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Pointers vs. values in parameters and return values
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When to use pointers [duplicate]
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Difference between returning a pointer and a value in initialization methods [duplicate]
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Why should constructor of Go return address?
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Why should I use a pointer ( performance)?
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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.
Related
I'm trying to expand my knowledge of Go's function pointers, and I have a question about what is and is not possible with passing functions as parameters in Go.
Let's say that I want to write a decorator() function that can wrap any existing function. For simplicity, let's limit this to functions that accept exactly one parameter and return exactly one value.
If I write a decorator that accepts func(interface{}) interface{} as it's argument, it will implicitly work as long as that function I pass in also accepts/returns an interface{} type (see funcA).
My question is--is there a way to convert an existing function of type func(string) string to a type of func(interface{}) interface{} so that it can also be passed into a decorator function without just wrapping it in a new anonymous function (see funcB)?
package main
import (
"fmt"
)
func decorate(inner func(interface{}) interface{}, args interface{}) interface {} {
fmt.Println("Before inner")
result := inner(args)
fmt.Println("After inner")
return result
}
func funcA(arg interface{}) interface{} {
fmt.Print("Inside A, with arg: ")
fmt.Println(arg)
return "This is A's return value"
}
func funcB(arg string) string {
fmt.Print("Inside B, with arg: ")
fmt.Println(arg)
return "This is B's return value"
}
func main() {
// This one works. Output is:
//
// Before inner
// Inside A, with arg: (This is A's argument)
// After inner
// This is A's return value
//
fmt.Println(decorate(funcA, "(This is A's argument)"))
// This doesn't work. But can it?
//fmt.Println(decorate(funcB, "(This is B's argument)"))
}
This is not possible. One reason for that is the mechanics of passing parameters differ from function to function, and using an interface{} arg does not mean "accept anything". For example, a function taking a struct as an arg will receive each member of that struct, but a function taking an interface{} containing that struct will receive two words, one containing the type of the struct, and the other containing a pointer to it.
So, without using generics, the only way to implement this is by using an adapter function.
Use the reflect package to handle functions with arbitrary argument and result types.
func decorate(inner interface{}, args interface{}) interface{} {
fmt.Println("Before inner")
result := reflect.ValueOf(inner).Call([]reflect.Value{reflect.ValueOf(args)})
fmt.Println("After inner")
return result[0].Interface()
}
Run the code on the playground.
Like the decorate function in the question, the function in this answer assumes one argument and one result. The function must be modified to handle other function types.
The OP should consider the tradeoffs between the anonymous wrapper function proposed in the question and the use of the reflect package here. Calling the function through the reflect API is slower than calling the function through the anonymous wrapper. There's also a loss of type safety with the reflect API. The anonymous wrapper function adds verbosity.
For the record, with Go 1.18 and the introduction of generics, the decorator function becomes almost trivial.
You may declare a type constraint as such:
type UnaryFunc[T any] interface {
func(T) T
}
The constraint itself is parametrized with T to allow for unary functions that take and return arbitrary types.
In the decorate function you then instantiate the constraint with a type parameter. The signature becomes:
decorate[T any, F UnaryFunc[T]](inner F, arg T) T
Thanks to type inference, you can just pass concrete arguments to the function, and both T and F will be unambiguous.
Example alternatives without a named constraint:
// accept and return T
decorate[T any](inner func(T) T, arg T) T
// only return T
decorate[T any](inner func() T) T
// return T and error
decorate[T any](inner func(T) (T, error), arg T) (T, error)
// N-ary function
decorate[T, U any](inner func(T, U) (T, error), argt T, argu U) (T, error)
The obvious limitation is that the interface constraint UnaryFunc specifies only functions that take and return exactly one arg of type T. You can't do otherwise, because the type set of an interface constraint may include types which support the same operations — and calling with one arg is not compatible with calling with N args.
The full program:
package main
import (
"fmt"
)
type UnaryFunc[T any] interface {
func(T) T
}
func decorate[T any, F UnaryFunc[T]](inner F, arg T) T {
fmt.Println("before inner")
result := inner(arg)
fmt.Println("after inner")
return result
}
func funcA(arg int) int {
fmt.Println("inside A with:", arg)
return arg
}
func funcB(arg string) string {
fmt.Println("inside B with:", arg)
return arg
}
func main() {
// this works
decorate(funcA, 200)
// this also works
decorate(funcB, "Func B")
}
Playground: https://go.dev/play/p/3q01NiiWsve
is there a way to convert an existing function of type func(string) string to a type of func(interface{}) interface{} so that it can also be passed into a decorator function without just wrapping it in a new anonymous function (see funcB)?
No. It's that simple: No.
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.
If I have function like this
func TestMethod ( d interface{} ) {
}
If I am calling this as
TestMethod("syz")
Is this pass by value or pass by pointer ?
To summarise some of the discussion in the comments and answer the question:
In go everything in Go is passed by value. In this case the value is an interface type, which is represented as a pointer to the data and a pointer to the type of the interface.
This can be verified by running the following snippet (https://play.golang.org/p/9xTsetTDfZq):
func main() {
var s string = "syz"
read(s)
}
//go:noinline
func read(i interface{}) {
println(i)
}
which will return (0x999c0,0x41a788), one pointer to the data and one pointer to the type of interface.
Updated: Answer and comments above are correct. Just a lite bit of extra information.
Some theory
Passing by reference enables function members, methods, properties,
indexers, operators, and constructors to change the value of the
parameters and have that change persist in the calling environment.
Little code sniped to check how function calls work in GO for pointers
package main_test
import (
"testing"
)
func MyMethod(d interface{}) {
// assume that we received a pointer to string
// here we reassign pointer
newStr := "bar"
d = &newStr
}
func TestValueVsReference(t *testing.T) {
data := "foo"
dataRef := &data
// sending poiner to sting into function that reassigns that pointer in its body
MyMethod(dataRef)
// check is pointer we sent changed
if *dataRef != "foo" {
t.Errorf("want %q, got %q", "bar", *dataRef)
}
// no error, our outer pointer was not changed inside function
// confirms that pointer was sent as value
}
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.
I am passing a pointer to a string, to a method which takes an interface (I have multiple versions of the method, with different receivers, so I am trying to work with empty interfaces, so that I don't end up with a ton of boilerplate madness. Essentially, I want to populate the string with the first value in the slice. I am able to see the value get populated inside the function, but then for some reason, in my application which calls it, tha value doesn't change. I suspect this is some kind of pointer arithmetic problem, but could really use some help!
I have the following interface :
type HeadInterface interface{
Head(interface{})
}
And then I have the following functions :
func Head(slice HeadInterface, result interface{}){
slice.Head(result)
}
func (slice StringSlice) Head(result interface{}){
result = reflect.ValueOf(slice[0])
fmt.Println(result)
}
and... here is my call to the function from an application which calls the mehtod...
func main(){
test := x.StringSlice{"Phil", "Jessica", "Andrea"}
// empty result string for population within the function
var result string = ""
// Calling the function (it is a call to 'x.Head' because I lazily just called th import 'x')
x.Head(test, &result)
// I would have thought I would have gotten "Phil" here, but instead, it is still empty, despite the Println in the function, calling it "phil.
fmt.Println(result)
}
*NOTE : I am aware that getting the first element doesn't need to be this complicated, and could be slice[0] as a straight assertion, but this is more of an exercise in reusable code, and also in trying to get a grasp of pointers, so please don't point out that solution - I would get much more use out of a solution to my actual problem here * :)
As you said in your NOTE, I'm pretty sure this doesn't have to be this complicated, but to make it work in your context:
package main
import (
"fmt"
"reflect"
)
type HeadInterface interface {
Head(interface{})
}
func Head(slice HeadInterface, result interface{}) {
slice.Head(result)
}
type StringSlice []string
func (slice StringSlice) Head(result interface{}) {
switch result := result.(type) {
case *string:
*result = reflect.ValueOf(slice[0]).String()
fmt.Println("inside Head:", *result)
default:
panic("can't handle this type!")
}
}
func main() {
test := StringSlice{"Phil", "Jessica", "Andrea"}
// empty result string for population within the function
var result string = ""
// Calling the function (it is a call to 'x.Head' because I lazily just called th import 'x')
Head(test, &result)
// I would have thought I would have gotten "Phil" here, but instead, it is still empty, despite the Println in the function, calling it "phil.
fmt.Println("outside:", result)
}
The hard part about working with interface{} is that it's hard to be specific about a type's behavior given that interface{} is the most un-specific type. To modify a variable that you pass as a pointer to a function, you have to use the asterisk (dereference) (for example *result) on the variable in order to change the value it points to, not the pointer itself. But to use the asterisk, you have to know it's actually a pointer (something interface{} doesn't tell you) so that's why I used the type switch to be sure it's a pointer to a string.