Go improper struct initialization? - pointers

Question

While coding I encountered a problem. When I use method of inner struct in goroutine, I can't see inner state like in this code.
package main
import (
"fmt"
"time"
)
type Inner struct {
Value int
}
func (c Inner) Run(value int) {
c.Value = value
for {
fmt.Println(c.Value)
time.Sleep(time.Second * 2)
}
}
type Outer struct {
In Inner
}
func (c Outer) Run() {
go c.In.Run(42)
for {
time.Sleep(time.Second)
fmt.Println(c.In)
}
}
func main() {
o := new(Outer)
o.Run()
}
Program printing:
from inner: {42}
from outer: {0}
from outer: {0}
from inner: {42}
from outer: {0}
from inner: {42}
from outer: {0}
from outer: {0}
Maybe it's pointer problem, but I don't know how resolve it.

Answer 1

The most obvious error in your code is that Inner.Run() has a value-receiver, which means it gets a copy of the Inner type. When you modify this, you modify the copy, and the caller won't see any change on the Inner value.
So first modify it to have a pointer-receiver:
func (c *Inner) Run(value int) {
// ...
}
If a method has a pointer-receiver, the address (pointer) of the value the method is called on will be passed to the method. And inside the method you will modify the pointed value, not the pointer. The pointer points to the same value that is present at the caller, so the same value is modified (and not a copy).
This change alone may make your code work. However, the output of your program is non-deterministic because you modify a variable (field) from one goroutine, and you read this variable from another goroutine too, so you must synchronize access to this field in some way.
One way to synchronize access is using sync.RWMutex:
type Inner struct {
m *sync.RWMutex
Value int
}
When you create your Outer value, initialize this mutex:
o := new(Outer)
o.In.m = &sync.RWMutex{}
Or in one line:
o := &Outer{In: Inner{m: &sync.RWMutex{}}}
And in Inner.Run() lock when you access the Inner.Value field:
func (c *Inner) Run(value int) {
c.m.Lock()
c.Value = value
c.m.Unlock()
for {
c.m.RLock()
fmt.Println(c.Value)
c.m.RUnlock()
time.Sleep(time.Second * 2)
}
}
And you also have to use the lock when you access the field in Outer.Run():
func (c Outer) Run() {
go c.In.Run(42)
for {
time.Sleep(time.Second)
c.In.m.RLock()
fmt.Println(c.In)
c.In.m.RUnlock()
}
}
Note:
Your example only changes Inner.Value once, in the beginning of Inner.Run. So the above code does a lot of unnecessary locks/unlocks which could be removed if the loop in Outer.Run() would wait until the value is set, and afterwards both goroutines could read the variable without locking. In general if the variable can be changed at later times too, the above presented locking/unlocking is required at each read/write.

Answer 2

The simplest way to resolve your issue is to use a pointer receiver in your Run function:
func (c *Inner) Run(value int) {
out = make(chan int)
c.Value = value
for {
fmt.Println(c.Value)
time.Sleep(time.Second * 2)
}
}
But another solution would be to use an out channel to which you can send the Inner struct value:
func (c Inner) Run(value int) {
out = make(chan int)
c.Value = value
for {
fmt.Println(c.Value)
time.Sleep(time.Second * 2)
out <- c.Value
}
}
Then in a separate goroutine to receive back the sent value:
for{
go func() {
c.In.Run(42)
<-out
fmt.Println(out)
}()
time.Sleep(time.Second)
}
Here is the full code:
package main
import (
"fmt"
"time"
)
type Inner struct {
Value int
}
var out chan int
func (c Inner) Run(value int) {
out = make(chan int)
c.Value = value
for {
fmt.Println(c.Value)
time.Sleep(time.Second * 2)
out <- c.Value
}
}
type Outer struct {
In Inner
}
func (c Outer) Run() {
for{
go func() {
c.In.Run(42)
<-out
fmt.Println(out)
}()
time.Sleep(time.Second)
}
}
func main() {
o := new(Outer)
o.Run()
}
https://play.golang.org/p/Zt_NAsM98_