当迭代golang无缓冲通道时,我遇到一个令人困惑的输出。通道定义为chan []int
。然后,我将两个片段推到通道[0 1]和[2 3]。但是,当我从通道中获取元素时,得到了[2 3]和[2 3]。为什么会这样?
package main
import "fmt"
import "sync"
func producer(jobs chan []int, wg *sync.WaitGroup) {
defer wg.Done()
a := make([]int, 2)
index := 0
for i := 0; i < 4; i++ {
a[index] = i
index++
if index == 2 {
index = 0
fmt.Printf("a: %+v\n", a)
jobs <- a
}
}
close(jobs)
}
func main() {
var wg sync.WaitGroup
wg.Add(1)
jobs := make(chan []int, 2)
go producer(jobs, &wg)
for job := range jobs {
fmt.Printf("job: %+v\n", job)
}
wg.Wait()
}
预期输出:
a: [0 1]
a: [2 3]
job: [0 1]
job: [2 3]
实际输出:
a: [0 1]
a: [2 3]
job: [2 3]
job: [2 3]
切片包含指向支持数组的指针。通过通道发送切片时,您正在发送对该后备阵列的引用,因此即使在接收端多次读取该切片时,您实际上也引用了同一共享后备阵列。
您可以为每次迭代创建一个新的切片并将其发送。每个切片将具有一个单独的支持阵列,并且将按预期工作。
略微修改了程序以更好地阅读。
package main
import "fmt"
import "sync"
func producer(jobs chan []int, wg *sync.WaitGroup) {
defer wg.Done()
a := make([]int, 2)
a[0] = 1
a[1] = 2
jobs <- a //We are passing memory location of slice ( is nature of slice ), so the values changing next line will affect here too
a[0] = 2
a[1] = 3
jobs <- a
close(jobs)
}
func main() {
var wg sync.WaitGroup
wg.Add(1)
jobs := make(chan []int, 2)
go producer(jobs, &wg)
for job := range jobs {
fmt.Printf("job: %+v\n", job)
}
wg.Wait()
}
我尝试过使用Array的同一程序,然后我们将得到您期望的结果,请参见下面的代码
package main
import "fmt"
import "sync"
func producer(jobs chan [2]int, wg *sync.WaitGroup) {
defer wg.Done()
var a[2]int
a[0] = 1
a[1] = 2
jobs <- a
a[0] = 2
a[1] = 3
jobs <- a
close(jobs)
}
func main() {
var wg sync.WaitGroup
wg.Add(1)
jobs := make(chan [2]int)
go producer(jobs, &wg)
for job := range jobs {
fmt.Printf("job: %+v\n", job)
}
wg.Wait()
}
Playground Link - Program using Slice
Playground Link - Program using Array
基本规则是切片通过引用传递