JS - assigning function to variable

Question

As we all know, this following will not run the a() function so the alert box will not appear

// 1st
function a() {
  alert('A!');
  return function() {
    alert('B!');
  };
};

and we know that the following code will run the a() function and alert box 'A!' will appear

// 2nd
function a() {
  alert('A!');
  return function() {
    alert('B!');
  };
};
a(); // calling function

However, if we run following code, the a() function will be called and alert box 'A!' will also appear, just like the second code above

// 3rd
function a() {
  alert('A!');
  return function() {
    alert('B!');
  };
};
var x = a(); // assigning function to new variable

QUESTION: why does this happen (on 3rd snippet)? we didn't call the a() function yet (my current understanding). Didn't we just assigning x to a() function?.

`()` this will call the function. And will assign the return value of that function to `x` — Rajaprabhu Aravindasamy, Apr 16 '16 at 12:22
Anything with parenthesis `()` will find the function/method of that name and will invoke it... No matter where you write! — Rayon, Apr 16 '16 at 12:22

score 21 · Accepted Answer · answered Apr 16 '16 at 12:23

21

Didn't we just assigning x to a() function?.

No, you assigned the returned value from a() to x.

If you don't want to call a, then do

var x = a;

and later do

x();

answered Apr 16 '16 at 12:23

gurvinder372

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score 6 · Answer 2 · answered Apr 16 '16 at 12:28

You're wrong. You are invoking the function:

var x = a(); // assigning function to new variable

This line of code says invoke a when you write a(). The parentheses indicate an invocation.

To assign a function to a variable you have to use just the name, such as:

var x = a;

or pass the name to a function f:

f(a)

As a counter-example you invoke it in this next line of code and pass to g not the function be the result of its execution:

g(a())

JS - assigning function to variable

2 Answers2

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