Question

var async = require('async');

function f1() {
    setTimeout(function(){ console.log('111');},300);
}

function f2() {
    setTimeout(function(){ console.log('222');},100);
}

async.series([f1,f2], function(err, result){
    console.log(result);
});

It prints:

111

No 222? Why?

Answer 1

f1() and f2() are expected to complete asynchronously, so they are given a callback to invoke when they are “done”. Try this instead:

var async = require('async');

function f1(done) {
    setTimeout(function(){ console.log('111'); done(); },300);
}

function f2(done) {
    setTimeout(function(){ console.log('222'); done(); },100);
}

async.series([f1,f2], function(err, result){
    console.log(result);
});

---

Edit: regarding the “why”, it’s because f1() gets invoked immediately, and so your initial setTimeout() is scheduled as expected. However, since you are never telling async that f1() is done, it never invokes f2().