The ||= operator only evaluates the right operand and assigns to the left if the left operand is falsy.
The ||= operator only evaluates the right operand and assigns to the left if the left operand is falsy.
x ||= y
Logical OR assignment short-circuits, meaning that x ||= y is equivalent to x || (x = y), except that the expression x is only evaluated once.
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:
const x = 1; x ||= 2;
Neither would the following trigger the setter:
const x = { get value() { return 1; }, set value(v) { console.log("Setter called"); }, }; x.value ||= 2;
In fact, if x is not falsy, y is not evaluated at all.
const x = 1; x ||= console.log("y evaluated"); // Logs nothing
If the "lyrics" element is empty, display a default value:
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.
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| Chrome | Edge | Firefox | Opera | Safari | Chrome Android | Firefox for Android | Opera Android | Safari on IOS | Samsung Internet | WebView Android | Deno | Node.js | ||
Logical_OR_assignment |
85 | 85 | 79 | 71 | 14 | 85 | 79 | 60 | 14 | 14.0 | 85 | 1.2 | 15.0.0 | |
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https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Operators/Logical_OR_assignment