[v3.5.x] destructure from `defineProps` with `withDefaults` throw an error

[zhaojjiang]

Vue version

3.5.5

Link to minimal reproduction

https://play.vuejs.org/#eNp9kctOwzAQRX9l8CatVFohWFVpEY9KwAIqQGLjTZROUhfHtuxJWynKvzNx1McCdeXM3DvxuZ5GPDg33tYopiINuVeOICDVDnRmypkUFKSYS6MqZz1BAx4LaKHwtoKExxJppMmtCQRVKGHW6YPkBbW28GO9Xl0lw5OlAUI+WvbtFK2fschqTWGwwkIZXHrrQtpIA9F2P4VAXplSmnY+GI7gqEwh6Q6+u+Wfp5Oemym5IKyczgi5AkjXN/OmiWRtm064il1lXE2wva7sCjVnZF0KmLCYTs7mxYjTM3ihyvEmWMNPFBGkyG3llEb/4UhxMCmmPVynZRx99xZ75GscHfr5GvPff/qbsO96Uiw9BvRblOKoUeZLpF5efL3jnr+PItPXmt0XxE8MVtcdY297rM2Ksc98kfY1Lpdf+jss9oQmHEJ1oJ2zjX4peOFPF6KfcG/Hd3GONyTaP1bly/I=

Steps to reproduce

create new project

// App.vue

<script setup lang="ts">
const { test } = withDefaults(defineProps<{
  test?: string
}>(), {
  test: 'test'
})
</script>

throw error:

error during build:
[vite:vue] [@vue/compiler-sfc] withDefaults() is unnecessary when using destructure with defineProps().
Prefer using destructure default values, e.g. const { foo = 1 } = defineProps(...).

D:/temp/vite/src/App.vue
1  |  <script setup lang="ts">
2  |  const { test } = withDefaults(defineProps<{
   |                   ^^^^^^^^^^^^
3  |    test?: string
4  |  }>(), {

What is expected?

It should not throw an error and shoule not break build, just keep it warning,

Or, if withDefaults not marked as deprecated, it should work without any warn or error messages

What is actually happening?

It breaks build

System Info

No response

Any additional comments?

No response

[Tofandel]

Can I ask, why is it unnecessary when using destructure with defineProps? How do you pass defaults in this case then?

[Justineo]

Can I ask, why is it unnecessary when using destructure with defineProps? How do you pass defaults in this case then?

const { foo = 1 } = defineProps(…)

[Tofandel]

But then they are still reactive right? If my understanding is correct there is some magic normally happening when using const { foo } = defineProps(); that makes foo reactive, or is it only something in nuxt?

I'm asking because now in the error message in question there is also Reactive destructure will be disabled when using withDefaults()

[zhaojjiang]

It means defineProps + withDefaults works same as before, --> destrcuture is non-reactive,
and destructure with default value const { foo = 1 } will keep reactive