I'm a bit perplexed by this go code. I have a struct (Outer) with an embedded struct (Inner), but when I initialize Outer, I intentionally leave the embedded struct uninitialized.
type Inner struct {
value int
}
func (i *Inner) MyFunc() string {
return "inner"
}
func (i *Inner) OnlyInner() string {
return "only inner stuff"
}
type Outer struct {
*Inner
}
func (o *Outer) MyFunc() string {
return "outer"
}
func main() {
// embedded struct is *not* initialized
o := &Outer{}
fmt.Println(o.Inner)
fmt.Println(o.Inner.MyFunc())
fmt.Println(o.Inner.OnlyInner())
//fmt.Println(o.Inner.value)
}
Output:
<nil>
inner
only inner stuff
And if I uncomment the last line (with o.Inner.value), I get a nil pointer dereference error.
What's up here? The effective go page says (https://golang.org/doc/effective_go.html#embedding):
When we embed a type, the methods of that type become methods of the outer type, but when they are invoked the receiver of the method is the inner type, not the outer one.
It seems like in my case, the inner type is <nil>, yet the method calls execute without problem. What's going on under the hood?
A method can be called with a nil receiver, as long as you do not dereference the receiver itself.
This means that the following works playground:
package main
import (
"fmt"
)
type foo struct {
val int
}
func (f *foo) Print() {
fmt.Println("Receiver:", f)
}
func (f *foo) PrintVal() {
fmt.Println("Val: ", f.val)
}
func main() {
var f *foo
f.Print()
//f.PrintVal()
}
f.Print() works without issues since we're just printing a pointer, we're not trying to dereference it.
However, f.PrintVal attempts to dereference a nil pointer, causing a panic.
When in doubt, remember that the methods in this example are equivalent to functions that take the receiver as first parameter:
func Print(f *foo)
func PrintVal(f *foo)
This is mentioned in the spec under method declarations:
The type of a method is the type of a function with the receiver as
first argument. For instance, the method Scale has type
func(p *Point, factor float64)
However, a function declared this way
is not a method.
This makes it clear that the receiver is nothing special, it can be nil as long as you don't dereference it.
The methods of the uninitialized struct are being called with a nil-receiver. If in the methods used that receiver you would get a panic. It is valid to call a method using a nil receiver, and the method could modify its behavior by checking if the receiver is nil.
Related
I have this code
package main
import "fmt"
type Foo struct {
Bar string
}
func (f *Foo) level4() {
fmt.Printf("Bar = %s\n", f.Bar)
}
func (f *Foo) level3() {
f.level4() // panics here, 2 levels down
}
func (f *Foo) level2() {
f.level3()
}
func (f *Foo) level1() {
f.level2()
}
type FooWrapper struct {
foo *Foo
}
func main() {
w := FooWrapper{}
w.foo.level1() // expected it to panic here, since foo is nil
}
As expected, running this gives
panic: runtime error: invalid memory address or nil pointer dereference
[signal SIGSEGV: segmentation violation code=0x1 addr=0x0 pc=0x47f454]
However, I expected the nil pointer dereference to happen at w.foo.level1(), since foo is nil. Instead, it calls levels 1, 2 and 3 and panics there. Why is this the case?
playground link
Why is this the case?
Because w.foo.level1() is valid statement and similarly f.level2(), f.level3(), f.level4() are also a valid statement.
try this
func (f *Foo) level1() {
println("Hello Go")
f.level2()
}
it will print Hello Go and call f.level2() and also see last call level 4
func (f *Foo) level4() {
println("Hello Go 4")
fmt.Printf("Bar = %s\n", f.Bar)
}
it will print Hello Go 4 but panic on next line it is showing the trace or you can say origin of error
The default value of any pointer type is nil.
As you are constructing your object using
w := FooWrapper{}
The value of foo will be assigned the default nil.
To resolve the problem you should pass a value for foo.
w := FooWrapper{foo: &Foo{Bar: "bar"}}
What I usually do to prevent this issue is to make some constructor function.
func NewFooWrapper() *FooWrapper {
return &FooWrapper{foo: &Foo{Bar: "bar"}}
}
w := NewFooWrapper()
Or in case the value of foo can't be hardcoded.
func NewFooWrapper(foo *Foo) *FooWrapper {
return &FooWrapper{foo: foo}
}
w := NewFooWrapper(&Foo{Bar: "bar"})
This makes it more explicit from a coding point of view preventing these errors in other places of code.
You did not initialize the FooWrapper in your main function correctly. Your foo field with type *Foo is nil inside w variable type of FooWrapper because you have not assigned a value for that and the default value for a pointer type variable is nil in Go
Simple correct initialization as follows
w := FooWrapper{
foo: &Foo{
Bar: "bar",
},
}
run in playground
output :
Bar = bar
The reason that panic doesn't come in level1(), level2(), level3() is because they are not using any Foo's fields inside those functions. And also in level4() panic occurs not because of the function call, but because it's using f.Bar in print.
In Go,
If the concrete value inside the interface itself is nil, the method will be called with a nil receiver.
You can find a simple example for this in https://go.dev/tour/methods/12
And also you can find a great explanation for this in Go specs documentation under Calls
A method call x.m() is valid if the method set of (the type of) x contains m and the argument list can be assigned to the parameter list of m. If x is addressable and &x's method set contains m, x.m() is shorthand for (&x).m()
This question already has answers here:
Pointers vs. values in parameters and return values
(5 answers)
When to use pointers [duplicate]
(1 answer)
Difference between returning a pointer and a value in initialization methods [duplicate]
(1 answer)
Why should constructor of Go return address?
(2 answers)
Why should I use a pointer ( performance)?
(3 answers)
Closed 1 year ago.
I am doing a tour of go language, and I have a question about pointers.
Example code (https://tour.golang.org/methods/19):
package main
import (
"fmt"
"time"
)
type MyError struct {
When time.Time
What string
}
func (e *MyError) Error() string {
return fmt.Sprintf("at %v, %s",
e.When, e.What)
}
func run() error {
return &MyError{
time.Now(),
"it didn't work",
}
}
func main() {
if err := run(); err != nil {
fmt.Println(err)
}
}
In this case it is using *MyError and &MyError, but I try to remove the * and & and it works correctly. Why are they using pointers in this example? What is the difference with normal variables? When should I use pointers or not?
"When should I use pointers?" is a very large question without a simple answer. Pointers are a way of passing a reference to a value, rather than a value itself, around, allowing you to modify the original value or "see" modifications to that value. It also prevents copying, which can be a performance improvement in very limited circumstances (do not pass pointers around all the time because it might be a performance improvement). Finally, pointers also let you represent "nothingness", which each pointer being able to be nil. This is both a blessing and a curse as you must check if each pointer is nil before accessing it, though.
In your specific example, the reason why returning &MyError works is because your Error() function operates on a value of *MyError (a pointer to MyError), rather than on a value of MyError itself. This means that *MyError implements the Error interface and is thus assignable to the error type, and so it can be returned from any function that expects an error as a return value.
Returning MyError wouldn't work on its own because MyError is not a *MyError. Go does some helpful things when dealing with function receivers: It will let you call any method on a MyError or *MyError if the receiver is *MyError, but it will only let you call methods on a *MyError if the type is *MyError - That is, Go will not "create" a pointer for you out of thin air.
If you were to remove * from func (e* MyError), you would be telling Go that Error() works on any instance of a MyError, which means that both *MyError and MyError would fulfill that contract. That's why both of the following are valid when you don't use a pointer receiver:
func (e MyError) Error() string {}
var _ error = MyError{} // Valid
var _ error = &MyError {}
In this particular case, using pointers will not make a difference. Here's one way to look at it:
In Go, all variables are passed by value. That means:
type T struct {...}
func f(value T) {..}
f(t)
Above, t is passed as value. That means when f is called, the compiler creates a copy of t and passed that to f. Any modifications f makes to that copy will not affect the t used to call f.
If you use pointers:
func f(value *T) {...}
f(&t)
Above, the compiler will create a pointer pointing to t, and pass a copy of that to f. If f makes changes to value, those changes will be made on the instance of t used to call f. In other words:
type T struct {
x int
}
func f(value T) {
value.x=1
}
func main() {
t:=T{}
f(t)
fmt.Println(t.x)
}
This will print 0, because the modifications made by f is done on a copy of t.
func f(value *T) {
value.x=1
}
func main() {
t:=T{}
f(&t)
fmt.Println(t.x)
}
Above, it will print 1, because the call to f changes t.
Same idea applies to methods and receivers:
type T struct {
x int
}
func (t T) f() {
t.x=1
}
func main() {
t:=T{}
t.f()
fmt.Println(t.x)
}
Above program will print 0, because the method modifies a copy of t.
func (t *T) f() {
t.x=1
}
func main() {
t:=T{}
t.f()
fmt.Println(t.x)
}
Above program will print 1, because the receiver of the method is declared with a pointer, and calling t.f() is equivalent to f(&t).
So, use pointers when passing arguments or when declaring methods if you want to modify the object, or if copying the object would be too expensive.
This is only a small part of the story about pointer arguments.
Version of Go
go version go1.11 darwin/amd64
Code 1:
package main
import "fmt"
type myintf interface {
GotU()
}
type esc struct {
i int
}
//func GotU(t esc);
func (e esc)GotU() {
e.i = 10
}
func TestFunc(it myintf) string {
it.GotU()
return "kk"
}
func main() {
var test esc
test.i = 9
TestFunc(test)
fmt.Println(test.i)
}
Code 2:
package main
import "fmt"
type myintf interface {
GotU()
}
type esc struct {
i int
}
func (e esc)GotU() {
e.i = 10
}
func TestFunc(it myintf) string {
it.GotU()
return "kk"
}
func main() {
var test esc
test.i = 9
TestFunc(&test)
fmt.Println(test.i)
}
Code 3:
package main
import "fmt"
type myintf interface {
GotU()
}
type esc struct {
i int
}
func (e *esc)GotU() {
e.i = 10
}
func TestFunc(it myintf) string {
it.GotU()
return "kk"
}
func main() {
var test esc
test.i = 9
TestFunc(test)
fmt.Println(test.i)
}
The outputs:
code 1 output: 9
code 2 output: 9
code 3 cannot be compiled due to a type mismatch
Since only func (e esc)GotU() implemented, why should both pieces of code work and deliver the same result?
It's kind of confusing for me to pass a pointer of struct to that function (TestFunc) to get the same answer.
The last code snippet has implemented a method receiver of pointer type. This will consider the situation if you want to modify the value of receiver.
func (e *esc) GotU() {
e.i = 10
}
In above case Since you are passing pointer type receiver on a method which is implementing the interface.
type myintf interface {
GotU()
}
So you need to pass address of struct in TestFunc. This is the reason you are getting type mismatch error, because you are passing variable of esc type while your method requires variable of *esc.
func main() {
var test esc
test.i = 9
TestFunc(&test)
fmt.Println(test.i)
}
Working example on Go playground
In Golang there are two ways to pass a method receiver.
func (s *MyStruct) pointerMethod() { } // method on pointer
func (s MyStruct) valueMethod() { } // method on value
For programmers unaccustomed to pointers, the distinction between
these two examples can be confusing, but the situation is actually
very simple. When defining a method on a type, the receiver (s in the
above examples) behaves exactly as if it were an argument to the
method. Whether to define the receiver as a value or as a pointer is
the same question, then, as whether a function argument should be a
value or a pointer. There are several considerations
First, and most important, does the method need to modify the receiver? If it does, the receiver must be a pointer. (Slices and maps act as references, so their story is a little more subtle, but for instance to change the length of a slice in a method the receiver must still be a pointer.) In the examples above, if pointerMethod modifies the fields of s, the caller will see those changes, but valueMethod is called with a copy of the caller's argument (that's the definition of passing a value), so changes it makes will be invisible to the caller.
The difference between the 1st and second version is, that you pass the struct directly in one version and the pointer to the struct in the other version. In this case these programs work the same, as the pointer also includes the all defined funcs on the struct.
But this does not work the other way around. You define the method GotU on the pointer receiver. The struct does not know about this func. If you would call
TestFunc(&test)
in the third program, it would compile but work differently than the other two: The output is: "10"
As the GotU is defined on the pointer receiver test is passed as reference and the modifications persist. In the other programs test is passed as value, i.e. it is copied, the copy is modified in GotU. When the func exits, the copy is discarded and the old version is still the same as before.
Assuming we have an understanding that,
For explicit method definition for type X, GO compiler implicitly defines the same method for type *X and vice versa, if I declare,
func (c Cat) foo(){
//do stuff_
}
and declare,
func (c *Cat) foo(){
// do stuff_
}
then GO compiler gives error,
Compile error: method re-declared
which indicates that, pointer method is implicitly defined and vice versa
In the below code,
package main
type X interface{
foo();
bar();
}
type Cat struct{
}
func (c Cat) foo(){
// do stuff_
}
func (c *Cat) bar(){
// do stuff_
}
func main() {
var c Cat
var p *Cat
var x X
x = p // OK; *Cat has explicit method bar() and implicit method foo()
x = c //compile error: Cat has explicit method foo() and implicit method bar()
}
GO compiler gives error,
cannot use c (type Cat) as type X in assignment:
Cat does not implement X (bar method has pointer receiver)
at x = c, because, implicit pointer methods satisfy interfaces, but implicit non-pointer methods do not.
Question:
Why implicit non-pointer methods do not satisfy interfaces?
Let's look into the language specification:
A type may have a method set associated with it. The method set
of an interface type is its interface. The method set of any other
type T consists of all methods declared with receiver type T. The
method set of the corresponding pointer type *T is the set of all
methods declared with receiver *T or T (that is, it also contains
the method set of T).
In your example, the method set of the interface type x is [foo(), bar()]. The method set of the type Cat is [foo()], and the method set of the type *Cat is [foo()] + [bar()] = [foo(), bar()].
This explains, why variable p satisfies the interface x, but variable c doesn't.
Method set
Following the spec:
The method set of any other named type T consists of all methods with receiver type T. The method set of the corresponding pointer type *T is the set of all methods with receiver *T or T (that is, it also contains the method set of T).
Method set definition sounds weird until you follow addressable and not addressable types concept.
Addressable and not addressable types
It is possible to call a pointer receiver method on a value if the value is of addressable type.
As with selectors, a reference to a non-interface method with a value receiver using a pointer will automatically dereference that pointer: pt.Mv is equivalent to (*pt).Mv.
As with method calls, a reference to a non-interface method with a pointer receiver using an addressable value will automatically take the address of that value: t.Mp is equivalent to (&t).Mp.
It is ok to call pointer receiver methods on values till you are dealing with addressable types (struct is addressable):
type Cat struct {}
func (c *Cat) bar() string { return "Mew" }
func main() {
var c Cat
c.bar()
}
Variables of interface type are not addressable
But not all Go types are addressable. Also variables referenced through interfaces are not addressable.
It is impossible to call pointer receiver on values of not addressable types:
type X interface {
bar() string
}
type Cat struct{}
func (c *Cat) bar() string { return "Mew" }
/* Note `cat` variable is not a `struct` type value but
it is type of `X` interface therefor it is not addressable. */
func CatBar(cat X) {
fmt.Print(cat.bar())
}
func main() {
var c Cat
CatBar(c)
}
So with the following error Go runtime prevents segment fault:
cannot use c (type Cat) as type X in assignment:
Cat does not implement X (bar method has pointer receiver)
Add a little to dev.bmax's answer.
type Cat struct{
}
func (c Cat) foo(){
// do stuff_
}
func (c *Cat) bar(){
// do stuff_
}
you can do
var c cat
c.bar() // ok to call bar(), since c is a variable.
but not
cat{}.bar() // not ok to call bar(), c is not a variable.
It's legal to call a *T method on an argument of type T so long as the argument is a variable; the compiler implicitly takes its address. But this is mere syntactic sugar: a value of type T does not posses all methods that a *T pointer does, and as a result it might satisfy fewer interfaces.
On the other hand, you can always call foo() with Cat or *Cat.
How about this?
package main
import (
"fmt"
)
type Growler interface{
Growl() bool
}
type Cat struct{
Name string
Age int
}
// *Cat is good for both objects and "references" (pointers to objects)
func (c *Cat) Speak() bool{
fmt.Println("Meow!")
return true
}
func (c *Cat) Growl() bool{
fmt.Println("Grrr!")
return true
}
func main() {
var felix Cat // is not a pointer
felix.Speak() // works :-)
felix.Growl() // works :-)
var ginger *Cat = new(Cat)
ginger.Speak() // works :-)
ginger.Growl() // works :-)
}
The following code works fine. Two methods operating on two different structs and printing a field of the struct:
type A struct {
Name string
}
type B struct {
Name string
}
func (a *A) Print() {
fmt.Println(a.Name)
}
func (b *B) Print() {
fmt.Println(b.Name)
}
func main() {
a := &A{"A"}
b := &B{"B"}
a.Print()
b.Print()
}
Shows the desired output in the console:
A
B
Now, if I change the method signature in the following way I get an compile error. I just move the receiver of the method to the arguments of the method:
func Print(a *A) {
fmt.Println(a.Name)
}
func Print(b *B) {
fmt.Println(b.Name)
}
func main() {
a := &A{"A"}
b := &B{"B"}
Print(a)
Print(b)
}
I can't even compile the program:
./test.go:22: Print redeclared in this block
previous declaration at ./test.go:18
./test.go:40: cannot use a (type *A) as type *B in function argument
Why is it that I can interchange struct types in the receiver, but not in the
arguments, when the methods have the same name and arity?
Because Go does not support overloading of user-defined functions on their argument types.
You can make functions with different names instead, or use methods if you want to "overload" on only one parameter (the receiver).
You can use type introspection. As a general rule, though, any use of the generic interface{} type should be avoided, unless you are writing a large generic framework.
That said, a couple of ways to skin the proverbial cat:
Both methods assume a Print() method is defined for both types (*A and *B)
Method 1:
func Print(any interface{}) {
switch v := any.(type) {
case *A:
v.Print()
case *B:
v.Print()
default:
fmt.Printf("Print() invoked with unsupported type: '%T' (expected *A or *B)\n", any)
return
}
}
Method 2:
type Printer interface {
Print()
}
func Print(any interface{}) {
// does the passed value honor the 'Printer' interface
if v, ok := any.(Printer); ok {
// yes - so Print()!
v.Print()
} else {
fmt.Printf("value of type %T passed has no Print() method.\n", any)
return
}
}
If it's undesirable to have a Print() method for each type, define targeted PrintA(*A) and PrintB(*B) functions and alter Method 1 like so:
case *A:
PrintA(v)
case *B:
PrintB(v)
Working playground example here.
You can not do function or method overloading in Go. You can have two methods with the same names in Go but the receiver of these methods must be of different types.
you can see more in this link .