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?
package main
import "fmt"
import "math/cmplx"
func max(a []complex128, base int, ans chan float64, index chan int) {
fmt.Printf("called for %d,%d\n",len(a),base)
maxi_i := 0
maxi := cmplx.Abs(a[maxi_i]);
for i:=1 ; i< len(a) ; i++ {
if cmplx.Abs(a[i]) > maxi {
maxi_i = i
maxi = cmplx.Abs(a[i])
}
}
fmt.Printf("called for %d,%d and found %f %d\n",len(a),base,maxi,base+maxi_i)
ans <- maxi
index <- base+maxi_i
}
func main() {
ans := make([]complex128,128)
numberOfSlices := 4
incr := len(ans)/numberOfSlices
tmp_val := make([]chan float64,numberOfSlices)
tmp_index := make([]chan int,numberOfSlices)
for i,j := 0 , 0; i < len(ans); j++{
fmt.Printf("From %d to %d - %d\n",i,i+incr,len(ans))
go max(ans[i:i+incr],i,tmp_val[j],tmp_index[j])
i = i+ incr
}
//After Here is it stops deadlock
maximumFreq := <- tmp_index[0]
maximumMax := <- tmp_val[0]
for i := 1; i < numberOfSlices; i++ {
tmpI := <- tmp_index[i]
tmpV := <- tmp_val[i]
if(tmpV > maximumMax ) {
maximumMax = tmpV
maximumFreq = tmpI
}
}
fmt.Printf("Max freq = %d",maximumFreq)
}
英文:
Why is there a deadlock even tho I just pass one and get one output from the channel?
package main
import "fmt"
import "math/cmplx"
func max(a []complex128, base int, ans chan float64, index chan int) {
fmt.Printf("called for %d,%d\n",len(a),base)
maxi_i := 0
maxi := cmplx.Abs(a[maxi_i]);
for i:=1 ; i< len(a) ; i++ {
if cmplx.Abs(a[i]) > maxi {
maxi_i = i
maxi = cmplx.Abs(a[i])
}
}
fmt.Printf("called for %d,%d and found %f %d\n",len(a),base,maxi,base+maxi_i)
ans <- maxi
index <- base+maxi_i
}
func main() {
ans := make([]complex128,128)
numberOfSlices := 4
incr := len(ans)/numberOfSlices
tmp_val := make([]chan float64,numberOfSlices)
tmp_index := make([]chan int,numberOfSlices)
for i,j := 0 , 0; i < len(ans); j++{
fmt.Printf("From %d to %d - %d\n",i,i+incr,len(ans))
go max(ans[i:i+incr],i,tmp_val[j],tmp_index[j])
i = i+ incr
}
//After Here is it stops deadlock
maximumFreq := <- tmp_index[0]
maximumMax := <- tmp_val[0]
for i := 1; i < numberOfSlices; i++ {
tmpI := <- tmp_index[i]
tmpV := <- tmp_val[i]
if(tmpV > maximumMax ) {
maximumMax = tmpV
maximumFreq = tmpI
}
}
fmt.Printf("Max freq = %d",maximumFreq)
}
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:
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:
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:
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
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
numberOfSlices := 4
incr := len(ans)/numberOfSlices
var tmp_val [4]chan float64
var tmp_index [4]chan int
for i := range tmp_val {
tmp_val[i] = make(chan float64)
tmp_index[i] = make(chan int)
}
for i,j := 0 , 0; i < len(ans); j++{
fmt.Printf("从 %d 到 %d [j:%d] - %d\n",i,i+incr,j,len(ans))
go maximumFunc(ans[i:i+incr],i,tmp_val[j],tmp_index[j])
i = i+ incr
}
英文:
I was wrong in making of the chan. Should have done
numberOfSlices := 4
incr := len(ans)/numberOfSlices
var tmp_val [4]chan float64
var tmp_index [4]chan int
for i := range tmp_val {
tmp_val[i] = make(chan float64)
tmp_index[i] = make(chan int)
}
for i,j := 0 , 0; i < len(ans); j++{
fmt.Printf("From %d to %d [j:%d] - %d\n",i,i+incr,j,len(ans))
go maximumFunc(ans[i:i+incr],i,tmp_val[j],tmp_index[j])
i = i+ incr
}
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