Logical OR assignment (||=)
The logical OR assignment (x ||= y
) operator only assigns if x
is falsy.
Try it
Syntax
js
x ||= y
Description
Logical OR assignment short-circuits, meaning that x ||= y
is equivalent to:
js
x || (x = y);
No assignment is performed if the left-hand side is not falsy, due to short-circuiting of the logical OR operator. For example, the following does not throw an error, despite x
being const
:
js
const x = 1;
x ||= 2;
Neither would the following trigger the setter:
js
const x = {
get value() {
return 1;
},
set value(v) {
console.log("Setter called");
},
};
x.value ||= 2;
In fact, if x
is truthy, y
is not evaluated at all.
js
const x = 1;
x ||= console.log("y evaluated");
// Logs nothing
Examples
Setting default content
If the "lyrics" element is empty, display a default value:
js
document.getElementById("lyrics").textContent ||= "No lyrics.";
Here the short-circuit is especially beneficial, since the element will not be updated unnecessarily and won't cause unwanted side-effects such as additional parsing or rendering work, or loss of focus, etc.
Note: Pay attention to the value returned by the API you're checking against. If an empty string is returned (a falsy value), ||=
must be used, so that "No lyrics." is displayed instead of a blank space. However, if the API returns null
or
undefined
in case of blank content, ??=
should be used instead.
Specifications
Specification |
---|
ECMAScript Language Specification # sec-assignment-operators |
Browser compatibility
BCD tables only load in the browser