<分区>
为什么下面的代码总是返回 2,1，而不是 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)
}
<分区>
为什么下面的代码总是返回 2,1，而不是 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)
}
最佳答案
如果你想知道顺序是什么，那就让你的程序包含顺序信息
这个例子使用一个函数闭包来生成一个序列
channel 返回两个数字的结构，其中一个是序列号
序列增量器在 go 例程中应该是安全的，因为序列计数器上有互斥锁
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\n", q)
received++
if received == 30 {
close(c)
}
}
}
第二个版本，从主循环调用序列，使序列号按照函数调用的顺序出现
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\n", q)
received++
if received == 30 {
close(c)
}
}
}
关于goroutine 值返回顺序，我们在Stack Overflow上找到一个类似的问题： https://stackoverflow.com/questions/50771812/