Go: Am I creating too many values? - go

Question

If I have a struct like this
type myStruct struct {
mystring string
myint int
}
and if I have a function that returns a new myStruct like this
func New() myStruct {
s := myStruct{}
s.mystring = "string"
s.myint = 1
return s
}
Because I first store it in the "s" variable before returning it, is my function actually making 2 myStruct values instead of one?
And if so, is it then a better practice to make sure I don't first store it in the variable?

Answer 1

The return statement will return a copy of the myStruct object value. If it is a small object then this is fine.
If you intend for the caller to be able to modify this object, and the struct will have methods that use a pointer as the receiver, then it makes more sense to return a pointer to your struct instead:
func New() *myStruct {
s := myStruct{}
s.mystring = "string"
s.myint = 1
return &s
}
You can see the copy happening when you compare the memory address of value vs pointer return types: http://play.golang.org/p/sj6mivYSHg
package main
import (
"fmt"
)
type myStruct struct {
mystring string
myint int
}
func NewObj() myStruct {
s := myStruct{}
fmt.Printf("%p\n", &s)
return s
}
func NewPtr() *myStruct {
s := &myStruct{}
fmt.Printf("%p\n",s)
return s
}
func main() {
o := NewObj()
fmt.Printf("%p\n",&o)
p := NewPtr()
fmt.Printf("%p\n",p)
}
0xf8400235a0 // obj inside NewObj()
0xf840023580 // obj returned to caller
0xf840023640 // ptr inside of NewPtr()
0xf840023640 // ptr returned to caller

Answer 2

I'm definitely not a Go expert (or even novice :) ), but as #max.haredoom mentioned, you can allocate variables in the function signature itself. In that way, you can also omit the s in the return:
package main
import "fmt"
type myStruct struct {
mystring string
myint int
}
func New() (s myStruct) {
s.mystring = "string"
s.myint = 1
return
}
func main() {
r := New()
fmt.Println(r)
}
// Outputs {string 1}
In the examples that I have come across in Effective Go, it does seem to be the most common way of doing things of this nature, but again, I am definitely not an authority on the subject (and will look for additional info on the actual performance).

Answer 3

I think I found the answer by using defer.
I updated the function so that there's a deferred modification to the myStruct value. This means it will happen after the return, but before it is received on the other end.
When I do this, the struct that is received by the caller does not show the updated value, so it appears as though I am indeed returning a copy.
func New() myStruct {
s := myStruct{}
defer func() {
s.mystring = "new value" // defer an update to the value
}()
s.mystring = "string"
s.myint = 1
return s
}
func main() {
b := New()
fmt.Println(b) // still shows the original value
}
http://play.golang.org/p/WWQi8HpDny