注意 - Go 中的新手。
我编写了一个多路复用器,它应该将一组通道的输出合并为一个。对建设性的批评感到满意。
func Mux(channels []chan big.Int) chan big.Int {
// Count down as each channel closes. When hits zero - close ch.
n := len(channels)
// The channel to output to.
ch := make(chan big.Int, n)
// Make one go per channel.
for _, c := range channels {
go func() {
// Pump it.
for x := range c {
ch <- x
}
// It closed.
n -= 1
// Close output if all closed now.
if n == 0 {
close(ch)
}
}()
}
return ch
}
我正在测试它:
func fromTo(f, t int) chan big.Int {
ch := make(chan big.Int)
go func() {
for i := f; i < t; i++ {
fmt.Println("Feed:", i)
ch <- *big.NewInt(int64(i))
}
close(ch)
}()
return ch
}
func testMux() {
r := make([]chan big.Int, 10)
for i := 0; i < 10; i++ {
r[i] = fromTo(i*10, i*10+10)
}
all := Mux(r)
// Roll them out.
for l := range all {
fmt.Println(l)
}
}
但我的输出很奇怪:
Feed: 0
Feed: 10
Feed: 20
Feed: 30
Feed: 40
Feed: 50
Feed: 60
Feed: 70
Feed: 80
Feed: 90
Feed: 91
Feed: 92
Feed: 93
Feed: 94
Feed: 95
Feed: 96
Feed: 97
Feed: 98
Feed: 99
{false [90]}
{false [91]}
{false [92]}
{false [93]}
{false [94]}
{false [95]}
{false [96]}
{false [97]}
{false [98]}
{false [99]}
所以我的问题:
- 我在 Mux 中做错了什么吗?
- 为什么我只能从我的输出通道获得最后 10 个?
- 为什么喂食看起来如此奇怪?(每个输入通道的第一个,所有最后一个通道,然后什么都没有)
- 有没有更好的方法来做到这一点?
我需要所有输入通道对输出通道具有平等的权利 - 即我不能从一个通道获得所有输出,然后从下一个通道获得所有输出,等等。
对于任何感兴趣的人 - 这是修复后的最终代码和正确(大概)使用sync.WaitGroup
import (
"math/big"
"sync"
)
/*
Multiplex a number of channels into one.
*/
func Mux(channels []chan big.Int) chan big.Int {
// Count down as each channel closes. When hits zero - close ch.
var wg sync.WaitGroup
wg.Add(len(channels))
// The channel to output to.
ch := make(chan big.Int, len(channels))
// Make one go per channel.
for _, c := range channels {
go func(c <-chan big.Int) {
// Pump it.
for x := range c {
ch <- x
}
// It closed.
wg.Done()
}(c)
}
// Close the channel when the pumping is finished.
go func() {
// Wait for everyone to be done.
wg.Wait()
// Close.
close(ch)
}()
return ch
}