JavaScript Promise then() ordering -

I'm still learning javascript promise s, and I came in a behavior that I Does not

  var o = $ (" # output "); Var w = function (s) {o.append (s + ""); } Var p = Promise.resolve (). Then (function () {w (0);}). Then (function () {w (1);}); P.then (function () {w (2); New promise back (function (r) {w (3); r ();}). Then (function () {w (4);})}}) .then (function () {w (5);}); P.then (function () {w (6);});  

  & lt; Script src = "https://ajax.googleapis.com/ajax/libs/jquery/2.1.1/jquery.min.js"></script> & Lt; Div id = "output" & gt; & Lt; / Div & gt;       
 I expect these statements to run in sequence - that is, this will be output  
  0 1 2 3 4 5 6  
 
 Instead, the output is 
 
  0 1 2 3 6 4 5  
 < P> Even remove the  promise , what I think, the opposite of the result  1  is before  2 , but there is  5  output before  6 . 
 
 Anyone explain this to me? 
 
 What I have seen is that reassigning  p  every time we expect that I will get the order. 
 
 
 
 
 You can see that the initial reason for  6  is that you have a series No, you are in the branch 
 
 when you call  p.then (). Then (). Then () , you have found a series of promises that the right sequence executed in  should  
 However, if you call  p.then (). Again (); P.then () , you have 2 promises related to  p  - essentially make a branch, and the second branch will be executed with the first. 
 
 You can make sure that you meet them together  p = p.then (). Again (); Pythen ();  
 
 FYI, you almost never want to branch, unless you bring them back together (like  Promise.all ), or deliberately a "fire And forgetting "Creating Branch"