Why does javascript function run each time?

Question

Why does the following javascript code run the way it does?

var plusOne = function(num){
  return num+1;
};
var total=plusOne(1);
console.log(total);

var total2=plusOne(3);
console.log(total2);

If I am correct

  var total=plusOne(1);
    console.log(total);

returns a value of 2 to the variable total AND to the variable plusOne which then logs to the console "2", but if the value of plusOne is now 2 than why does

var total2=plusOne(3);
    console.log(total2);

return as a value of 4 because isn't the actual code that it is executing actually

 var total2=2(3);
        console.log(total2);

returns a value of 2 to the variable total AND to the variable plusOne -- no, it does nothing to plusOne. Why would it? Is plusOne on the left hand side of an assignment? — Jon
– Jon, Commented Jan 3, 2014 at 20:21
Would people please stop saying "reference" or "pointer"? Neither of these exist in JavaScript - the terms are pulled in from C/Java/etc, are sometimes used incorrectly (or in an overloaded manner) and don't impart a conceptual advantage. Instead, simply the plusOne expression is evaluated to a function-object (objects, including functions, are first-class values) which is then invoked with the () operator. This is also exactly why/how an IIFE works in JavaScript. — user2864740
– user2864740, Commented Jan 3, 2014 at 20:36

OptimusCrime · Accepted Answer · 2014-01-03 20:24:43Z

1

No.

Javascript does not work this way. Actually, I can't think of any programming-languages that function this way. plusOne is just a pointer to the function.

When you execute the first line the value 2 is stored in total, but nothing happens to plusOne at all.

When you execute the second line, Javascript does not care that the function was called eariler.

answered Jan 3, 2014 at 20:24

OptimusCrime

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1 Comment

user2864740 Over a year ago

JavaScript has no "pointers" or "references" in this sense. Simply, plusOne evaluates to a function-object which is invoked via the () operator.

Kevin Bowersox · Accepted Answer · 2014-01-03 20:22:06Z

0

plusOne is the reference to the function. It only refers to the function and does not store the value it returns. Each call the function referenced by plusOne is independent of prior calls to the function.

The code is functionally equivalent to:

function plusOne(num){
  return num+1;
};
var total=plusOne(1);
console.log(total);

var total2=plusOne(3);
console.log(total2);

answered Jan 3, 2014 at 20:22

Kevin Bowersox

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Comments

Hrishikesh · Accepted Answer · 2014-01-03 20:25:32Z

0

plusOne isnt even assigned the value of the first calculation. plusOne being there or not there doesnt hold any significance. total is assigned the value 2. So, first print is going to be 2 and the second has to be 4 because you pass in 3 during the next function call.

answered Jan 3, 2014 at 20:25

Hrishikesh

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Comments

Akhil · Accepted Answer · 2014-01-03 20:27:35Z

0

plusOne is of type function not a number where as total is an number So plusOne doesnot hold reference to a number but to the function itself.

http://tech.deepumohan.com/2013/07/javascript-pass-function-as-parameter.html

answered Jan 3, 2014 at 20:27

Akhil

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swallace · Accepted Answer · 2014-01-03 20:38:24Z

0

You have declared a function called plusOne() which takes in a single parameter: namely the number that you are supplying when calling the function. plusOne() although assigned as a variable points to a bucket in memory which will return 1+ the number you pass as a parameter. It will not store the previous call to the function's parameter.

There are multiple ways of declaring functions in javascript: read this for reference

answered Jan 3, 2014 at 20:38

swallace

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Comments

Tony · Accepted Answer · 2014-01-03 20:41:20Z

0

so here is an example showing why and what happens during certain changes. Follow this JSFiddle link and click Run and read the comments. http://jsfiddle.net/pGcrL/

Below is the js fiddle code you will find at the link.

var plusOne = function(num){
  return num+1;
};
$('#tempconsoleoutput').html(typeof plusOne);
//Since plusOne is a function you can not make it a variable.
//You must say something like plusOne = plusOne(1);
plusOne = plusOne(1);

$('#tempconsoleoutput2').html(typeof plusOne);

var total=plusOne(1);//With the plusOne now being a number
//you will find that there is an exception in your console now.
console.log(total);

var total2=plusOne(3);
console.log(total2);

answered Jan 3, 2014 at 20:41

Tony

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