I wrote some odd code, but I'm not sure why it works and what I can learn from it. I have a slice type build from another struct. I made a function on the slice type to modify itself. To do this, I seem to have to throw around *'s a little much.
I'm trying to learn about pointers in Go and would like a little help. Here's an example (http://play.golang.org/p/roU3MEeT3q):
var ClientNames = []string {"Client A", "Client B", "ClientC"}
type InvoiceSummaries []InvoiceSummary
type InvoiceSummary struct {
Client string
Amt int
}
func (summaries *InvoiceSummaries) BuildFromAbove() {
for _, name := range ClientNames {
*summaries = append(*summaries, InvoiceSummary{name, 100})
}
}
My question is: What is the purpose for each of these * and why am I not using any &?
What is the purpose for each of these * ?
By making the method receiver as pointer, you could easily change the property of the object. I think that's one of the benefit. This example below will prove it.
package main
import "fmt"
type someStruct struct {
someVar int
}
func (s someStruct) changeVal1(newVal int) {
s.someVar = newVal
}
func (s *someStruct) changeVal2(newVal int) {
s.someVar = newVal
}
func main() {
s := someStruct{0}
fmt.Println(s) // {0}
s.changeVal1(3)
fmt.Println(s) // {0}
s.changeVal2(4)
fmt.Println(s) // {4}
(&s).changeVal2(5)
fmt.Println(s) // {5}
}
and why am I not using any &?
Pointer method receiver is quite special, it can also be called from non-pointer struct object. Both of s.changeVal2(4) and (&s).changeVal2(5) are valid & will affect the value of someVar.
Example http://play.golang.org/p/sxCnCD2D6d
You have to use a pointer for the receiver - (summaries *InvoiceSummaries) - because otherwise the argument is passed by value, having a pointer means you pass a reference to the value instead. If not for that, then you couldn't modify the collection at all.
Inside of the methods body you have use * because it is the dereferncing operator and returns the value at the address. Ampersand (&) is the opposite, it gives the address of a value.
Nothing wrong with your code but normally addresses to slices aren't used. A slice is a small struct that gophers are normally happy to pass by value. If a method or function is creating a new slice, the gopher is happy to return the new slice, by value again, as the return value.
Of course passing a slice by value doesn't guarantee anything about the backing store remaining unchanged when the method/function returns. So it can't be used as a way of guaranteeing the data elements of the slice haven't mutated.
Related
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
}
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.
when I compile my code, I get the following error message, not sure why it happens. Can someone help me point why? Thank you in advance.
cannot use px.InitializePaxosInstance(val) (type PaxosInstance) as
type *PaxosInstance in assignment
type Paxos struct {
instance map[int]*PaxosInstance
}
type PaxosInstance struct {
value interface{}
decided bool
}
func (px *Paxos) InitializePaxosInstance(val interface{}) PaxosInstance {
return PaxosInstance {decided:false, value: val}
}
func (px *Paxos) PartAProcess(seq int, val interface{}) error {
px.instance[seq] = px.InitializePaxosInstance(val)
return nil
}
Your map is expecting a pointer to a PaxosInstance (*PaxosInstance), but you are passing a struct value to it. Change your Initialize function to return a pointer.
func (px *Paxos) InitializePaxosInstance(val interface{}) *PaxosInstance {
return &PaxosInstance {decided:false, value: val}
}
Now it returns a pointer. You can take the pointer of a variable using & and, should you need the struct value itself, dereference it again with *.
After a line like
x := &PaxosInstance{}
or
p := PaxosInstance{}
x := &p
the value type of x is *PaxosInstance. And if you ever need to, you can dereference it back into a PaxosInstance struct value with
p = *x
You usually do not want to pass structs around as actual values, because Go is pass-by-value, which means it will copy the whole thing. Using struct values with maps and slices often results in logic errors because a copy is made should you iterate them or otherwise reference them except via index. It depends on your use-case, but your identifier Instance would infer that you would want to avoid duplications and such logic errors.
As for reading the compiler errors, you can see what it was telling you. The type PaxosInstance and type *PaxosInstance are not the same.
The instance field within the Paxos struct is a map of integer keys to pointers to PaxosInstance structs.
When you call:
px.instance[seq] = px.InitializePaxosInstance(val)
You're attempting to assign a concrete (not pointer) PaxosInstance struct into an element of px.instance, which are pointers.
You can alleviate this by returning a pointer to a PaxosInstance in InitializePaxosInstance, like so:
func (px *Paxos) InitializePaxosInstance(val interface{}) *PaxosInstance {
return &PaxosInstance{decided: false, value: val}
}
or you could modify the instance field within the Paxos struct to not be a map of pointers:
type Paxos struct {
instance map[int]PaxosInstance
}
Which option you choose is up to your use case.
For anyone else pulling their hair out: check your imports.
Not sure when it started happening, but my Visual Studio Code + gopls setup will occasionally insert an import line that references my vendored dependencies path instead of the original import path. I usually won't catch this until I start polishing code for release, or an error like this one pops up.
In my case this caused two otherwise identical types to not compare equally. Once I fixed my imports this resolved the error.
I have a struct type with a *int64 field.
type SomeType struct {
SomeField *int64
}
At some point in my code, I want to declare a literal of this (say, when I know said value should be 0, or pointing to a 0, you know what I mean)
instance := SomeType{
SomeField: &0,
}
...except this doesn't work
./main.go:xx: cannot use &0 (type *int) as type *int64 in field value
So I try this
instance := SomeType{
SomeField: &int64(0),
}
...but this also doesn't work
./main.go:xx: cannot take the address of int64(0)
How do I do this? The only solution I can come up with is using a placeholder variable
var placeholder int64
placeholder = 0
instance := SomeType{
SomeField: &placeholder,
}
Note: the &0 syntax works fine when it's a *int instead of an *int64. Edit: no it does not. Sorry about this.
Edit:
Aparently there was too much ambiguity to my question. I'm looking for a way to literally state a *int64. This could be used inside a constructor, or to state literal struct values, or even as arguments to other functions. But helper functions or using a different type are not solutions I'm looking for.
The Go Language Specification (Address operators) does not allow to take the address of a numeric constant (not of an untyped nor of a typed constant).
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.
For reasoning why this isn't allowed, see related question: Find address of constant in go. A similar question (similarly not allowed to take its address): How can I store reference to the result of an operation in Go?
0) Generic solution (from Go 1.18)
Generics are added in Go 1.18. This means we can create a single, generic Ptr() function that returns a pointer to whatever value we pass to it. Hopefully it'll get added to the standard library. Until then, you can use github.com/icza/gog, the gog.Ptr() function (disclosure: I'm the author).
This is how it can look like:
func Ptr[T any](v T) *T {
return &v
}
Testing it:
i := Ptr(2)
log.Printf("%T %v", i, *i)
s := Ptr("abc")
log.Printf("%T %v", s, *s)
x := Ptr[any](nil)
log.Printf("%T %v", x, *x)
Which will output (try it on the Go Playground):
2009/11/10 23:00:00 *int 2
2009/11/10 23:00:00 *string abc
2009/11/10 23:00:00 *interface {} <nil>
Your other options (prior to Go 1.18) (try all on the Go Playground):
1) With new()
You can simply use the builtin new() function to allocate a new zero-valued int64 and get its address:
instance := SomeType{
SomeField: new(int64),
}
But note that this can only be used to allocate and obtain a pointer to the zero value of any type.
2) With helper variable
Simplest and recommended for non-zero elements is to use a helper variable whose address can be taken:
helper := int64(2)
instance2 := SomeType{
SomeField: &helper,
}
3) With helper function
Note: Helper functions to acquire a pointer to a non-zero value are available in my github.com/icza/gox library, in the gox package, so you don't have to add these to all your projects where you need it.
Or if you need this many times, you can create a helper function which allocates and returns an *int64:
func create(x int64) *int64 {
return &x
}
And using it:
instance3 := SomeType{
SomeField: create(3),
}
Note that we actually didn't allocate anything, the Go compiler did that when we returned the address of the function argument. The Go compiler performs escape analysis, and allocates local variables on the heap (instead of the stack) if they may escape the function. For details, see Is returning a slice of a local array in a Go function safe?
4) With a one-liner anonymous function
instance4 := SomeType{
SomeField: func() *int64 { i := int64(4); return &i }(),
}
Or as a (shorter) alternative:
instance4 := SomeType{
SomeField: func(i int64) *int64 { return &i }(4),
}
5) With slice literal, indexing and taking address
If you would want *SomeField to be other than 0, then you need something addressable.
You can still do that, but that's ugly:
instance5 := SomeType{
SomeField: &[]int64{5}[0],
}
fmt.Println(*instance2.SomeField) // Prints 5
What happens here is an []int64 slice is created with a literal, having one element (5). And it is indexed (0th element) and the address of the 0th element is taken. In the background an array of [1]int64 will also be allocated and used as the backing array for the slice. So there is a lot of boilerplate here.
6) With a helper struct literal
Let's examine the exception to the addressability requirements:
As an exception to the addressability requirement, x [in the expression of &x] may also be a (possibly parenthesized) composite literal.
This means that taking the address of a composite literal, e.g. a struct literal is ok. If we do so, we will have the struct value allocated and a pointer obtained to it. But if so, another requirement will become available to us: "field selector of an addressable struct operand". So if the struct literal contains a field of type int64, we can also take the address of that field!
Let's see this option in action. We will use this wrapper struct type:
type intwrapper struct {
x int64
}
And now we can do:
instance6 := SomeType{
SomeField: &(&intwrapper{6}).x,
}
Note that this
&(&intwrapper{6}).x
means the following:
& ( (&intwrapper{6}).x )
But we can omit the "outer" parenthesis as the address operator & is applied to the result of the selector expression.
Also note that in the background the following will happen (this is also a valid syntax):
&(*(&intwrapper{6})).x
7) With helper anonymous struct literal
The principle is the same as with case #6, but we can also use an anonymous struct literal, so no helper/wrapper struct type definition needed:
instance7 := SomeType{
SomeField: &(&struct{ x int64 }{7}).x,
}
Use a function which return an address of an int64 variable to solve the problem.
In the below code we use function f which accepts an integer and
returns a pointer value which holds the address of the integer. By using this method we can easily solve the above problem.
type myStr struct {
url *int64
}
func main() {
f := func(s int64) *int64 {
return &s
}
myStr{
url: f(12345),
}
}
There is another elegant way to achieve this which doesn't produce much boilerplate code and doesn't look ugly in my opinion. In case I need a struct with pointers to primitives instead of values, to make sure that zero-valued struct members aren't used across the project, I will create a function with those primitives as arguments.
You can define a function which creates your struct and then pass primitives to this function and then use pointers to function arguments.
type Config struct {
Code *uint8
Name *string
}
func NewConfig(code uint8, name string) *Config {
return &Config{
Code: &code,
Name: &name,
}
}
func UseConfig() {
config := NewConfig(1, "test")
// ...
}
// in case there are many values, modern IDE will highlight argument names for you, so you don't have to remember
func UseConfig2() {
config := NewConfig(
1,
"test",
)
// ...
}
If you don't mind using third party libraries, there's the lo package which uses generics (go 1.18+) which has the .ToPtr() function
ptr := lo.ToPtr("hello world")
// *string{"hello world"}
this code works fine but the temp var used to call the function feels clunky
package main
import "fmt"
type Foo struct {
name string
value int
}
// SetName receives a pointer to Foo so it can modify it.
func (f *Foo) SetName(name string) {
f.name = name
}
var users = map[string]Foo{}
func main() {
// Notice the Foo{}. The new(Foo) was just a syntactic sugar for &Foo{}
// and we don't need a pointer to the Foo, so I replaced it.
// Not relevant to the problem, though.
//p := Foo{}
users["a"] = Foo{value: 1}
x := users["a"]
x.SetName("Abc")
users["a"] = x
fmt.Println(users)
}
http://play.golang.org/p/vAXthNBfdP
Unfortunately no. In Go typically pointers are transparent, and values get auto-addressed when you call pointer methods on them. You managed to find one of the few cases where they aren't. That case is map storage -- values in maps are not considered addressable. That is, you can never do val := &map[key].
When you have a value val := Typ{} and methods defined on *Typ, when you try to call val.Method() Go will super secretly do (&val).Method(). Since you can't do &map[key], then this doesn't work so that temporary variable dance you do is the only way.
As for why that's the case, the internals of a map are considered a bit secret to the user, since it's a hashmap it reserves the right to reallocate itself, shuffle around data, etc, allowing you to take the address of any value undermines that. There have been proposals considered to allow this specific case to work (that is: calling a method with a pointer receiver on it), since the fix is so easy, but none have been accepted yet. It may be allowed someday, but not right now.
Following Jsor’s detailed explanation: if you really need to call methods of map values, it seems the only way for now is to use pointers for values.
var users = make(map[string]*Foo)
func main() {
users["a"] = &Foo{value: 1}
users["a"].SetName("Abc")
fmt.Println(users["a"])
}
But that loses you, precisely, the ability to meaningfully print them (values are just memory addresses now). You’d need to write a custom printing function for *Foo:
func (f *Foo) String() string {
return fmt.Sprintf("%v", *f)
}
http://play.golang.org/p/6-y2ewdnre