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 int
for 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 int
for 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 main
import (
"fmt"
"sync"
)
func main() {
var ch [4]chan []string
for 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] <- ch1
case 1:
ch[i] <- ch2
case 2:
ch[i] <- ch3
case 3:
ch[i] <- ch4
default:
}
}
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 main
import (
"fmt"
"sync"
)
func main() {
var ch [4]chan []string
for 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] <- ch1
case 1:
ch[i] <- ch2
case 2:
ch[i] <- ch3
case 3:
ch[i] <- ch4
default:
}
}
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.
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