g.GetVal() + 1
}
return g.GetVal()
}
If I call these methods from inside another type method, this will generate the error:
func (m MyTypeA) OtherThing(x int) {
SomeStuff(m, x)
}
Either of the following calls will work:
func (m *MyTypeA) OtherThing(x int) {
SomeStuff(m, x)
}
func (m MyTypeA) OtherThing(x int) {
SomeStuff(&m, x)
}
Extend from above answers (Thanks for all of your answers)
I think it would be more instinctive to show all the methods of pointer / non pointer struct.
Here is the playground code.
https://play.golang.org/p/jkYrqF4KyIf
To summarize all the example.
Pointer struct type would include all non pointer / pointer receiver methods
Non pointer struct type would only include non pointer receiver methods.
For embedded struct
non pointer outer struct + non pointer embedded struct => only non pointer receiver methods.
non pointer outer struct + pointer embedded struct / pointer outer struct + non pointer embedded struct / pointer outer struct + pointer embedded struct => all embedded methods
I have a function
func doStuff(inout *interface{}) {
...
}
the purpose of this function is to be able to treat a pointer of any type as input.
But when I want to call it with a the pointer of a struct I have an error.
type MyStruct struct {
f1 int
}
When calling doStuff
ms := MyStruct{1}
doStuff(&ms)
I have
test.go:38: cannot use &ms (type *MyStruct) as type **interface {} in argument to doStuff
How can I cast &ms to be compatible with *interface{}?
There is no such thing as a "pointer to an interface" (technically, you can use one, but generally you don't need it).
As seen in "what is the meaning of interface{} in golang?", interface is a container with two words of data:
one word is used to point to a method table for the value’s underlying type,
and the other word is used to point to the actual data being held by that value.
So remove the pointer, and doStuff will work just fine: the interface data will be &ms, your pointer:
func doStuff(inout interface{}) {
...
}
See this example:
ms := MyStruct{1}
doStuff(&ms)
fmt.Printf("Hello, playground: %v\n", ms)
Output:
Hello, playground: {1}
As newacct mentions in the comments:
Passing the pointer to the interface directly works because if MyStruct conforms to a protocol, then *MyStruct also conforms to the protocol (since a type's method set is included in its pointer type's method set).
In this case, the interface is the empty interface, so it accepts all types anyway, but still.
It seems golang does not have the pointer operator -> as C and C++ have. Now let's say I have a function looks something like this: myfun(myparam *MyType), inside the function, if I want to access the member variables of MyType, I have to do (*myparam).MyMemberVariable. It seems to be a lot easier to do myparam->MyMemberVariable in C and C++.
I'm quite new to go. Not sure if I'm missing something, or this is not the right way to go?
Thanks.
In Go, both -> and . are represented by .
The compiler knows the types, and can dereference if necessary.
package main
import "fmt"
type A struct {
X int
}
func main() {
a0, a1 := A{42}, &A{27}
fmt.Println(a0.X, a1.X)
}
You can do myparam.MyMemberValue, pointers are automatically dereferenced
Go spec:
Selectors automatically dereference pointers to structs. If x is a pointer to a struct, x.y is shorthand for (x).y; if the field y is also a pointer to a struct, x.y.z is shorthand for ((*x).y).z, and so on. If x contains an anonymous field of type *A, where A is also a struct type, x.f is shorthand for (*x.A).f.
Hummm, this automatic dereferencing can be very confusing (for old programmers like me)
If you learn programmation with GOLANG, no problem, it is practical.
If you pass from C to GOLANG, this will seem strange, you will probably prefer (at the beginning at least) to keep the "(*my_var).my_field" expression instead of "my_var.myfield"
If you pass from GOLANG to C, you will get many compilation errors.
Goes uses -> for passing data by using channels.
package main
import "fmt"
type Person struct {
Name string
}
func (p *Person) printName() {
p.Name = "brian"
}
func main() {
// obj
brian := Person{""}
// prints obj default value
fmt.Println("No pointer", brian.Name)
// it access the method with the pointer
brian.printName()
// prints the result
fmt.Println("With a pointer", brian.Name)
}
I am new to GoLang, coming from the Delphi, C++ world - admittedly very excited about this language, which I think is destined to become "the next big thing".
I am trying to get a handle around how the Go parser and compiler handle pointers and references - can't seem to find any place where some clear rules are laid out.
In the below code sample for example, the return type *list.List and the local variable l are pointer types and require the pointer symbol * in their declarations, but they don't have to be dereferenced in use: l.PushBack(i). But in this same code the input parameter value *int64 is declared as a pointer and has to be dereferenced to be used properly: var i int64 = *value / 2
I assume that this is because list.List is a reference type, so the dereferencing is implicit when used, while int64 is a value type and must be handled just as any pointer to a value type, as in C++ for example: It must be dereferenced.
What is confusing to me is that even though *list.List has to be declared as a pointer type using *, when using the list instance, dereferencing is not required. This had me quite confused initially. Is that "just the way it is", or am I missing something?
Sample:
func GetFactors(value *int64) *list.List {
l := list.New()
l.PushBack(*value)
var i int64 = *value / 2
for ; i > 1; i-- {
if *value%i == 0 {
l.PushBack(i)
}
}
return l
}
All of the methods for a List have *List receivers: (http://golang.org/pkg/container/list/)
func (l *List) Back() *Element
func (l *List) Front() *Element
func (l *List) Init() *List
...
func (l *List) Remove(e *Element) interface{}
In your example l is of type *List, so there's no need to dereference them.
Suppose, instead, that you had something like this:
type A struct{}
func (a A) X() {
fmt.Println("X")
}
func (a *A) Y() {
fmt.Println("Y")
}
You are allowed to write:
a := A{}
a.X()
a.Y() // == (&a).Y()
Or you can do the following:
a := &A{}
a.X() // same like == (*a).X()
a.Y()
But it only works for method receivers. Go will not automatically convert function arguments. Given these functions:
func A(x *int) {
fmt.Println(*x)
}
func B(y int) {
fmt.Println(y)
}
This is invalid:
A(5)
You have to do this:
var x int
A(&x)
This is also invalid:
var y *int
B(y)
You have to do this:
B(*y)
Unlike C# or Java, when it comes to structs, Go does not make a distinction between reference and value types. A *List is a pointer, a List is not. Modifying a field on a List only modifies the local copy. Modifying a field on a *List modifies all "copies". (cause they aren't copies... they all point to the same thing in memory)
There are types which seem to hide the underlying pointer (like a slice contains a pointer to an array), but Go is always pass by value.