Skipping optional function parameters in JavaScript

问题

Could you please point me to the nice way of skipping optional parameters in JavaScript.

For example, I want to throw away all opt_ parameters here:

goog.net.XhrIo.send(url, opt_callback, opt_method, opt_content, {\'Cache-Control\': \'no-cache\'}, opt_timeoutInterval)

回答1:

Solution:

goog.net.XhrIo.send(url, undefined, undefined, undefined, {'Cache-Control': 'no-cache'})

You should use undefined instead of optional parameter you want to skip, because this 100% simulates the default value for optional parameters in JavaScript.

Small example:

myfunc(param);

//is equivalent to

myfunc(param, undefined, undefined, undefined);

Strong recommendation: use JSON if you have a lot of parameters, and you can have optional parameters in the middle of the parameters list. Look how this is done in jQuery.

回答2:

Short answer

The safest bet is undefined, and should work almost ubiquitously. Ultimately, though, you cannot trick the function being called into thinking you truly omitted a parameter.

If you find yourself leaning towards using null just because it's shorter, consider declaring a variable named _ as a nice shorthand for undefined:

(function() { // First line of every script file
    "use strict";
    var _ = undefined; // For shorthand
    // ...
    aFunction(a, _, c);
    // ...
})(); // Last line of every script

Details

First, know that:

• typeof undefined evaluates to "undefined"
• typeof null evaluates to "object"

So suppose a function takes an argument that it expects to be of type "number". If you provide null as a value, you're giving it an "object". The semantics are off.¹

As developers continue to write increasingly robust javascript code, there's an increasing chance that the functions you call explicitly check a parameter's value for undefined as opposed to the classic if (aParam) {...}. You'll be on shaky ground if you continue to use null interchangeably with undefined just because they both happen to coerce to false.

Be aware, though, that it is in fact possible for a function to tell if a parameter was actually omitted (versus being set to undefined):

f(undefined); // Second param omitted
function f(a, b) {
    // Both a and b will evaluate to undefined when used in an expression
    console.log(a); // undefined
    console.log(b); // undefined
    // But...
    console.log("0" in arguments); // true
    console.log("1" in arguments); // false
}

---

Footnotes

1. While undefined also isn't of type "number", it's whole job is to be a type that isn't really a type. That's why it's the value assumed by uninitialized variables, and the default return value for functions.

回答3:

Just pass null as parameter value.

Added: you also can skip all consequent optional parameters after the last that you want to pass real value (in this case you may skip opt_timeoutInterval parameter at all)

来源：https://stackoverflow.com/questions/8356227/skipping-optional-function-parameters-in-javascript