在此示例中渠道如何工作？

Question

This is is an example of prime number sieve

package main

func Generate(ch chan<- int) {
  for i := 2; ; i++ {
    ch <- i
  }
}

func Filter(in <-chan int, out chan<- int, prime int) {
  for {
    i := <-in
    if i%prime != 0 {
      out <- i
    }
  }
}

func main() {
  ch := make(chan int)
  go Generate(ch)
  for i := 0; i < 10; i++ {
    prime := <-ch
    print(prime, "
")
    ch1 := make(chan int)
    go Filter(ch, ch1, prime)
    ch = ch1
  }
}

What I understand is this line of code

prime := <-ch

the channel is waiting for an input and assigns to prime. So, why all the numbers are not printed when the next statement is called

print(prime, "
")

If I remove these last 3 lines

ch1 := make(chan int)
go Filter(ch, ch1, prime)
ch = ch1

then all numbers are printed from 2 to 11. What does this line do ch = ch1?

Thanks

Answer 1

Your code's output is :

2
3
5
7
11
13
17
19
23
29

So the procedure is like this :

i=0,

after prime := <-ch , prime=2, ch = {3};

after go Filter(ch, ch1, prime) , mark as Filter0,in function Filter0 channel in will be 3,4,5,6 ... and channel out will be 3,5,7 ...;

After ch = ch1, So ch = {3}, which will be 3,5,7.

i=1,

after prime := <-ch ,prime= 3, ch ={5},why is there 5 in ch?'Cause now ch is the ch1 in last loop ;

after go Filter(ch, ch1, prime) , mark as Filter1,in function Filter1 channel in will be 5,7,9,11 ... and channel out will be 5,7,11 ...;

After ch = ch1, So ch = {3}, which will be 5,7,11.

i=2, the same .

That how it output.