erlang 如何在爱尔朗或 Elixir 中产生n-k光进程?

gzszwxb4  于 2022-12-08  发布在  Erlang
关注(0)|答案(1)|浏览(178)

In Go I can create goroutines like this (EDITED as reported by kelu-thatsall 's answer):

// test.go
package main

import (
        "fmt"
        "os"
        "strconv"
        "sync"
        "runtime"
)

func main() {
        var wg sync.WaitGroup
        if len(os.Args) < 2 {
                os.Exit(1)
        }
        k, ok := strconv.Atoi(os.Args[1])
        if ok != nil {
                os.Exit(2)
        }
        wg.Add(k * 1000)
        for z := 0; z < k*1000; z++ {
            go func(x int) {
                    defer wg.Done()
                    fmt.Println(x)
            }(z)
            if z%k == k-1 {
                // @mattn: avoid busy loop, so Go can start processing like BEAM do 
                runtime.Gosched() 
            }
        }
        wg.Wait()
}

The result in Go 1.8.0 (64-bit):

# shell    
$ go build test.go ; for k in 5 50 500 5000 50000 500000; do echo -n $k; time ./test $k > /dev/null; done

5
CPU: 0.00s      Real: 0.00s     RAM: 2080KB
50
CPU: 0.06s      Real: 0.01s     RAM: 3048KB
500
CPU: 0.61s      Real: 0.12s     RAM: 7760KB
5000
CPU: 6.02s      Real: 1.23s     RAM: 17712KB # 17 MB
50000
CPU: 62.30s     Real: 12.53s    RAM: 207720KB # 207 MB
500000
CPU: 649.47s    Real: 131.53s   RAM: 3008180KB # 3 GB

What's the equivalent code in Erlang or Elixir? (EDITED as reported by patrick-oscity 's comment)
What I've tried so far is the following:

# test.exs
defmodule Recursion do
  def print_multiple_times(n) when n <= 1 do
    spawn fn -> IO.puts n end
  end

  def print_multiple_times(n) do
    spawn fn -> IO.puts n end
    print_multiple_times(n - 1)
  end
end

[x]=System.argv()
{k,_}=Integer.parse(x)
k=k*1000
Recursion.print_multiple_times(k)

The result in elixir 1.4.2 (erts-8.2.2):

# shell    
$ for k in 5 50 500 5000 50000 ; do echo -n $k; time elixir --erl "+P 90000000" test.exs $k > /dev/null; done

5
CPU: 0.53s      Real: 0.50s     RAM: 842384KB # 842 MB
50
CPU: 1.50s      Real: 0.62s     RAM: 934276KB # 934 MB
500
CPU: 11.92s     Real: 2.53s     RAM: 1675872KB # 1.6 GB
5000
CPU: 122.65s    Real: 20.20s    RAM: 4336116KB # 4.3 GB
50000
CPU: 1288.65s   Real: 209.66s   RAM: 6573560KB # 6.5 GB

But I'm not sure if the two are equivalent. Are they ?

  • EDIT* Shortened version as mudasobwa 's comment does not give correct output
# test2.exs
[x]=System.argv()
{k,_}=Integer.parse(x)
k=k*1000
1..k |> Enum.each(fn n -> spawn fn -> IO.puts n end end)

The result for k in 5 50 500 5000 50000 ; do echo -n $k; time elixir --erl "+P 90000000" test.exs $k | wc -l ; done :

5
CPU: 0.35s      Real: 0.41s     RAM: 1623344KB # 1.6 GB
2826 # does not complete, this should be 5000
50
CPU: 1.08s      Real: 0.53s     RAM: 1691060KB # 1.6 GB
35062
500
CPU: 8.69s      Real: 1.70s     RAM: 2340200KB # 2.3 GB
373193
5000        
CPU: 109.95s    Real: 18.49s    RAM: 4980500KB # 4.9 GB
4487475
50000
erl_child_setup closed
Crash dump is being written to: erl_crash.dump...Command terminated by signal 9
CPU: 891.35s    Real: 157.52s   RAM: 24361288KB # 24.3 GB

Not testing 500m for elixir because it took too long and +P 500000000 argument is bad number of processes

vcudknz3

vcudknz31#

I'm sorry guys but I'm not convinced that this code in Go is really working as expected. I'm not an expert, so please correct me if I'm wrong. First of all it prints z which it seems is a current value of it in global scope (usually k*1000 ) https://play.golang.org/p/a4TJyjKBQh

// test.go
package main
import (
  "fmt"
  "time"
)

func main() {

  for z:=0; z<1000; z++ {
    go func(x int) { // I'm passing z to the function with current value now
      fmt.Println(x) 
    }(z)
  }

  time.Sleep(1 * time.Nanosecond)

}

And also if I comment out Sleep the program will exit before even starting any goroutines (at least it doesn't print out the results). I would be happy to know if I'm doing something wrong, but from this simple example it seems the problem is not with Elixir , but Go code provided. Some Go gurus out there?
I've also run some test on my local machine:

go run test.go 500 | wc -l
72442 # expected 500000
go run test.go 5000 | wc -l
76274 # expected 5000000

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