Does a function call in Javascript result in any object creation?

Question

let's say,

function f(){}

f();

When the function f() is defined a function object gets created in memory (along with it's prototype object)

Will the execution of the function further create any new objects? If yes, what it is ?

EDIT:

I am trying to understand the module pattern implemented in Javascript

function f(){
    var name=""; 
    out = {
      getName: function(){return name;}, 
      setName: function(newName){name = newName}
    }; 
    return out;}; 

var x= f(); 

var y = f();

Calling x.setName("foo"); does not affect the value of y.getName()

Where actually is the variable name stored for x and y if not in any object?

Constructor mode of function execution will definitely create a new object. Will the non-constructor mode of function execution i.e. f() create any new object? — Gourav Kumar Shaw
– Gourav Kumar Shaw, Commented Feb 9, 2017 at 11:31
should be easy to check using chrome's profiling instruments — alebianco
– alebianco, Commented Feb 9, 2017 at 11:34
From what I'm told, placing a () next to a function/routine is the closing of a stream, after declaring f opened it. — Joshua
– Joshua, Commented Feb 9, 2017 at 11:44

Fanyo SILIADIN · Accepted Answer · 2017-02-09 12:58:31Z

1

Your edited question is totally different from the first version.. in this case you explicitly create a new object every time you execute f. The returned object is different everytime. so if you do

var x= f(); 

var y = f();

x is different from y.

answered Feb 9, 2017 at 12:58

Fanyo SILIADIN

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2 Comments

Gourav Kumar Shaw Over a year ago

I apologize. Now I understand, I didn't ask the right question. I will ask a new question. Thank you for the help

Fanyo SILIADIN Over a year ago

never mind, you're welcome! mark the answer as resolved if it is :)

Fanyo SILIADIN · Accepted Answer · 2017-02-09 11:54:34Z

1

Indeed the fonction definition will create an object with the variables in its scope and add that object to its scope chain. when executing, it invokes its scope chain an add it to the current scope (the scope chain of the function in which it is called, which already exist). So no calling a function doesn't create any object

edited Feb 9, 2017 at 11:54

answered Feb 9, 2017 at 11:36

Fanyo SILIADIN

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2 Comments

Gourav Kumar Shaw Over a year ago

I am trying to understand the module pattern implemented in Javascript .........function f(){var name=""; out = {getName: function(){return name;}, setName: function(newName){name = newName}}; return out;}; var x= f(); var y = f(); ..............Calling x.setName("foo"); does not affect the value of y.getName() ..... where actually is the variable name stored for x and y if not in any object

Gourav Kumar Shaw Over a year ago

I am trying to understand the module pattern implemented in Javascript

function f(){ var name="";  out = {getName: function(){return name;}, setName: function(newName){name = newName}};  return out;};   var x= f();   var y = f();

Calling x.setName("foo"); does not affect the value of y.getName() ..... where actually is the variable name stored for x and y if not in any object

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