英文:
What is the correct way to write a distinct channel in Go?
问题
我是一名Go语言初学者。
我正在尝试找到一种简单的方法来实现一个只输出不同值的通道。
我想要做的是这样的：
package exampleimport ("fmt""testing")func TestShouldReturnDistinctValues(t *testing.T) {c := make(chan int)go func() {c <- 1c <- 1c <- 2c <- 2c <- 3close(c)}()seen := make(map[int]bool)for e := range c {if !seen[e] {seen[e] = true// 只打印1、2和3。fmt.Println(e)}}}
如果我使用一个映射来记住先前的值，我应该担心内存泄漏吗？
英文:
I am a beginner in go.
I am trying to figure out an easy way to implement a channel that only output distinct values.
What I want to do is this:
package exampleimport ("fmt""testing")func TestShouldReturnDistinctValues(t *testing.T) {var c := make([]chan int)c <- 1c <- 1c <- 2c <- 2c <- 3for e := range c {// only print 1, 2 and 3.fmt.println(e)}}
Should I be concern about memory leak here if I were to use a map to remember previous values?
答案1
得分: 7
你真的不能这样做，你必须以某种方式跟踪这些值，map[int]struct{}可能是最节省内存的方法。
一个简单的示例：
func UniqueGen(min, max int) <-chan int {m := make(map[int]struct{}, max-min)ch := make(chan int)go func() {for i := 0; i < 1000; i++ {v := min + rand.Intn(max)if _, ok := m[v]; !ok {ch <- vm[v] = struct{}{}}}close(ch)}()return ch}
英文:
You really can't do that, you'd have to keep a track of the values somehow, a map[int]struct{} is probably the most memory efficient way.
A simple example:
func UniqueGen(min, max int) <-chan int {m := make(map[int]struct{}, max-min)ch := make(chan int)go func() {for i := 0; i < 1000; i++ {v := min + rand.Intn(max)if _, ok := m[v]; !ok {ch <- vm[v] = struct{}{}}}close(ch)}()return ch}
答案2
得分: 3
我以前做过类似的事情，只是我的问题是按升序输出输入。你可以通过添加一个中间的goroutine来实现这个功能。这里有一个示例：
package mainfunc main() {input, output := distinct()go func() {input <- 1input <- 1input <- 2input <- 2input <- 3close(input)}()for i := range output {println(i)}}func distinct() (input chan int, output chan int) {input = make(chan int)output = make(chan int)go func() {set := make(map[int]struct{})for i := range input {if _, ok := set[i]; !ok {set[i] = struct{}{}output <- i}}close(output)}()return}
你可以参考这个示例来解决你的问题。
英文:
I have done similar things before, except my problem was output inputs in ascending order. You can do this by adding a middle go routine. Here is an example:
package mainfunc main() {input, output := distinct()go func() {input <- 1input <- 1input <- 2input <- 2input <- 3close(input)}()for i := range output {println(i)}}func distinct() (input chan int, output chan int) {input = make(chan int)output = make(chan int)go func() {set := make(map[int]struct{})for i := range input {if _, ok := set[i]; !ok {set[i] = struct{}{}output <- i}}close(output)}()return}
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