Invoking a function of an object inside a for loop - JavaScript

Question

I have an object called foo:

function Foo(){
  this.conceal = function(){ ... };
};

and a method called toggleView, which iterates through a collection of foo and calls conceal() method on each instance:

function toggleView(foos){
  for(a in foos){
    foos[a].conceal();
  }
}

and, of course, foos[a].conceal(); returns: conceal is not a function

There's not enough code here to see what's wrong. If foos really is an array, you should use a plain for loop with a numeric index, or else .forEach(), to iterate. However, what you've got should work unless foos is not what you say it is. — Pointy
– Pointy, Commented May 30, 2016 at 16:05
The loop is visiting a property that doesn't hold an instance of Foo. Note that for..in loops aren't strictly limited to indexes. Related: Why is using “for…in” with array iteration a bad idea? — Jonathan Lonowski
– Jonathan Lonowski, Commented May 30, 2016 at 16:08
Try to invoke a function, outside a loop first, to see if there is any problem. — Walle Cyril
– Walle Cyril, Commented May 30, 2016 at 16:21

Roko C. Buljan · Accepted Answer · 2016-05-30 16:16:03Z

1

function Foo(){
  this.conceal = function( item ) { 
    console.log(item); // "a" "b" "c"
  };
};

function toggleView(foos) {
  var myFoo = new Foo();      // Create new instance (here or outside this fn)
  for(a in foos){
    myFoo.conceal(foos[a]);   // use your instance and pass as items arg. reference
  }
}

toggleView(["a","b","c"]);

Example making toggleView a prototype of Foo

function Foo(){
  this.conceal = function( item ) { 
    console.log(item); // "a" "b" "c"
  };
}

Foo.prototype.toggleView = function(foos) {
  for(a in foos){
    this.conceal(foos[a]);        // invoke .conceal Method using `this` reference
  }
}

var myFoo = new Foo();            // Create new instance
myFoo.toggleView(["a","b","c"]);

edited May 30, 2016 at 16:16

answered May 30, 2016 at 16:10

Roko C. Buljan

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Comments

Gar · Accepted Answer · 2016-05-30 16:09:37Z

0

You should at some point call the function Foo so that the conceal becomes a "function".

answered May 30, 2016 at 16:09

Gar

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