Explain expression: Object followed by Array

Question

I saw a javascript expression like this:

var foo = {...}[...];

can anyone explain what that means?

Best Regards

Update:

this is a code example:

var ENV_PRODUCTION = {
  production: true,
  development: false
}[process.ENV.NODE_ENV|| 'development'];

Please provide a working example, empty object and empty array (var a = {}[]) gives syntax error, so it's not valid. — Jakub Jankowski
– Jakub Jankowski, Commented Oct 13, 2016 at 8:09
Was it actually an array (as in [1,2,3]), or just something that would evaluate to a property name? You've simplified the expression too much. — James Thorpe
– James Thorpe, Commented Oct 13, 2016 at 8:09
var values = {key1: 'value1', key2: 'value2' }; var key = 'key1'; var foo = values[key];? — Yury Tarabanko
– Yury Tarabanko, Commented Oct 13, 2016 at 8:11
x = { foo: 42, bar: 'baz' }['foo'] is an immediately invoked object literal. — Nina Scholz
– Nina Scholz, Commented Oct 13, 2016 at 9:22

Jonathan Newton · Accepted Answer · 2016-10-13 12:46:40Z

4

It will return the inner value by key value see:

var foo = {
 "1": "Jam",
"2":"Stuff"}[1];

var bar = {
 "1": "Jam",
"2":"Stuff"}[2];

var foo = {
 "1": "Jam",
"2":"Stuff"}[1];

var bar = {
 "1": "Jam",
"2":"Stuff"}[2];

var jam = {
 "1": "Jam",
"B":"Other Stuff"}['B'];

var foobar = {
 "...": "...",
"B":"Other Stuff"}['...'];

console.log(foo, bar, jam, foobar)

Further documentation

edited Oct 13, 2016 at 12:46

answered Oct 13, 2016 at 8:11

Jonathan Newton

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Comments

Vladimir M · Accepted Answer · 2016-10-13 08:11:31Z

1

Its a reference to element of the object.

Try in console:

var x = {a:1}['a'];

x = 1

answered Oct 13, 2016 at 8:11

Vladimir M

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Comments

gurvinder372 · Accepted Answer · 2016-10-13 08:11:28Z

0

This is a spread operator as per ES6

The spread syntax allows an expression to be expanded in places where multiple arguments (for function calls) or multiple elements (for array literals) or multiple variables (for destructuring assignment) are expected.

answered Oct 13, 2016 at 8:11

gurvinder372

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4 Comments

James Thorpe Over a year ago

I suspect the ... in this case is just a simplification by the OP and is meant to represent other stuff, rather than be the spread operator.

gurvinder372 Over a year ago

@JamesThorpe oh..ok. I guess then OP should show the exact code.

Hitmands Over a year ago

I think isn't destructuring, probably @Newton answer is correct. Could you provide an example of your assertion?

gurvinder372 Over a year ago

@Hitmands sorry, not sure if I can give a relevant answer unless you share the exact code that you saw.

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