javascript variable assignment from chain of functions

Question

Trying to dumb this one down a bit:

    //first named function
    function tryMe(x){
        tryMeAgain(x);
    };

    //second named function
    function tryMeAgain(y){
        return y;
    };

    //assign the result of first function (which will actually be the result of the second function) to a var
    var testTry = tryMe('worth a shot');

    console.log(testTry); //undefined! But I would like 'testTry' to return 'worth a shot'

So i've got two questions:

Why is this?
How to I assign 'testTry' properly?

EDIT BELOW: So the responses all make sense and I think my problem may lay somewhere else in my code. My attempt to simplify the question may have overlooked another piece of the puzzle. I'm including a new version and hoping you all can shed some light:

  var runtime = (function(){

        var $jq = jQuery.noConflict(true);

        function Runtime(){
            this.$jq = $jq;
        }

        Runtime.prototype.method1 = function( _value, _callback ){
            setTimeout(function(){ console.log('dependency1_resolved');
                _callback.apply(this, [{valIs:_value}]);
            }.bind(this), (Math.random() * 1000));
        };

        Runtime.prototype.method2 = function( _value, _callback ){
            var self = this;
            setTimeout(function(){ console.log('dependency2_resolved');
                _callback.apply(self, [{differntValIs:3}]);
            }.bind(this), (Math.random() * 1000));
        };

        Runtime.prototype.method3 = function( _value, _callback ){
            setTimeout(function(){ console.log('dependency3_resolved');
                _callback.apply(this, [{valIs:_value['differntValIs'] *= 4}]);
            }.bind(this), (Math.random() * 1000));
        };

        return new Runtime();
    })();


  runtime.initialize(function( $ ){
    function firstCalc(firstInput){
        return runtime.method1(firstInput,secondCalc);
    };
    function secondCalc(secondInput){
        return runtime.method2(secondInput, thirdCalc);
    };
    function thirdCalc(thirdInput){
        return runtime.method3(thirdInput, fourthCalc);
    };
    function fourthCalc(ourResult){
        //console.log( ourResult );
        return ourResult;
    };

    var _value = firstCalc(4); //this is undefined!!
});

@IsmailBadawi thanks for the quick response. I've updated the question with some more code as my issue seems to lie somewhere else — abigwonderful
– abigwonderful, Commented Aug 13, 2015 at 3:32
I'd suggest reading some basic tutorials about how asynchronous things work in JS. None of your routines return anything--nor can they, because they work in the future--future routines cannot return present values. In addition, using callbacks is not such a great idea, consider using promises. Also, as a matter of protocol, instead of essentially replacing your original question with a new one, you should close the first question and write a new one. — user663031
– user663031, Commented Aug 13, 2015 at 4:01
thatnks @torazaburo. I understand that the 'routines' don't return anything. This whole chunk of code is a bit of an exercise (in futility?) I'm trying to fulfill a set of 'challenges'. In this case I can't edit the 'routines' and can only play with named functions (firstCalc through fourthCalc). Do you have a suggestion for how a series of callbacks setup in a such a way could return a value from the final callback? — abigwonderful
– abigwonderful, Commented Aug 13, 2015 at 13:34
agreed with @torazaburo. I definitely need to revisit callbacks and explore the new hotness of promises. — abigwonderful
– abigwonderful, Commented Aug 16, 2015 at 23:51

Arun P Johny · Accepted Answer · 2015-08-13 03:12:22Z

1

You are returning the value from tryMeAgain but not from tryMe method, so the default value returned is undefined

//first named function
function tryMe(x) {
  return tryMeAgain(x);//need to return the value returned by tryMeAgain to the caller of tryMe
};

//second named function
function tryMeAgain(y) {
  return y;
};

//assign the result of first function to a var
var testTry = tryMe('worth a shot');

snippet.log(testTry);

<!-- Provides the `snippet` object, see http://meta.stackexchange.com/a/242144/134069 -->
<script src="http://tjcrowder.github.io/simple-snippets-console/snippet.js"></script>

answered Aug 13, 2015 at 3:12

Arun P Johny

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1 Comment

abigwonderful Over a year ago

This is a very clear answer, but didn't actually solve my problem. I've updated the question with some new code and am curious what you might have to offer

Sindhoo Oad · Accepted Answer · 2015-08-13 03:54:36Z

0

your whole code is right just with minor mistake.

  function tryMe(x){
             tryMeAgain(x);
         };

in this you are just calling the function but not returning that value who is calling it, then definitely

 var testTry = tryMe('worth a shot');

will be undefined

change it to

function tryMe(x){
     return  tryMeAgain(x);
 };

EDIT

 function firstCalc(firstInput){
        return runtime.method1(firstInput,secondCalc);
    };

your firstCalc is calling secondCalc and secondCalc calling thirdCalc, but what you are missing is your firstCalc , secondCalc and thirdCalc are functions and your passing these values as variable which does not have any value call them as function.

function firstCalc(firstInput){
        var secondCalc = secondCalc(thiValueYouWantToCalculate);
            return runtime.method1(firstInput,secondCalc);
        };

edited Aug 13, 2015 at 3:54

answered Aug 13, 2015 at 3:40

Sindhoo Oad

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2 Comments

abigwonderful Over a year ago

thanks @honeybee. I've updated the question as I dumbed down my true issue with the 'tryme' example. any thoughts?

abigwonderful Over a year ago

thanks for the update. I'm not quite following you there. tried as you described and getting new 'not a function' errors

leo.fcx · Accepted Answer · 2015-08-13 03:13:30Z

0

Default return value of a function is undefined unless you state otherwise. And that is what is happening in your code, testTry returns undefined

answered Aug 13, 2015 at 3:13

leo.fcx

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1 Comment

abigwonderful Over a year ago

thanks @leo-fcx. I've updated the question a bit with some more 'sophisticated' functions. Your answer makes sense, but how can I apply it to the series of callbacks so that the final value is actually defined - in this case = 12?

domehead100 · Accepted Answer · 2015-08-13 04:24:15Z

0

Let's follow the function calls starting with:

var _value = firstCalc(4);

What is the result of firstCalc(4)? Well, firstCalc() returns the result of runtime.method1().

What's the result of runtime.method1(4, secondCalc)? It's undefined, because runtime.method1() simply does a setTimeout to call secondCalc() later, then it returns. There is no explicit return statement, so the default behavior is to return undefined.

That's where your initial expression "var _value = firstCalc(4)" ends.

The setTimeouts cause the other functions to be called after random delays, and indeed inside of fourthCalc() the variable ourResult should be { valIs: 12 }, but this occurs later, after the setTimeouts have caused the various calc functions to be called. Also, the "return ourResult" in fourthCalc returns ourResult to... whom? No one; there is nothing assigning the result of fourthCalc to anything.

Try this:

var _value = firstCalc(4); //this is undefined!!
setTimeout(function() { console.log(_value); }, 5000);

answered Aug 13, 2015 at 4:24

domehead100

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1 Comment

abigwonderful Over a year ago

thanks for looking at this @domehead100. setting a timeout on the console.log(_value) doesn't help. I think I get what you're saying re: there not being an explicit return value, but how do I create one WITHOUT editing the declared method1,method2 and method3. I want the series of callbacks to execute and return the final value (12)

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