在Go函数通道中的死锁问题

huangapple
go评论122阅读模式
英文:

Deadlock in go function channel

问题

为什么即使我只从通道中传递一个值并获得一个输出，仍然会出现死锁？

  1. package main
  3. import "fmt"
  4. import "math/cmplx"
  6. func max(a []complex128, base int, ans chan float64, index chan int) {
  7.    
  8.     fmt.Printf("called for %d,%d\n",len(a),base)
  9.     
  10.     maxi_i := 0
  11.     maxi := cmplx.Abs(a[maxi_i]);
  13.     for i:=1 ; i< len(a) ; i++ {
  14.         if cmplx.Abs(a[i]) > maxi {
  15.             maxi_i = i
  16.             maxi = cmplx.Abs(a[i])
  17.         }
  18.     }
  19.     
  20.     fmt.Printf("called for %d,%d and found %f %d\n",len(a),base,maxi,base+maxi_i)
  21.     
  22.     ans <- maxi
  23.     index <- base+maxi_i
  24. }
  26. func main() {
  27.     ans := make([]complex128,128)
  28.     
  29.     numberOfSlices := 4
  30.     incr := len(ans)/numberOfSlices
  31.     tmp_val := make([]chan float64,numberOfSlices)
  32.     tmp_index := make([]chan int,numberOfSlices)
  33.     for i,j := 0 , 0; i < len(ans); j++{
  34.         fmt.Printf("From %d to %d - %d\n",i,i+incr,len(ans))
  35.         go max(ans[i:i+incr],i,tmp_val[j],tmp_index[j])
  37.         i = i+ incr
  38.     }
  39. //After Here is it stops deadlock
  40.     maximumFreq := <- tmp_index[0]
  41.     maximumMax := <- tmp_val[0]
  42.     for i := 1; i < numberOfSlices; i++ {
  44.         tmpI := <- tmp_index[i]
  45.         tmpV := <- tmp_val[i]
  47.         if(tmpV > maximumMax ) {
  48.             maximumMax = tmpV
  49.             maximumFreq = tmpI
  50.         }
  51.     }
  53.     fmt.Printf("Max freq = %d",maximumFreq)
  54.     
  55. }
英文:

Why is there a deadlock even tho I just pass one and get one output from the channel?

  1. package main
  3. import &quot;fmt&quot;
  4. import &quot;math/cmplx&quot;
  6. func max(a []complex128, base int, ans chan float64,  index chan int) {
  7.    
  8.     fmt.Printf(&quot;called for %d,%d\n&quot;,len(a),base)
  9.     
  10.     maxi_i := 0
  11.     maxi := cmplx.Abs(a[maxi_i]);
  13.     for i:=1 ; i&lt; len(a) ; i++ {
  14.         if cmplx.Abs(a[i]) &gt; maxi {
  15.             maxi_i = i
  16.             maxi = cmplx.Abs(a[i])
  17.         }
  18.     }
  19.     
  20.     fmt.Printf(&quot;called for %d,%d and found %f %d\n&quot;,len(a),base,maxi,base+maxi_i)
  21.     
  22.     ans &lt;- maxi
  23.     index &lt;- base+maxi_i
  24. }
  26. func main() {
  27.     ans := make([]complex128,128)
  28.     
  29.     numberOfSlices := 4
  30.     incr := len(ans)/numberOfSlices
  31.     tmp_val := make([]chan float64,numberOfSlices)
  32.     tmp_index := make([]chan int,numberOfSlices)
  33.     for i,j := 0 , 0; i &lt; len(ans); j++{
  34.         fmt.Printf(&quot;From %d to %d - %d\n&quot;,i,i+incr,len(ans))
  35.         go max(ans[i:i+incr],i,tmp_val[j],tmp_index[j])
  37.         i = i+ incr
  38.     }
  39. //After Here is it stops deadlock
  40.     maximumFreq := &lt;- tmp_index[0]
  41.     maximumMax := &lt;- tmp_val[0]
  42.     for i := 1; i &lt; numberOfSlices; i++ {
  44.         tmpI := &lt;- tmp_index[i]
  45.         tmpV := &lt;- tmp_val[i]
  47.         if(tmpV &gt; maximumMax ) {
  48.             maximumMax = tmpV
  49.             maximumFreq = tmpI
  50.         }
  51.     }
  53.     fmt.Printf(&quot;Max freq = %d&quot;,maximumFreq)
  54.     
  55. }

答案1

得分: 2

对于阅读这个问题并且可能想知道为什么他的代码在这里失败的人，这里有一个解释。

当他像这样构建他的通道切片时：

  1. tmp_val := make([]chan float64,numberOfSlices)

他创建了一个通道切片，其中每个索引都是通道的零值。通道的零值是nil，因为通道是引用类型，一个nil通道会永远阻塞发送操作，而且由于nil通道中从来没有任何东西，它也会永远阻塞接收操作。因此，你会得到一个死锁。

当footy将他的代码更改为使用循环逐个构建每个通道时：

  1. tmp_val[i] = make(chan float64)

他构建了非nil通道，一切都很好。

英文:

For those reading this question and perhaps wondering why his code failed here's an explanation.

When he constructed his slice of channels like so:

  1. tmp_val := make([]chan float64,numberOfSlices)

He made slice of channels where every index was to the channels zero value. A channels zero value is nil since channels are reference types and a nil channel blocks on send forever and since there is never anything in a nil channel it will also block on recieve forever. Thus you get a deadlock.

When footy changes his code to construct each channel individually using

  1. tmp_val[i] = make(chan float64)

in a loop he constructs non-nil channels and everything is good.

答案2

得分: 1

我在制作chan时犯了错误。应该这样做

  1. numberOfSlices := 4
  2. incr := len(ans)/numberOfSlices
  3. var tmp_val [4]chan float64
  4. var tmp_index [4]chan int
  6. for i := range tmp_val {
  7.     tmp_val[i] = make(chan float64)
  8.     tmp_index[i] = make(chan int)
  9. }
  11. for i,j := 0 , 0; i < len(ans); j++{
  12.     fmt.Printf("从 %d 到 %d [j:%d] - %d\n",i,i+incr,j,len(ans))
  13.     go maximumFunc(ans[i:i+incr],i,tmp_val[j],tmp_index[j])
  15.     i = i+ incr
  16. }
英文:

I was wrong in making of the chan. Should have done

  1. numberOfSlices := 4
  2. incr := len(ans)/numberOfSlices
  3. var tmp_val [4]chan float64
  4. var tmp_index [4]chan int
  6. for i := range tmp_val {
  7.     tmp_val[i] = make(chan float64)
  8.     tmp_index[i] = make(chan int)
  9. }
  11. for i,j := 0 , 0; i &lt; len(ans); j++{
  12.     fmt.Printf(&quot;From %d to %d [j:%d] - %d\n&quot;,i,i+incr,j,len(ans))
  13.     go maximumFunc(ans[i:i+incr],i,tmp_val[j],tmp_index[j])
  15.     i = i+ incr
  16. }

huangapple
  • 本文由 发表于 2013年5月17日 06:11:45
  • 转载请务必保留本文链接：https://go.coder-hub.com/16598353.html
  • deadlock
  • go
  • goroutine
  • parallel-processing
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