英文:
How to allocate an array of channels
问题
如何创建一个通道数组？
**例如：**用一个大小为5的通道数组替换以下五行代码：
var c0 chan int = make(chan int);var c1 chan int = make(chan int);var c2 chan int = make(chan int);var c3 chan int = make(chan int);var c4 chan int = make(chan int);
英文:
How to create an array of channels?
For example: replace the following five lines with an array of channels, with a size of 5:
var c0 chan int = make(chan int);var c1 chan int = make(chan int);var c2 chan int = make(chan int);var c3 chan int = make(chan int);var c4 chan int = make(chan int);
答案1
得分: 77
语句var chans [5]chan int将分配一个大小为5的数组，但所有的通道都将是nil。
一种方法是使用切片字面量：
var chans = []chan int {make(chan int),make(chan int),make(chan int),make(chan int),make(chan int),}
如果你不想重复自己，你需要遍历它并初始化每个元素：
var chans [5]chan intfor i := range chans {chans[i] = make(chan int)}
英文:
The statement var chans [5]chan int would allocate an array of size 5, but all the channels would be nil.
One way would be to use a slice literal:
var chans = []chan int {make(chan int),make(chan int),make(chan int),make(chan int),make(chan int),}
If you don't want to repeat yourself, you would have to iterate over it and initialize each element:
var chans [5]chan intfor i := range chans {chans[i] = make(chan int)}
答案2
得分: 2
c := make(map[int]chan int)
for i := 1; i <= 5; i++ {
c[i] = make(chan int)
}
for _,v := range c {
fmt.Println(v)
}
你可以这样创建，使用切片和通道
英文:
c := make(map[int]chan int)for i := 1; i <= 5; i++ {c[i] = make(chan int)}for _,v := range c {fmt.Println(v)}
You can create like that, use slice and channel
答案3
得分: 0
[]chan[]string的示例。它可以扩展到所有类型的情况。
package mainimport ("fmt""sync")func main() {var ch [4]chan []stringfor i := range ch {ch[i] = make(chan []string, 1)}ch1 := []string{"this", "that", "who"}ch2 := []string{"one", "two", "three"}ch3 := []string{"four", "five", "six"}ch4 := []string{"seven", "eight", "nine"}wg := sync.WaitGroup{}wg.Add(1)for i := 0; i < 4; i++ {switch i {case 0:ch[i] <- ch1case 1:ch[i] <- ch2case 2:ch[i] <- ch3case 3:ch[i] <- ch4default:}}wg.Done()for i := 0; i < 4; i++ {fmt.Println(<-ch[i])}wg.Wait()}
英文:
Example for []chan[]string. it can be extended for all type of cases.
package mainimport ("fmt""sync")func main() {var ch [4]chan []stringfor i := range ch {ch[i] = make(chan []string, 1)}ch1 := []string{"this", "that", "who"}ch2 := []string{"one", "two", "three"}ch3 := []string{"four", "five", "six"}ch4 := []string{"seven", "eight", "nine"}wg := sync.WaitGroup{}wg.Add(1)for i := 0; i < 4; i++ {switch i {case 0:ch[i] <- ch1case 1:ch[i] <- ch2case 2:ch[i] <- ch3case 3:ch[i] <- ch4default:}}wg.Done()for i := 0; i < 4; i++ {fmt.Println(<-ch[i])}wg.Wait()}
答案4
得分: 0
如果是可变长度数组，在make([]chan int, N)之后使用for循环进行初始化。
c := make([]chan int, 5) // [nil, nil, nil, nil, nil] (尚未初始化)for i := 0; i < len(c); i++ {c[i] = make(chan int)}
英文:
If variable length array, initialize with for loop after make([]chan int, N).
c := make([]chan int, 5) // [nil, nil, nil, nil, nil] (doesn't initialize yet)for i := 0; i < len(c); i++ {c[i] = make(chan int)}
答案5
得分: -5
我认为在这种情况下，你可以使用带缓冲的通道。
通道可以被缓冲。通过将缓冲长度作为第二个参数传递给 make 函数来初始化一个带缓冲的通道：
ch := make(chan int, 5)
当缓冲区满时，向带缓冲的通道发送数据会阻塞。当缓冲区为空时，接收数据会阻塞。
https://tour.golang.org/concurrency/3
英文:
I think you can use buffered channels in this case.
Channels can be buffered. Provide the buffer length as the second argument to make to initialize a buffered channel:
ch := make(chan int, 5)
Sends to a buffered channel block only when the buffer is full. Receives block when the buffer is empty.
通过集体智慧和协作来改善编程学习和解决问题的方式。致力于成为全球开发者共同参与的知识库，让每个人都能够通过互相帮助和分享经验来进步。
评论