如何在Erlang中实现以下循环？

chy5wohz 于 2022-12-08 发布在 Erlang

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I have the following pseudocode:

for ( int i = 0; i < V ; i++ )
{
  for( int j = 0 ; j < V ; j++ )
  {
    if( ( i != j ) && ( tuple {i,j} belong to E ) )
    {
      R[i] := {i,j};
    }
  }
}

I want to parallelise this code using erlang .
How can I achieve the same thing using Erlang? I am new to Erlang...
Edit:
I know that the following code runs both the calls to say/2 concurrently:

-module(pmap).

-export([say/2]).

say(_,0) ->
  io:format("Done ~n");

say(Value,Times) ->
  io:format("Hello ~n"),
  say(Value,Times-1).

start_concurrency(Value1, Value2) ->
  spawn(pmap, say, [Value1, 3]),
  spawn(pmap, say, [Value2, 3]).

However, here we are hardcoding the functions. So, suppose I want to call say 1000 times, do I need to write spawn(pmap, say, [Valuex, 3]) 1000 times? I can use recursion, but won't it be giving a sequential performance?
Edit:
I tried the following code, where I aim to create 3 threads, where each thread wants to run a say function. I want to run these 3 say functions concurrently(Please comment in the box for more clarification):

-module(pmap).

-export([say/1,test/1,start_concurrency/1]).

say(0) ->
  io:format("Done ~n");

say(Times) ->
  io:format("Hello ~p ~n",[Times]),
  say(Times-1).

test(0) ->
  spawn(pmap, say, [3]);

test(Times) ->
  spawn(pmap, say, [3]),
  test(Times-1).

start_concurrency(Times) ->
  test(Times).

Is this code correct?

erlang

来源：https://stackoverflow.com/questions/64727038/how-can-i-implement-the-following-loop-in-erlang

1条答案

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bkkx9g8r1#

I want to run these 3 say functions concurrently. Is this code correct?
You can get rid of your start_concurrency(N) function because it doesn't do anything. Instead, you can call test(N) directly.
I aim to create 3 threads
In erlang, you create processes .
In erlang, indenting is 4 spaces--not 2.
Don't put blank lines between multiple function clauses for a function definition.
If you want to see concurrency in action, then there has to be some waiting in the tasks you are running concurrently. For example:

-module(a).
-compile(export_all).

say(0) ->
    io:format("Process ~p finished.~n", [self()]);
say(Times) ->
    timer:sleep(rand:uniform(1000)),  %%perhaps waiting to receive data from an http request
    io:format("Hello ~p from process ~p~n",[Times, self()]),
    say(Times-1).

loop(0) ->
    spawn(a, say, [3]);
loop(Times) ->
    spawn(a, say, [3]),
    loop(Times-1).

In the shell:

3> c(a).     
a.erl:2: Warning: export_all flag enabled - all functions will be exported
{ok,a}

4> a:loop(3).
<0.84.0>
Hello 3 from process <0.82.0>
Hello 3 from process <0.81.0>
Hello 2 from process <0.82.0>
Hello 3 from process <0.83.0>
Hello 2 from process <0.81.0>
Hello 3 from process <0.84.0>
Hello 2 from process <0.83.0>
Hello 1 from process <0.81.0>
Process <0.81.0> finished.
Hello 1 from process <0.82.0>
Process <0.82.0> finished.
Hello 2 from process <0.84.0>
Hello 1 from process <0.83.0>
Process <0.83.0> finished.
Hello 1 from process <0.84.0>
Process <0.84.0> finished.

5>

If there is no random waiting in the tasks that you are running concurrently, then the tasks will complete sequentially:

-module(a).
-compile(export_all).

say(0) ->
    io:format("Process ~p finished.~n", [self()]);
say(Times) ->
    %%timer:sleep(rand:uniform(1000)),
    io:format("Hello ~p from process ~p~n",[Times, self()]),
    say(Times-1).

loop(0) ->
    spawn(a, say, [3]);
loop(Times) ->
    spawn(a, say, [3]),
    loop(Times-1).

In the shell:

5> c(a).     
a.erl:2: Warning: export_all flag enabled - all functions will be exported
{ok,a}

6> a:loop(3).
Hello 3 from process <0.91.0>
Hello 3 from process <0.92.0>
Hello 3 from process <0.93.0>
Hello 3 from process <0.94.0>
<0.94.0>
Hello 2 from process <0.91.0>
Hello 2 from process <0.92.0>
Hello 2 from process <0.93.0>
Hello 2 from process <0.94.0>
Hello 1 from process <0.91.0>
Hello 1 from process <0.92.0>
Hello 1 from process <0.93.0>
Hello 1 from process <0.94.0>
Process <0.91.0> finished.
Process <0.92.0> finished.
Process <0.93.0> finished.
Process <0.94.0> finished.

7>

When there is no random waiting in the tasks that you are running concurrently, then concurrency provides no benefit.

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如何在Erlang中实现以下循环？

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