How to resolve Illegal Invocation Type Error in this case?

Question

While solving a programming challenge, I wrote a function which was intended to take a function as an argument and return a function as well. The returned function was meant to execute the argument function (which was passed to first function). The function's code :-

function func1(f){
  let func2 = function(){
    if(/*a condition based on a global variable*/){
      f();
    }
  }
  return func2;
}

This is currently not working and it raises an Illegal Invocation Type Error. I saw this question but I don't know how to relate it's answers to my code. So, my questions are :-

Why isn't my code working?
What can I do to make it work?

EDIT

I'm invoking the function like this :-

var someFunc = func1(alert);
someFunc("foo");
someFunc("bar");

^. In any cases there are many limitations here, like function arguments and returned value. — Denys Séguret, May 16 '16 at 13:01
check that for a sec https://jsfiddle.net/3masLhgw/ this should work. just add your condition — Neta Meta, May 16 '16 at 13:14
@NetaMeta Why putting in a comment what was in an answer 5 minutes before... — Denys Séguret, May 16 '16 at 13:17
@Densys it wasnt in the answer when i posted it, and its slightly different .... also for String what your trying to do can be accomplished by using .bind() https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Global_Objects/Function/bind — Neta Meta, May 16 '16 at 13:18

score 2 · Accepted Answer · answered May 16 '16 at 13:10

You need to handle

the context of the function call
the arguments
the returned value

Here's an implementation:

function func1(f, context){
  let func2 = function(){
    if( some condition ){
      return f.apply(context, arguments);
    } // there probably should be some "else" behavior...
  }
  return func2;
}

With some example uses:

var fun = func1(console.log, console);
fun("A", 25); // logs "A", 25
fun = func1(alert);
fun("foo");   // alerts "foo"

score -1 · Answer 2 · answered May 16 '16 at 13:07

-1

Why don't you use var instead of let

function func1(f){
  var func2 = function(){
    if(/*a condition based on a global variable*/){
      f();
    }
  }
  return func2;
}

answered May 16 '16 at 13:07

Mohit Bhardwaj

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How to resolve Illegal Invocation Type Error in this case?

2 Answers2