Function as an argument to a function in Javascript [duplicate]

Question

I'm trying to implement a function that takes another function as an argument, and returns a new version of that function that can only be called once.

Subsequent calls to the resulting function should have no effect (and should return undefined).

For example:

logOnce = once(console.log) 
logOnce("foo") // -> "foo" 
logOnce("bar") // -> no effect

Do you have a working fiddle to illustrate your question? Because as is, we cannot understand it... — sp00m
– sp00m, Commented Oct 28, 2014 at 8:52
here function once has the argument console.log, which is assigned to to the logOnce. So if i call logOnce("foo") it should return me same functionality as console.log("foo") does. But this should be invoked only once. — Vishwa
– Vishwa, Commented Oct 28, 2014 at 9:00

Federico Nieri · Accepted Answer · 2014-10-28 09:18:11Z

1

You can use a flag on the function obeject you are passing as argument

function once(func){
  return function(){
    if(!func.performed){
      func.apply(this,arguments);
      func.performed = true;
    }    
  }
}

var logOnce = once(console.log);
logOnce("Test 1");
logOnce("Test 2");

answered Oct 28, 2014 at 9:18

Federico Nieri

362 bronze badges

Sign up to request clarification or add additional context in comments.

2 Comments

Vishwa Over a year ago

thanks for the answer but it should run only once, second time should do nothing(or return undefined).

Federico Nieri Over a year ago

The second time it doesn't log infact

Collectives™ on Stack Overflow

Function as an argument to a function in Javascript [duplicate]

1 Answer 1

2 Comments

Linked

Hot Network Questions

Collectives™ on Stack Overflow

1 Answer 1

2 Comments

Linked

Related