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 typeT
andM
is in the method set of typeT
,x.M
is called a method value. The method valuex.M
is a function value that is callable with the same arguments as a method call ofx.M
. The expressionx
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 typeT
, andMp
, 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)
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
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.