This question already has an answer here:
Why following codes always return 2,1, not 1,2.
func test(x int, c chan int) {
c <- x
}
func main() {
c := make(chan int)
go test(1, c)
go test(2, c)
x, y := <-c, <-c // receive from c
fmt.Println(x, y)
}
</div>
If you want to know what the order is, then let your program include ordering information
This example uses a function closure to generate a sequence
The channel returns a struct of two numbers, one of which is a sequence order number
The sequence incrementer should be safe across go routines as there is a mutex lock on the sequence counter
package main
import (
"fmt"
"sync"
)
type value_with_order struct {
v int
order int
}
var (
mu sync.Mutex
)
func orgami(x int, c chan value_with_order, f func() int) {
v := new(value_with_order)
v.v = x
v.order = f()
c <- *v
}
func seq() func() int {
i := 0
return func() int {
mu.Lock()
defer mu.Unlock()
i++
return i
}
}
func main() {
c := make(chan value_with_order)
sequencer := seq()
for n := 0; n < 10; n++ {
go orgami(1, c, sequencer)
go orgami(2, c, sequencer)
go orgami(3, c, sequencer)
}
received := 0
for q := range c {
fmt.Printf("%v
", q)
received++
if received == 30 {
close(c)
}
}
}
second version where the sequence is called from the main loop to make the sequence numbers come out in the order that the function is called
package main
import (
"fmt"
"sync"
)
type value_with_order struct {
v int
order int
}
var (
mu sync.Mutex
)
func orgami(x int, c chan value_with_order, seqno int) {
v := new(value_with_order)
v.v = x
v.order = seqno
c <- *v
}
func seq() func() int {
i := 0
return func() int {
mu.Lock()
defer mu.Unlock()
i++
return i
}
}
func main() {
c := make(chan value_with_order)
sequencer := seq()
for n := 0; n < 10; n++ {
go orgami(1, c, sequencer())
go orgami(2, c, sequencer())
go orgami(3, c, sequencer())
}
received := 0
for q := range c {
fmt.Printf("%v
", q)
received++
if received == 30 {
close(c)
}
}
}