What is this in javascript: “var var1 = var1 || []”

Question

I just want to increase my core javascript knowledge.

Sometimes I see this statement but I don't know what it does:

var var1 = var1 || [];

What does it means and/or what's it for, and how do you use it?

Thank you.

Answer 1

Basically, it looks to see if a variable var1 already exists and is "truthy". If it is, it assigns the local var1 variable its value; if not, it gets assigned an empty array.

This works because the JavaScript || operator returns the value of the first truthy operand, or the last one, if none are truthy. var1 || var2 var1 || var2 returns var1 if it's truthy, or var2 otherwise.

Here are some examples:

var somevar;
somevar = 5 || 2; // 5
somevar = 0 || 2; // 2
somevar = 0 || null; // null

Values that aren't "truthy": false , 0 , undefined , null , "" (empty string), and NaN . Empty arrays and objects are considered truthy in JavaScript, unlike in some other languages.

Answer 2

如果它的布尔表示为假（例如它尚未初始化），它会为var1指定一个空数组。

Answer 3

基本上，如果var1为NULL或false ，则var1将设置为空array 。

Answer 4

The logical operators in JavaScript actually evaluate to one of the two objects. When you use a || b a || b it evaluates to b if a is false, or to a if a is true. Thus a || [] a || [] will be a if a is any value that is true, or [] if a is any value that is false.

It's much more obvious to use if (!a) { a = [] };

Answer 5

Javascript or (||) works a bit differently to some other languages, it returns the first "truthy" value instead of a boolean. This is being used in this case to say "Set the value of var1 to var1 , but if that value is "falsey" set it to [] ".

This is often used to set a "default" value to a variable that may or may not be set already, such as an argument to a function.

Answer 6

The || operator evaluates the first of its operands that is "truthy".

[] is an empty array. ( [ "Hi!" ] is an array of one string)

Therefore, the expression x || [] x || [] evaluates to x if it's "truthy" or an empty array if it isn't.

This allows the var1 parameter to be optional.

Answer 7

The statement assigns an empty array to var1.

---

longer answer and explanation:

This happens because var1 is not initialized at that time. Non-initialized is a falsy value.

take this statement:

var1 = var1 || [];

If var1 is not initialized it becomes an empty array, it if is an empty array nothing happens as its assigned to an empty array, if var1 is false , null , or any other value that javascript things of as false , it becomes an empty array, if var1 is anything other value, nothing happens as it is assigned to itself. (thanks pst for the link).

In short, its a stupid statement that's neither readable nor useful, but you're smart for wanting to know what it means. :)

Answer 8

While this has been pointed out as 'being a silly statement', I present the following two counters:

(Just to keep people on their toes and reinforce some of the "finer details" of JavaScript.)

1)

var is the variable is already local . Eg

function x (y) {
  var y = y || 42 // redeclaration warning in FF, however it's "valid"
  return y
}
x(true) // true
x()     // 42

2)

var is function-wide annotation (it is "hoisted" to the top) and not a declaration at the point of use.

function x () {
  y = true
  var y = y || 42
}
x()      // true

I do not like code like either of the preceding, but...

Because of the hoisting, and allowed re-declarations, the code in the post has these semantics:

var var1
if (!var1) {
  var1 = []
}

Edit I am not aware of how "strict mode" in Ed.5 influences the above.