CODEWITHSUNDEEP
javascriptnode.jsoptional-parameters
daaanipm
daaanipm

Reputation: 273

javascript unassigned default function parameters

In the following function:

foo = function(a){
    if (!a) a = "Some value";
    // something done with a
    return a;
}

When "a" is not declared I want to assign a default value for use in the rest of the function, although "a" is a parameter name and not declared as "var a", is it a private variable of this function? It does not seem to appear as a global var after execution of the function, is this a standard (i.e. consistent) possible use?

Upvotes: 4

Views: 2757

Answers (3)

unloco
unloco

Reputation: 7320

It's a private variable within the function scope. it's 'invisible' to the global scope.
As for your code you better write like this

foo = function(a){
    if (typeof a == "undefined") a = "Some value";
    // something done with a
    return a;
}

Because !a can be true for 0, an empty string '' or just null.

Upvotes: 4

Matthew
Matthew

Reputation: 15962

Parameters always have private function scope.

var a = 'Hi';
foo = function(a) {
    if (!a) a = "Some value";
        // something done with a
        return a;
    };
console.log(a); // Logs 'Hi'
console.log('Bye'); // Logs 'Bye'

Upvotes: 0

kemiller2002
kemiller2002

Reputation: 115488

Yes, in this context a has a scope inside the function. You can even use it to override global variables for a local scope. So for example, you could do function ($){....}(JQuery); so you know that $ will always be a variable for the JQuery framework.

Upvotes: 0

Related Questions