Question

Normally, I see function closure achieved by the form

var closedF = (function()
{
    return function(){/* return value */}
})();

so that calling closedF() returns the return value from the inner function. But I want to create closedF using a function declaration (the above is a function expression) so that it is defined at parse time. I.e.

function closedF()
{
    return function(){/* return value */}
}

but this doesn’t work because when calling closedF(), it returns the inner function as opposed to the return value of the inner function. Note: with the above declaration I could use closedF()(), but that’s just inelegant.

Is this possible?

p.s. As is usually the case there are many ways for me to solve my particular programming problem, but I actually want someone to show me that closed-over functions aren’t “second-class citizen” functions in JS.

Answer 1

Your problem has not necessarily anything to do with function expression vs function declaration.

In the both cases you showed, you don’t get the return value of the inner function.

You could get it by just calling the inner function and return it’s value:

function closedF() {
    return (function(){
        /* return value */
    }()); // <- calling the inner function here
}

The disadvantage is that the inner function will always be defined anew when calling the outer function. You could prevent this by caching the inner function as a property of the outer function:

function closedF() {
    var func = closedF.__func ||  (closedF.__func = function(){
        /* return value */
    });
    return func();
}

However, this might be more confusing than helpful.

Easier is to execute the outer function immediately and assign the returned function to a variable, as @James Long shows in his answer.

Update:

You could also do this:

(function() {
    window.closedF = function() {

    };
}());

though both are function expressions