How are variables declared in constructor functions in JavaScript treated?

Question

function jQueryValidatorWrapper(formId, rules, messages) {
    var validator = $("form:visible[id='" + formId + "']").validate({
                         onchange: false,
                         messages: messages,
                         rules: rules
                     });

    this.validate = function (hideErrors) {
        var showErrorMessage = hideErrors ? false : true;

        // What does 'validator' refer to?
        var result = validator.form();
    };
}

When I execute this,

var validatorObj = new jQueryValidatorWrapper('testForm', [], []);
validatorObj.validate();

The jQueryValidatorWrapper function had only one method declared with this, so when the constructor executed, it simply created an object with the validate method.

What happens to validator variable declared inside jQueryValidatorWrapper? It is not prefixed with this so it is not part of the object being constructed.

Is the validator variable a global? or is it part of the closure which is the validate method?

validator is not a global variable... it will exists in the closure though — Arun P Johny
– Arun P Johny, Commented Oct 14, 2014 at 5:17
A constructor function is still a function. There is no difference between how local variables are handled or scope works. — Felix Kling
– Felix Kling, Commented Oct 14, 2014 at 5:19

Troy Gizzi · Accepted Answer · 2014-10-14 05:17:45Z

4

The local validator variable is not global; it's merely accessible in the validate method because of the closure.

It can be considered a "private member" of the jQueryValidatorWrapper object, per Douglas Crockford's article here: http://javascript.crockford.com/private.html

Same goes for the 3 constructor parameters.

answered Oct 14, 2014 at 5:17

Troy Gizzi

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3 Comments

Developer Over a year ago

So, if I were to create multiple instances of jQueryValidatorWrapper with different data for the 3 constructor parameters each time, would the closure environment be modified for the previous instances?

Troy Gizzi Over a year ago

Nope, they each get their own copy. Here's a fiddle to illustrate that: jsfiddle.net/sck02myt

RobG Over a year ago

@Lakeland-FL—each time a function is called, an entirely new execution context is created so an entirely new closure is also created in this instance.

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