Calling function from array using setTimeout in Javascript

Question

I am creating two function name call and call2 and putting this two functions in array x. In another function called timer i am trying to call this function every 2 seconds. But its giving error expected an assigment or function call instead show and expression.

And also i dont want this functions to run when i create an array. this my code http://jsbin.com/IMiVadE/2/

function call(name)
{
  console.log("Hello " + name);
}

function call2()
{
  console.log("Hello world");
}

var x = [call("Nakib"), call2()];
var i = 0;

function timer(x, i)
{
  x[i];
  i++;
  if(i<x.length)
     window.setTimeout(timer(x, i), 2000);

}
timer(x, i);

Sanjo · Accepted Answer · 2013-10-06 08:37:12Z

1

You have some mistakes in your code:

call2() calls the function. Correct: call2 is the reference to the function.
x[i] accesses the value (the function reference). You need x[i]() to execute the function.

Here is your code working: http://jsbin.com/IMiVadE/6/edit

sayHello(name) is a function that generates new functions that output a specific "Hello ..." text.

function sayHello(name)
{
  return function () {
    console.log("Hello " + name);
  }
}

function timer(x, i)
{
  x[i]();
  if (i < x.length - 1) {
    setTimeout(function () {
      timer(x, i + 1);
    }, 2000);
  }
}

var x = [sayHello("Nakib"), sayHello("world")];

setTimeout(function () {
  timer(x, 0);
}, 2000);

answered Oct 6, 2013 at 8:37

Sanjo

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Comments

GolezTrol · Accepted Answer · 2013-10-06 08:27:43Z

0

setTimeout needs a function without parameter. So you can wrap the recursive call to timer in an anonymous function to fix that:

window.setTimeout(function(){timer(x, i)}, 2000);

Also, the first line in your timer function, consisting only of x[i];, is useless, although it probably isn't the cause of your problem.

answered Oct 6, 2013 at 8:27

GolezTrol

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Comments

Jivings · Accepted Answer · 2013-10-06 08:28:32Z

0

The setTimeout function takes a function as a parameter, you are executing the timer function before it is passed and since timer doesn't return anything, undefined is being passed to the timeout;

window.setTimeout(timer(x, i), 2000); // is the same as...
window.setTimeout(undefined, 2000);

It should be;

window.setTimeout(function() { timer(x, i) }, 2000); // or...
window.setTimeout(timer.bind(this, x, i), 2000);

I'm not sure if this is intentional, but you are doing the same thing with your array;

var x = [call("Nakib"), call2()];

This will execute the functions and their results will be stored in the array. Is this what you want?

answered Oct 6, 2013 at 8:28

Jivings

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2 Comments

Nakib Over a year ago

I dont want its result to be stored in an array but i want that function to be stored so i can call then in my timer() function.

Jivings Over a year ago

@Nakib Okay, well you need to have another think about how your code works, and take a look at how first class functions work. Come back if you still need help afterwards. I hope this was helpful :)

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