Calling Function only once

Question

Could anyone help me with this troubleshooting?

var something = (function() {
  var executed = false;
  return function() {
    if (!executed) {
      executed = true;
      alert("Hi");
    }
  };
})(); //Removing the following calls the function does not execute with parenthesis

something(); //Happens
something(); //Nothing happens

My Approach:

var only = (function once() {
  alert("Kaixo");
  only = false;
})(); //That parenthesis calls the function

only(); //Calls the function 
only(); //Nothing happens

If my example is run before the original, breaks. Continuing of this topic: Function in javascript that can be called only once

You are resetting `only` to `false`, so it doesn't run again because `false` isn't a function. — Get Off My Lawn, May 03 '19 at 15:57
@GetOffMyLawn only is undefined because the once function returns nothing. That's because `only` is set to `false`, *then* to the value returned by the called function, `undefined`. http://i.stack.imgur.com/ado5W.png — Maxime Launois, May 03 '19 at 16:01
Could anyone explain me the first case? I don't understand, first, why the parenthesis does not call the function, and second why does not the second call happen? Thank you — Xabier Aparicio Muñoa, May 03 '19 at 16:34

Maheer Ali · Accepted Answer · 2019-05-03T16:07:13.480

4

You are setting the only to the value returnted by function i.e undefined.Even if you don't call it directly it will not work because false can't be called.

You can set the function to another dummy function which have nothing.

function only(){
   alert("Kaixo");
   only = function(){}
}

only(); //Calls the function 
only();

edited May 03 '19 at 16:07

answered May 03 '19 at 15:58

Maheer Ali

35,834
5
42
73

Thank you did not notice it :S I try now the following one: Setting return false in place of only=false; Does work (Calls twice). Is it correct? – Xabier Aparicio Muñoa May 03 '19 at 16:14
@XabierAparicioMuñoa Do you mean that what will happen if you use `return false` instead of `only = function(){}`? – Maheer Ali May 03 '19 at 16:18
Maheer Ali Yes! I got confused by the secuence. I checked the error and throws me the same error. – Xabier Aparicio Muñoa May 03 '19 at 16:24

arkeros · Answer 2 · 2019-05-03T16:05:19.503

3

Here only is a function, not a boolean. So, instead of overriding it to false, override it to an empty function. See:

var only = (function once() {
    alert("Kaixo");
    only = function () {};
}); // Don't call the function!

only(); // Calls the function 
only(); // Nothing happens

edited May 03 '19 at 16:05

answered May 03 '19 at 15:59

arkeros

177
3

1

Be interesting to know why this has been down-voted. Yes, it's basically the same answer as @Maheer Ali, but it was answered pretty much the same time, and he is a new contributor so it's expected that it will take longer to post an answer. – Keith May 03 '19 at 16:04
if the code works its answer valid, maybe no the most optime answer, but still right, always is good see all ways to do something, some people dont see that – Black Hole May 03 '19 at 16:10

score 0 · Answer 3 · answered May 03 '19 at 16:04

var only = (function once() {
  var executed = false; // Inits the flag;
  return function() {
    if (!executed) {      
      executed = true; //Set the flag
      alert("Kaixo");
    }
}  

})(); //That parenthesis calls the function

only(); //Calls the function 
only(); //Nothing happens

Calling Function only once

3 Answers3