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@miguel-martinr miguel-martinr commented May 31, 2025

Fix what I think is probably a typo in the example for the destructuring explanation. The docs states:

Note that if you destructure a reactive value, the references are not reactive — as in normal JavaScript, they are evaluated at the point of destructuring

... but then, in the example, the destructured variables are not actually used:

let { done, text } = todos[0];

// this will not affect the value of `done`
todos[0].done = !todos[0].done;  // <=== Not using the destrucutred variables, thus this actually updates the values correctly

I think that the example should be this instead:

let { done, text } = todos[0];

// this will not affect the value of `done`
done = !done;  // <=== Actually uses the destructured values

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Fix what I think is probably a typo in the destructuring explanation
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@Rich-Harris
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No, the whole point is that mutating todos[0].done doesn't change the value of done because it's already been evaluated

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2 participants