CODEWITHSUNDEEP
javascriptself-invoking-function
sabu
sabu

Reputation: 2057

Self invoking function in JavaScript

I have a self invoking function like this:

var f = (function f(){ return "123"; }, 
         function g(){ return 2; },
         function h(){ return "test"; })();

typeof f;

typeof f is always the type of what is returned in the last function definition. Like if h is last, then it is "string", but if I remove h and have g as last, then "number".

Could someone explain why?

Upvotes: 0

Views: 409

Answers (3)

kevingessner
kevingessner

Reputation: 18985

Let's break this down.

The comma operator in Javascript evaluates several expressions, and returns the last one:

>>> "a", 1
1
>>> 1, "a"
"a"

So when you take three anonymous functions and string them together with commas, it evaluates to the last one:

>>> (function f(){ return "123"; }, function g(){ return 2; }, function h(){ return "test"; })
function h(){ return "test"; }

Evaluating that result executes the function, returning "test".

Whichever function is last in the comma-separated list will be executed, and decide the overall return value.

Upvotes: 2

kzh
kzh

Reputation: 20598

The comma operator returns the last item. What you are doing is like:

var f = function(){}, function(){}, "string";

Which will make f a string, because only the last function is being called.

Upvotes: 1

user1106925
user1106925

Reputation:

Because the functions are separated by the , comma operator.

This evaluates the separated expressions, and returns the result of the last expression.

var x = ("a", "b", "c");

console.log(x); // "c"

So in your case, the last function is returned, as the result of the enclosing () group, and that's the one invoked by the trailing () function call.

   // result from group---v   v---invoked
var f = (func1, func2, func3)()

Upvotes: 6

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