CODEWITHSUNDEEP
dartdart-async
richard
richard

Reputation: 3299

Strange async behaviour

I am trying to understand how Dart performs async calls.

Given the following code: 

main() {
  test("test", () async{
    await myCall( );
    print( "done");
  });
}

Future myCall( ) async{

  ["0","1","2"].forEach( (item) async{
    print( "about to callback");
    await futurePrint( item);
    print( "done  callback");
  });
  print( "done mycall");
  return new Future.value();
}

Future futurePrint( String str){
  print( "string = ${str}");
  return new Future.value();
}

I would expect the output to be

about to future print 
string = 0 
done future print 
about to future print 
string = 1
done future print 
about to future print 
string = 2 
done future print 
done mycall
done

But the code actual output is

print( "done mycall");
about to future print
string = 0
about to future print
string = 1
about to future print
string = 2
done
done future print
done future print
done future print

Why does the 'done myCall' occur as the first line printed, not the last? Why do the 'done future print's occur after the 'done' print?

How do I get myCall to only return when the array iteration is complete?

Upvotes: 1

Views: 65

Answers (1)

Günter Zöchbauer
Günter Zöchbauer

Reputation: 657158

["0","1","2"].forEach(<callback>) doesn't wait for futures returned by <callback>.

If you do

Future myCall( ) async {

  for(item in ["0","1","2"]){
    print( "about to callback");
    await futurePrint( item);
    print( "done  callback");
  });
  print( "done mycall");
  return new Future.value();
}

it should do what you expect.

Upvotes: 1

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