Passing in NULL as a parameter in ES6 does not use the default parameter when one is provided

Question

Is there a known reason why passing in null as a parameter in ES6 does not use the default parameter when one is provided?

function sayHello(name = "World") {
    console.log("Hello, " + name + "!");
}

sayHello("Jim");     // Hello, Jim!
sayHello(undefined); // Hello, World!
sayHello(null);      // Hello, null!

Answer 1

This is not that obvious

I've read some comments of why undefined is completely different than null and that's why it explains the current behavior of default parameters.

One could argue that explicitly passing undefined should not trigger the default value substitution because when I have a function:

const f = (x = 'default') => console.log(x);

I would like it to print "default" when I run it as:

f();

but I would like it to print "undefined" when I explicitly run it as:

f(undefined);

because otherwise why would I use f(undefined) in the first place? Clearly my intention here is to provide some argument instead of leaving it out.

Examples to the contrary

Now, consider this function:

const g = (...a) => console.log(JSON.stringify(a));

When I use it as:

g();

I get: []

But when I use it as:

g(undefined);

I get: [null]

which clearly demonstrates that:

1. passing undefined is not the same as not passing an argument at all
2. sometimes null can be a default value instead of undefined

TC39 decision

Some background on the current behavior of the default parameters can be seen in the July 24 2012 Meeting Notes by TC39:

• https://github.com/rwaldron/tc39-notes/blob/master/es6/2012-07/july-24.md

Incidentally, it shows that explicitly passing undefined originally did not trigger the default value in the first draft and there was a discussion about whether or not it should do that. So as you can see the current behavior was not so obvious to the TC39 members as it now seems to be to people who comment here.

Other languages

That having been said, the decision of what should and what should not trigger the default value substitution is completely arbitrary at the end of the day. Even having a separate undefined and null can be though of as quite strange if you think about it. Some language have only undefined (like undef in Perl), some have only null (like Java), some languages use equivalents of false or an empty list or array for that (like Scheme where you can have an empty list or #f (false) but there is no equivalent of null that would be distinct from both an empty list and a false value) and some languages don't even have equivalents of null, false or undefined (like C which uses integers instead of true and false and a NULL pointer which is actually a normal pointer pointing to address 0 - making that address inaccessible even when mapped by any code that tests for null pointers).

What you can do

Now, I can understand your need to substitute default values for null. Unfortunately this is not a default behavior but you can make a simple function to help you:

const N = f => (...a) => f(...a.map(v => (v === null ? undefined : v)));

Now every time you want defaults substituted for null values you can use it like this. E.g. if you have this function from one of the examples above:

const f = (x = 'default') => console.log(x);

it will print "default" for f() and f(undefined) but not for f(null). But when you use the N function defined above to define the f function like this:

const f = N((x = 'default') => console.log(x));

now f() and f(undefined) but also f(null) prints "default".

If you want somewhat different behavior, e.g. substituting default values for empty strings - useful for environment variables that can sometimes be set to empty strings instead of not existing, you can use:

const N = f => (...a) => f(...a.map(v => (v === '' ? undefined : v)));

If you want all falsy values to be substituted you can use it like this:

const N = f => (...a) => f(...a.map(v => (v || undefined)));

If you wanted empty objects to be substituted you could use:

const N = f => (...a) => f(...a.map(v => (Object.keys(v).length ? v : undefined)));

and so on...

Conclusion

The point is that it's your code and you know what should be the API of your functions and how the default values should work. Fortunately JavaScript is powerful enough to let you easily achieve what you need (even if that is not the default behavior of default values, so to speak) with some higher order function magic.

Answer 2

That's just how default parameters are defined in the spec. See MDN: (emphasis mine)

Default function parameters allow formal parameters to be initialized with default values if no value or undefined is passed.

null is neither no value, nor undefined, so it doesn't trigger the default initialization.

Answer 3

null is a value that won't trigger the default value to be used, the default values will be used when the argument is undefined.

Answer 4

ramdajs version code

// import ramda

// infra code
const isEmptyOrNil = (value) => isEmpty(value) || isNil(value)
const nullChange = (defaultValue) => (value) => isEmptyOrNil(value)? defaultValue: value

// single null
const _somef = (value) => value
const somefWithDefault = (defaultValue) => pipe(nullChange(defaultValue), _somef)
const somef = somefWithDefault('myValue')

somef(null)

// many args
const _somef2 = (value, value2, value3) => value + value2 + value3
const nullChangeMyValue = nullChange('myValue')
const somef2 = useWith(_somef2, [nullChangeMyValue, nullChangeMyValue, nullChangeMyValue])

somef2(null, undefined, 1234)

// many args2 version2
const nullChangeWith = (defaultValue) => nullChange(defaultValue)
const arrayNullChangeWith = (defaultValue) => (array) => array.map(nullChangeWith(defaultValue))
const arg2Array = (...args) => args
const funcApplyDefaultValue = (defaultValue) => (func) => pipe(
  arg2Array, arrayNullChangeWith(defaultValue), apply(func)
)

const v2somef2 = funcApplyDefaultValue(8)(_somef2)

v2somef2(1,2,null)

gits: https://gist.github.com/otwm/3a6358e53ca794cc2e57ade4af91d3bb