How to use yield with my own functions?

Question

I am new to the Generator concept. My understanding is that if a function returns a Promise, then it can be used with yield. So I have a very small node.js script that looks like this:

Q.fcall(function*(){
 var url = "mongodb://" + config.host + ":" + config.port + "/" + config.db;
 var db = yield MongoClient.connect( url );
 var data = yield makeRequest();
 console.log( data );
 db.close();
});
function makeRequest(){
 var deferred = Q.defer();
 request({
 "method" : "GET",
 "url" : "....",
 "headers" : {
 "Accept" : "application/json",
 "user_key" : "...."
 }
 },function(err,data){
 if( err ){
 deferred.reject( err );
 }else{
 deferred.resolve( data );
 }
 });
 return deferred.promise;
} 

I know this works because I am porting it from the callback hell style to generator style. However, I do not see the data in console.log.

What do I need to change to make this work?

• Felix Kling: "My understanding is that if a function returns a Promise, then it can be used with yield." Generators (and yield) have nothing to do with promises. They can be used together to implement something like async/await, but yield itself doesn't treat promises differently from any other value.
• Felix Kling: This should answer your question: What happens when promise is yielded in javascript?
• An SO User: @FelixKling thanks ! I was reading this stackoverflow.com/questions/17516952/… when you posted that :)
• An SO User: @FelixKling I did a little digging around and changed Q.fcall( ) to co() and it worked. So what does co exactly do? :)
• jmunsch: @LittleChild as far as I understand co it basically keeps calling passedInGenerator.next() until the {done: true} object is returned or an error occurs.

Answer 1

Q.fcall is not suited to work with generator functions. You need to use the dedicated generator methods for that, in your case Q.spawn:

Q.spawn(function*() {
 var url = "mongodb://" + config.host + ":" + config.port + "/" + config.db;
 var db = yield MongoClient.connect( url );
 try {
 var data = yield makeRequest();
 console.log( data );
 } finally {
 db.close();
 }
});

Without a runner, the generator won't be automatically advanced asynchronously.

Answer 2

Consider using co that can wrap generators and make the callable. It returns promise.

const co = require('co');
const promisedGenerator = co.wrap(function* () {
 yield whatever1();
 yield whatever2();
 return databaseRequest(); 
});
promisedGenerator()
 .then(databaseResult => {
 console.log(databaseResult);
 });