I have been trying to figure out why one particular line will compile, while another won't.. Here's a distilled version:
type A *string
type B *string
func TakeStringPointer(a *string) {
fmt.Println("something: %s\n", *a)
}
func TakeA(a A) {
fmt.Println("something else: %s\n", *a)
}
func Sample() {
aa := "asdf"
var pA A
var pB B
var pString *string
pA = &aa
pB = &aa
pString = &aa
TakeStringPointer(pString)
TakeStringPointer(pA)
TakeStringPointer(pB)
TakeA(pA)
TakeA(pB) // Does not compile
TakeA(pString) // Does compile
}
as far as I understand it, TakeA(pB) and TakeA(pString) should either both work, or both not work...
A value x is assignable to a variable of type T if:
x’s type is identical to T.
x’s type V and T have identical underlying types…
is in the go spec. For me, I would expect both to compile, as both A and B have identical underlying types. (Or, neither would, as *string is not the same thing as A, since we have a type declaration).
What's going on here?
x’s type V and T have identical underlying types…
You quoted the spec and elided the important part. In full, that part of the spec reads:
x's type V and T have identical underlying types and at least one of V or T is not a defined type.
What you have is not a type alias, it is a defined type. (Type aliases are of the form type A = B.) Therefore a function that takes the defined type B cannot take the defined type A; it can take B, or it can take B's underlying type.
Related
I am little consfused about value type struct wrapped by pointer type struct.
Example:
package main
import (
"fmt"
)
type A struct {
id int
B
}
func (a *A) setId(val int) {
a.id = val
}
type B struct {
name string
}
func (b B) setNameViaValue(val string) {
b.name = val
}
func (b *B) setNameViaPointer(val string) {
b.name = val
}
func main() {
a := new(A)
a.setId(1)
a.setNameViaValue("valuename")
fmt.Println(a)
a.setNameViaPointer("pointername")
fmt.Println(a)
}
I would expect that referencing through pointer type A struct(which addresses concrete memory) that wraps B value type struct will set inner value no matter what kind of refence to B is used (B/*B). This is also related to type definition. Can anyone also explain what is different when I define it like this? Is there any usecase scenario?
type A struct {
id int
*B
}
Why doesn't setNameViaValue end up setting the name
If you declare a method with a value (non-pointer) receiver, then that method cannot "modify" the receiver cause it will actually receive a copy.
That's why the "name" is not set as you expected with the setNameViaValue method.
If you want to be able to set the name of B in that way, only setNameViaPointer is an option.
You can read more about the differences between method and value receivers here:
https://tour.golang.org/methods/8
Why can you actually invoke the inner struct methods on the outer struct
That's because you "embedded" the inner struct in the outer one.
When you include a struct type name without giving it a field name, then all of the "inner" struct methods and fields are "promoted" to the outer one.
That means you can call a.setNameViaValue and it will be equivalent to doing a.B.setNameViaValue.
Is there any difference if I embedded as a pointer
If you define A this way:
type A struct {
id int
*B
}
Then:
The "setNameViaValue" will still not work (that's only related to the method being defined over a value instead of a pointer receiver, it has nothing to do with how A references B).
You will need to initialize B explicitly when creating an A object, otherwise *B will end up being nil. (if you reference it as a value, it will be initialized as an empty B)
You can change what *B points to later on (if it was a value, then it will always reference the same "value")
Is it possible to export only a subset of methods implemented by an embedded struct?
Is this a very go-unlike way to reduce copy - and - pasting of code and there is a more idiomatic way to do this?
type A struct {
}
func (a *A) Hello() {
fmt.Println("Hello!")
}
func (a *A) World() {
fmt.Println("World!")
}
type B struct {
A
}
type C struct {
A
}
func main() {
b := B{}
c := C{}
// B should only export the Hello - function
b.Hello()
// C should export both Hello - and World - function
c.Hello()
c.World()
}
This is how embedding works, there's nothing you can do about it. (Actually there is, see dirty trick at the end.)
What you want may be achieved with interfaces though. Make your structs unexported, (B => b and C => c), and create "constructor" like functions, which return interface types, containing only the methods you wish to publish:
type b struct {
A
}
type c struct {
A
}
type Helloer interface {
Hello()
}
type HelloWorlder interface {
Helloer
World()
}
func NewB() Helloer {
return &b{}
}
func NewC() HelloWorlder {
return &c{}
}
You might want to call the interfaces and functions different, this is just for demonstration.
Also note that while the returned Helloer interface does not include the World() method, it is still possible to "reach" it using type assertion, e.g.:
h := NewB() // h is of type Helloer
if hw, ok := h.(HelloWorlder); ok {
hw.World() // This will succeed with the above implementations
}
Try this on the Go Playground.
Dirty trick
If a type embeds the type A, (fields and) methods of A that get promoted will become part of the method set of the embedder type (and thus become methods of type A). This is detailed in Spec: Struct types:
A field or method f of an anonymous field in a struct x is called promoted if x.f is a legal selector that denotes that field or method f.
The focus is on the promotion, for which the selector must be legal. Spec: Selectors describes how x.f is resolved:
The following rules apply to selectors:
For a value x of type T or *T where T is not a pointer or interface type, x.f denotes the field or method at the shallowest depth in T where there is such an f. If there is not exactly one f with shallowest depth, the selector expression is illegal.
[...]
What does this mean? Simply by embedding, B.World will denote the B.A.World method as that is at the shallowest depth. But if we can achieve so that B.A.World won't be the shallowest, the type B won't have this World() method, because B.A.World won't get promoted.
How can we achieve that? We may add a field with name World:
type B struct {
A
World int
}
This B type (or rather *B) will not have a World() method, as B.World denotes the field and not B.A.World as the former is at the shallowest depth. Try this on the Go Playground.
Again, this does not prevent anyone to explicitly refer to B.A.World(), so that method can be "reached" and called, all we achieved is that the type B or *B does not have a World() method.
Another variant of this "dirty trick" is to exploit the end of the first rule: "If there is not exactly one f with shallowest depth". This can be achieved to also embed another type, another struct which also has a World field or method, e.g.:
type hideWorld struct{ World int }
type B struct {
A
hideWorld
}
Try this variant on the Go Playground.
I have an interface that defines one parameter to have type func(interface{}, proto.Message) interface{} and I'm trying to pass something of type func reduceMsg(a interface{}, b proto.Message) []*PersistentData to it. This results in the following compiler error:
Cannot use reduceMsg (type func(a interface{}, b proto.Message) []*PersistentData as type func(interface{}, proto.Message) interface{}
What is the reason for this error, and how can I work around it? It seems like returning a more specific type than interface{} should be legal. Here's a simple complete example that illustrates the issue:
package main
import "fmt"
func main() {
var t func() interface{} = func() []string { return []string{} }
fmt.Println(t)
}
The type of the object is the whole function signature. If the signature don't match, then it's not the same type and can't be assigned that way.
Anything can be assigned to the empty interface, because all types satisfy the interface, but in your problem neither type is the empty interface, you just have a function that returns an empty interface.
Not because a part of the function can be assigned to another it makes it the same. The type is the whole function signature. I think it's the same logic behind not being able to assign an int to an int8. You can cast them if you want, but for go, they are separate types and you need to deal with making the necessary conversions to be able to assign them.
What you can do is change your second function signature to return an empty interface like this:
func(interface{}, proto.Message) interface{}
func reduceMsg(a interface{}, b proto.Message) interface{} {
var a []*PersistentData
// do something here
return a
}
This way the function signature is the same, so it's consider the same type and you are returning an []*PersistentData. Of course you will need to do a type assertion before using it as such because the program will treat it as an {}interface because that is the type that the function returned.
Referencing the spec,
In assignments, each value must be assignable to the type of the operand to which it is assigned, with the following special cases:
Any typed value may be assigned to the blank identifier.
If an untyped constant is assigned to a variable of interface type or the blank identifier, the constant is first converted to its default type.
If an untyped boolean value is assigned to a variable of interface type or the blank identifier, it is first converted to type bool.
Assignability
A value x is assignable to a variable of type T ("x is assignable to T") in any of these cases:
x's type is identical to T.
x's type V and T have identical underlying types and at least one of V or T is not a named type.
T is an interface type and x implements T.
x is a bidirectional channel value, T is a channel type, x's type V and T have identical element types, and at least one of V or T is not a named type.
x is the predeclared identifier nil and T is a pointer, function, slice, map, channel, or interface type.
x is an untyped constant representable by a value of type T.
In general, Go doesn't allow you to implicitly convert values from one type to another, with the exception of being able to use concrete-typed objects as though they were interfaces (that they implement).
In this particular case, since your function doesn't actually return an interface{}, the compiler would have to do some extra work to wrap up the return value as an interface{} and return it; if you really want to accomplish what you're trying you can do this explicitly yourself:
type Foo struct {
X int
}
func create(x int) Foo {
return Foo{X: x}
}
func main() {
var f func(int) interface{} = func(x int) interface{} {
return create(x)
}
}
which is basically doing (explicitly) the wrapping operation that you want the runtime to do implicitly.
Sorry for the ambiguous title. I'm not getting a compiler error when I believe that I should, based on creating a new type and a function that takes an argument of that type.
The example:
package search
//Some random type alias
type Search string
//Takes a string and returns Search
func NewSearch(s string) Search {
return Search(s)
}
//This is where things are getting weird
//Returns an int just for arbitrary testing
func PrintSearch(s Search) int{
return 5
}
Now my assumption would be, if I created an object using NewSearch, I would be able to pass it to PrintSearch and have everything run as expected, but if I passed PrintSearch a primitive string, it should not compile. I am not experiencing this behavior.
The main code:
package main
import (
"fmt"
".../search" //no need to type the whole path here
)
func main() {
SearchTerm := search.NewSearch("Test")
StringTerm := "Another test"
fmt.Println(search.PrintSearch(SearchTerm)) // This should print 5
fmt.Println(search.PrintSearch(StringTerm)) // This should throw a compiler error, but it is not
}
It seems like if I write the type and the function in the same package as main, everything works as I'd expect? As in, it throws a compiler error. Is there something I've missed about cross-package type coercion?
We can simplify this example a bit further (playground):
package main
type Foo string
type Bar int
func main() {
var f Foo = "foo"
var b Bar = 1
println(f, b)
}
This is explained in the spec's assignability section.
A value x is assignable to a variable of type T ("x is assignable to
T") in any of these cases:
x's type is identical to T.
x's type V and T have identical underlying types and at least one of V or T is not a named type.
T is an interface type and x implements T.
x is a bidirectional channel value, T is a channel type, x's type V and T have identical element types, and at least one of V or T is not a named type.
x is the predeclared identifier nil and T is a pointer, function, slice, map, channel, or interface type.
x is an untyped constant representable by a value of type T.
In the documentation for effective Go it states:
As we saw with ByteSize, methods can be defined for any named type
(except a pointer ...
type ByteSlice []byte
func (slice ByteSlice) Append(data []byte) []byte {
// Body exactly the same as above
}
It then goes on to provide an example with a pointer as receiver:
func (p *ByteSlice) Append(data []byte) {
slice := *p
// Body as above, without the return.
*p = slice
}
Doesn't that contradict ? Or does it mean that this isn't valid:
type ByteSlice []byte
type Pb *ByteSlice
func (p Pb) Append(data []byte) []byte {
}
Though it looks just like a typedef!
Named pointer types could lead to ambiguities like so:
type T int
func (t *T) Get() T {
return *t + 1
}
type P *T
func (p P) Get() T {
return *p + 2
}
func F() {
var v1 T
var v2 = &v1
var v3 P = &v1
fmt.Println(v1.Get(), v2.Get(), v3.Get())
}
In the last case (esp. v3) per the current spec, it's ambiguous which Get() method should be called. Yes, there's no reason that the method resolution couldn't be defined, but it's one more detail to add to the language for something that already has a solution.
A named type that is a pointer isn't the same as a pointer to a named type.
This is what the language reference says:
That parameter section must declare a single parameter, the receiver.
Its type must be of the form T or *T (possibly using parentheses)
where T is a type name. The type denoted by T is called the receiver
base type; it must not be a pointer or interface type and it must be
declared in the same package as the method.
That's clear enough: if T denotes a pointer type, then you can't use it as a method receiver.
Note the careful language of the reference here: the phrase "Its type must be of the form T or T*" refers to the syntax-level specification of the method receiver (that is, the "form" of the type). That's different from the phrase "The type denoted by T", where it's talking about the type that T names (that is, what the type "denotes").