Map of methods in Go - methods

I have several methods that I'm calling for some cases (like Add, Delete, etc..). However over time the number of cases is increasing and my switch-case is getting longer. So I thought I'd create a map of methods, like Go map of functions; here the mapping of functions is trivial. However, is it possible to create a map of methods in Go?
When we have a method:
func (f *Foo) Add(a string, b int) { }
The syntax below create compile-time error:
actions := map[string]func(a, b){
"add": f.Add(a,b),
}
Is it possible to create a map of methods in Go?

Yes. Currently:
actions := map[string]func(a string, b int){
"add": func(a string, b int) { f.Add(a, b) },
}
Later: see the go11func document guelfi mentioned.

There is currently no way to store both receiver and method in a single value (unless you store it in a struct). This is currently worked on and it may change with Go 1.1 (see http://golang.org/s/go11func).
You may, however, assign a method to a function value (without a receiver) and pass the receiver to the value later:
package main
import "fmt"
type Foo struct {
n int
}
func (f *Foo) Bar(m int) int {
return f.n + m
}
func main() {
foo := &Foo{2}
f := (*Foo).Bar
fmt.Printf("%T\n", f)
fmt.Println(f(foo, 42))
}
This value can be stored in a map like anything else.

I met with a similar question.
How can this be done today, 9 years later:
the thing is that the receiver must be passed to the method map as the first argument. Which is pretty unusual.
package main
import (
"fmt"
"log"
)
type mType struct {
str string
}
func (m *mType) getStr(s string) {
fmt.Println(s)
fmt.Println(m.str)
}
var (
testmap = make(map[string]func(m *mType, s string))
)
func main() {
test := &mType{
str: "Internal string",
}
testmap["GetSTR"] = (*mType).getStr
method, ok := testmap["GetSTR"]
if !ok {
log.Fatal("something goes wrong")
}
method(test, "External string")
}
https://go.dev/play/p/yy3aR_kMzHP

You can do this using Method Expressions:
https://golang.org/ref/spec#Method_expressions
However, this makes the function take the receiver as a first argument:
actions := map[string]func(Foo, string, int){
"add": Foo.Add
}
Similarly, you can get a function with the signature func(*Foo, string, int) using (*Foo).Add

If you want to use pointer to type Foo as receiver, like in:
func (f *Foo) Add(a string, b int) { }
then you can map string to function of (*Foo, string, int), like this:
var operations = map[string]func(*Foo, string, int){
"add": (*Foo).Add,
"delete": (*Foo).Delete,
}
Then you would use it as:
var foo Foo = ...
var op string = GetOp() // "add", "delete", ...
operations[op](&foo, a, b)
where GetOp() returns an operation as string, for example from a user input.
a and b are your string and int arguments to methods.
This assumes that all methods have the same signatures. They can also have return value(s), again of the same type(s).
It is also possible to do this with Foo as receiver instead of *Foo. In that case we don't have to de-reference it in the map, and we pass foo instead of &foo.

Related

Function types in Go - particular type casting to more general type

What cast / assertion need I do in Go in order to pass to a function expecting a generic function like func(interface{}) interface{}, a more specific function like func(int) int instead?
For example, in code like this, fooA can be passed to MakeExclamer, but not fooB:
func MakeExclamer(foo func (interface{}) interface{}, n int) func () {
return func() {
fmt.Printf("%v!!!", foo(n))
}
}
func fooA(x interface{}) interface{} {
return x.(int)*2
}
func fooB(x int) int {
return x * 10
}
func main() {
exclamerA := MakeExclamer(fooA, 12)
exclamerA()
exclamerB := MakeExclamer(fooB, 66)
// >> cannot use fooB (type func(int) int) as type func(interface {}) interface {} in argument to MakeExclamer
exclamerB()
}
(Go Playground link: https://play.golang.org/p/xGzfco0IAG)
I'm not interested much in alternative code structure patterns, since this is how I want it to work: a specific function should be passed to a general function transformer (accepting function of type Any -> Any) that will return another general function (Any -> Any). This may not be idiomatic in Go, but it is the pattern that I want my code to follow.
To use type assertions, every possible type must be enumerated in MakeExclamer:
func MakeExclamer(fn interface{}, arg interface{}) func() {
switch fn := fn.(type) {
case func(int) int:
return func() {
fmt.Printf("%v!!!\n", fn(arg.(int)))
}
case func(interface{}) interface{}:
return func() {
fmt.Printf("%v!!!\n", fn(arg))
}
default:
panic("not supported")
}
}
To accept a function of any type, the fn argument is declared as type interface{}. The code uses a type switch to handle the different function types.
playground example
Reflection can be used to write a more general function.
func MakeExclamer(fn interface{}, arg interface{}) func() {
fnr := reflect.ValueOf(fn)
argr := reflect.ValueOf(arg)
return func() {
resultr := fnr.Call([]reflect.Value{argr})
fmt.Printf("%v!!!\n", resultr[0].Interface())
}
}
playground example
First things first : When it comes to typing in Go, everything is theoretically possible. That's because even though the compiler does a lot of checks at compile-time, it is possible to change the runtime... at runtime. So-called runtime hacks, where you dynamically manipulate runtime structs that you're NOT supposed to handle.
Now, you have an interesting question, whose answer doesn't include the need to use the 'unsafe' package. However, the way I found of generalizing a function involves heavy reflection.
How to call a function (via reflection) ?
The documentation for the reflect package can be found here.
So, like all elements in Golang, functions have a Type. Without going through all fields, functions do take an array of arguments and produce an array of results. It is possible to investigate the Type of arguments and results through the In(int) and Out(int) method.
func investigate(fn interface{}) {
fnType := reflect.TypeOf(fn)
for idx := 0; idx < fnType.NumIn(); idx ++ {
fmt.Printf("Input arg %d has type %v\n", idx, fnType.In(idx))
}
for idx := 0; idx < fnType.NumOut(); idx ++ {
fmt.Printf("Output arg %d has type %v\n", idx, fnType.Out(idx))
}
}
We won't use this code. However, two important things are to be noted at this point :
The generic type under which a function can be passed around without caring about its type is interface{}. Something like "func(interface{}) interface{}" is not a generalization of a function, it is already a concrete type. Hence, "func(interface{}) interface{}" is not a generalization of "func(int) int", those are two different function types entirely. This is why you can't use type assertions/cast to convert from one function type to another.
A function can be represented as something that takes an input array and produces and output array.
Now, in order to call a function, you have to get not its Type, but its Value. Once you get its value, you can call it using an array of arguments, which must all be Values.
The prototype is:
func (v Value) Call(in []Value) []Value
Using this method, it is possible to call any function.
The code
So, the only thing you need is to convert whichever arguments array you have to an array of Values, then you will be able to call your function.
Here is your code:
package main
import (
"fmt"
"reflect"
)
func MakeExclamer(foo interface{}, n int) func() {
exclamer := generalize(foo, n)
return func() {
fmt.Printf("%v!!!\n", exclamer())
}
}
func fooA(x interface{}) interface{} {
return x.(int) * 2
}
func fooB(x int) int {
return x * 10
}
func generalize(implem interface{}, args ...interface{}) func() interface{} {
valIn := make([]reflect.Value, len(args), len(args))
fnVal := reflect.ValueOf(implem)
for idx, elt := range args {
valIn[idx] = reflect.ValueOf(elt)
}
ret := func() interface{} {
res := fnVal.Call(valIn)
// We assume the function produces exactly one result
return res[0].Interface()
}
return ret
}
func main() {
exclamerA := MakeExclamer(fooA, 12)
exclamerA()
exclamerB := MakeExclamer(fooB, 66)
exclamerB()
}
Playground
The important bit is the generalize function which makes the translation between your arguments and an array of Values, then returns a new function whith all parameters already filled.
Do not hesitate if you need any precision !

Return values of function as input arguments to another

If I have
func returnIntAndString() (i int, s string) {...}
And I have:
func doSomething(i int, s string) {...}
Then I can do the following successfully:
doSomething(returnIntAndString())
However, let's say I want to add another argument to doSomething like:
func doSomething(msg string, i int, s string) {...}
Go complains when compiling if I call it like:
doSomething("message", returnIntAndString())
With:
main.go:45: multiple-value returnIntAndString() in single-value context
main.go:45: not enough arguments in call to doSomething()
Is there a way to do this or should I just give up and assign the return values from returnIntAndString to some references and pass msg and these values like doSomething(msg, code, str) ?
It's described here in the spec. It requires the inner function to return the correct types for all arguments. There is no allowance for extra parameters along with a function that returns multiple values.
As a special case, if the return values of a function or method g are
equal in number and individually assignable to the parameters of
another function or method f, then the call f(g(parameters_of_g)) will
invoke f after binding the return values of g to the parameters of f
in order. The call of f must contain no parameters other than the call
of g, and g must have at least one return value. If f has a final ...
parameter, it is assigned the return values of g that remain after
assignment of regular parameters.
func Split(s string, pos int) (string, string) {
return s[0:pos], s[pos:]
}
func Join(s, t string) string {
return s + t
}
if Join(Split(value, len(value)/2)) != value {
log.Panic("test fails")
}
If those specific conditions are not met, then you need to assign the return values and call the function separately.
I had the same question. The best solution I could come up with is creating types or structs for my desired extra parameters and writing methods for them like this:
package main
import (
"fmt"
)
type Message string
type MessageNumber struct {
Message string
Number int
}
func testfunc() (foo int, bar int) {
foo = 4
bar = 2
return
}
func (baz Message) testfunc2(foo int, bar int) {
fmt.Println(foo, bar, baz)
}
func (baz MessageNumber) testfunc3(foo int, bar int) {
fmt.Println(foo, bar, baz.Number, baz.Message)
}
func main() {
Message("the answer").testfunc2(testfunc())
MessageNumber{"what were we talking about again?", 0}.testfunc3(testfunc())
fmt.Println("Done. Have a day.")
}
The output looks like this:
user#Frodos-Atari-MEGA-STE:~/go/test$ go run main.go
4 2 the answer
4 2 0 what were we talking about again?
Done. Have a day.

How to convert a user-defined named type/struct to it's anonymous type

The original question
I have the following Go code. I would like to handle Foo a struct or Bar a type as a string. With "handle" I mean that I would like to convert/cast/whatever it's underlaying value to the (real) type string. I have a workaround, but I find it unintuitive in the case of a struct.
Going for a Type (instead of a struct) seems the better approach. I don't need to maintain any state, nor do I have any need for "inheriting" type specific functionality, so in my case it should work. However a call to type.String() causes stack recursion. I'm mostly curious if I'm not missing something (obvious).
package main
import "fmt"
type Foo struct {
string
}
func (f *Foo) String() string {
return f.string
}
type Bar string
func (b *Bar) String() string {
return fmt.Sprintf("%s", b) // Cannot use: return string(b) here.
}
func main() {
a := Foo{"a"}
var b Bar
b = "b"
fmt.Printf("A is: %s\n", a) // Doesn't call a.String() ?
//fmt.Printf("A is: %s\n", string(a)) // Doesn't work
fmt.Printf("A is: %s\n", a.string) // workaround A
fmt.Printf("A is: %s\n", a.String()) // workaround B, required if I want to use it in a different package
fmt.Printf("B is: %s\n", b) // Calls b.String()
fmt.Printf("B is: %s\n", string(b))
//fmt.Printf("B is: %s\n", b.String()) // Causes a stack overflow
}
Output:
A is: {a}
A is: a
A is: a
B is: b
B is: b
Code on Go's Playground: https://play.golang.org/p/zgrKao4cxa
The behaviour is from Go version 1.5.2
Answers
The following are short examples based on the answers of my original questions. Also, the following post helped in understanding and reasoning of the subject: Value receiver vs. Pointer receiver in Golang?
In case of a type, the following works:
type MyString string
func (b MyString) String() string {
return string(b)
}
Go's Playground link: https://play.golang.org/p/H12bteAk8D
In case of a struct, the following works:
package main
import "fmt"
type MyString struct {
string
someState int
}
func (m MyString) String() string {
return string(m.string)
}
func main() {
// The verbose version:
//var a MyString = MyString{string: "a", someState: 1}
a := MyString{"a", 1}
fmt.Printf("A is: %s\n", a)
fmt.Printf("A is: %s\n", a.String())
}
Go's Playground link: https://play.golang.org/p/GEKeY4rmB8
You've made a pointer receivers for your String methods, but you are working with values, not pointers to them, so it wouldn't apply. You need to switch to pointers or change String methods signatures:
package main
import "fmt"
type Foo struct {
string
}
func (f Foo) String() string {
return "My " + f.string
}
type Bar string
func (b Bar) String() string {
return fmt.Sprintf("My %s", string(b))
}
func main() {
a := Foo{"a"}
var b Bar = "b"
fmt.Printf("A is: %s\n", a)
fmt.Printf("B is: %s\n", b)
}
Please, be careful with receivers type:
The rule about pointers vs. values for receivers is that value methods
can be invoked on pointers and values, but pointer methods can only be
invoked on pointers
Oh, and one more thing. fmt.Sprintf("%s", b) will call String method if it's defined. So, you'll get a recursion.

How do I declare a function pointer to a method in Go

I am trying to create function pointer to a function that has a method receiver. However, I can't figure out how to get it to work (if it is possible)?
Essentially, I have the following:
type Foo struct {...}
func (T Foo) Bar bool {
...
}
type BarFunc (Foo) func() bool // Does not work.
The last line of the code gives the error
syntax error: unexpected func, expecting semicolon or newline
If you want to create a function pointer to a method, you have two ways. The first is essentially turning a method with one argument into a function with two:
type Summable int
func (s Summable) Add(n int) int {
return s+n
}
var f func(s Summable, n int) int = (Summable).Add
// ...
fmt.Println(f(1, 2))
The second way will "bind" the value of s (at the time of evaluation) to the Summable receiver method Add, and then assign it to the variable f:
s := Summable(1)
var f func(n int) int = s.Add
fmt.Println(f(2))
Playground: http://play.golang.org/p/ctovxsFV2z.
Any changes to s after f is assigned will have no affect on the result: https://play.golang.org/p/UhPdYW5wUOP
And for an example more familiar to those of us used to a typedef in C for function pointers:
package main
import "fmt"
type DyadicMath func (int, int) int // your function pointer type
func doAdd(one int, two int) (ret int) {
ret = one + two;
return
}
func Work(input []int, addthis int, workfunc DyadicMath) {
for _, val := range input {
fmt.Println("--> ",workfunc(val, addthis))
}
}
func main() {
stuff := []int{ 1,2,3,4,5 }
Work(stuff,10,doAdd)
doMult := func (one int, two int) (ret int) {
ret = one * two;
return
}
Work(stuff,10,doMult)
}
https://play.golang.org/p/G5xzJXLexc
I am very likely off-target (just started on Golang) but what if you create a pointer then examine type:
pfun := Bar
fmt.Println("type of pfun is:", reflect.TypeOf(pfun))
then it seems that you can declare the type of pointer correctly:
https://play.golang.org/p/SV8W0J9JDuQ

Using function names as parameters

In Go, you can pass functions as parameters like callFunction(fn func). For example:
package main
import "fmt"
func example() {
fmt.Println("hello from example")
}
func callFunction(fn func) {
fn()
}
func main() {
callFunction(example)
}
But is it possible to call a function when it's a member of a struct? The following code would fail, but gives you an example of what I'm talking about:
package main
import "fmt"
type Example struct {
x int
y int
}
var example Example
func (e Example) StructFunction() {
fmt.Println("hello from example")
}
func callFunction(fn func) {
fn()
}
func main() {
callFunction(example.StructFunction)
}
(I know what I'm trying to do in that example is a little odd. The exact problem I have doesn't scale down to a simple example very well, but that's the essence of my problem. However I'm also intrigued about this from an academic perspective)
Methods (which are not "members of a struct" but methods of any named type, not only structs) are first class values. Go 1.0.3 didn't yet implemented method values but the tip version (as in the comming Go 1.1) has support method values. Quoting the full section here:
Method values
If the expression x has static type T and M is in the method set of type T, x.M is called a method value. The method value x.M is a function value that is callable with the same arguments as a method call of x.M. The expression x is evaluated and saved during the evaluation of the method value; the saved copy is then used as the receiver in any calls, which may be executed later.
The type T may be an interface or non-interface type.
As in the discussion of method expressions above, consider a struct type T with two methods, Mv, whose receiver is of type T, and Mp, whose receiver is of type *T.
type T struct {
a int
}
func (tv T) Mv(a int) int { return 0 } // value receiver
func (tp *T) Mp(f float32) float32 { return 1 } // pointer receiver
var t T
var pt *T
func makeT() T
The expression
t.Mv
yields a function value of type
func(int) int
These two invocations are equivalent:
t.Mv(7)
f := t.Mv; f(7)
Similarly, the expression
pt.Mp
yields a function value of type
func(float32) float32
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.Mv is equivalent to (&t).Mv.
f := t.Mv; f(7) // like t.Mv(7)
f := pt.Mp; f(7) // like pt.Mp(7)
f := pt.Mv; f(7) // like (*pt).Mv(7)
f := t.Mp; f(7) // like (&t).Mp(7)
f := makeT().Mp // invalid: result of makeT() is not addressable
Although the examples above use non-interface types, it is also legal to create a method value from a value of interface type.
var i interface { M(int) } = myVal
f := i.M; f(7) // like i.M(7)
Go 1.0 does not support the use of bound methods as function values. It will be supported in Go 1.1, but until then you can get similar behaviour through a closure. For example:
func main() {
callFunction(func() { example.StructFunction() })
}
It isn't quite as convenient, since you end up duplicating the function prototype but should do the trick.
I fixed your compile errors.
package main
import "fmt"
type Example struct {
x, y float64
}
var example Example
func (e Example) StructFunction() {
fmt.Println("hello from example")
}
func callFunction(fn func()) {
fn()
}
func main() {
callFunction(example.StructFunction)
}
Output:
hello from example
To add to #zzzz great answer (and the one given at https://golang.org/ref/spec#Method_values) here is an example that creates a method value from a value of an interface type.
package main
import "fmt"
type T struct{}
func (T) M(i int) { fmt.Println(i) }
func main() {
myVal := T{}
var i interface{ M(int) } = myVal
f := i.M
f(7) // like i.M(7)
}

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