Why do you need to wrap a Immediately invoked function expression IIFE in parenthesis JavaScript

Question

I have the following code that wont run (I am trying to have 0 global variables).

function() {
  var createworker = function() {
    //private implemenation 
    var count = 0;
    var task1 = function() {
      count += 1;
      console.log("task1 " + count);
    };
    var task2 = function() {
      count += 1;
      console.log("task2 " + count);
    };

    return {
      // these are the aliases or the public api 
      // revealing module pattern mode 
      job1: task1,
      job2: task2
    };
  };

  var result = 2 + 2;
  var worker = createworker();

  worker.job1();
  worker.job2();
  worker.job2();
  worker.job2();
  worker.job2();
}();

JavaScript doesn't like that for some reason. But the following example or just wrapping the anonymous function in () allows its it call itself. What is going on here and why is this the case?

(function() {
  var createworker = function() {
    //private implemenation 
    var count = 0;
    var task1 = function() {
      count += 1;
      console.log("task1 " + count);
    };
    var task2 = function() {
      count += 1;
      console.log("task2 " + count);
    };

    return {
      // these are the aliases or the public api 
      // revealing module pattern mode 
      job1: task1,
      job2: task2
    };
  };

  var result = 2 + 2;
  var worker = createworker();

  worker.job1();
  worker.job2();
  worker.job2();
  worker.job2();
  worker.job2();
}());

Function declarations require names, function expressions don't. `(...)` forces the function to be evaluated as a function expression. — Felix Kling, Jun 09 '16 at 21:02
@NinaScholz—you can you anything that means the *function* keyword isn't at the start of the statement, `!` is also used but `()` is generally preferred. — RobG, Jun 09 '16 at 21:05

score 1 · Accepted Answer · answered Jun 09 '16 at 21:03

1

Wrapping it in parenthesis is how you make it an expression. The result of the expression is the last argument, and since there is only one argument in there (the function) that is the value of the expression.

Once evaluated, the value of the expression (the function) is then called using the following set of parenthesis.

That is why it is wrapped in parenthesis, to create the actual expression.

answered Jun 09 '16 at 21:03

Travis J

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That really has nothing to do with wrapping it in parenthesis. – Travis J Jun 09 '16 at 22:12

Why do you need to wrap a Immediately invoked function expression IIFE in parenthesis JavaScript

1 Answers1