javascript - 当没有参数时，为什么vue计算函数可以破坏参数的分配？

Question

new Vue({
      el: "#app",
       data: {
         value: "text",
       },
       computed:{
        all: function({value}){
          return value
        }
       }
      });

<script src="https://cdnjs.cloudflare.com/ajax/libs/vue/2.5.17/vue.js"></script>

<div id="app">
  {{ value }}
  <br>
  {{ all }}
</div>

在Vue文档上没有看到此用法。
但是我在项目中看到了这一点，为什么可以使用这种用法？

   computed:{
    all: function({value}){
      return value
    }
   }

首先，我认为当函数没有得到参数时，将“this”作为参数。见下文。
但是是错误的。那么为什么计算没有得到参数，对参数进行解构分配仍然有效？

var value = "couldn't fly";
function ref ({value}){
  console.log(value)
}

ref({
  value : "could fly"
})

// didn't get window
try{
  ref();
}catch(e){
  console.log('error')
}


// ===================================================


class cal {
  constructor(){
    value : "couldn't walk"
  }
  ref({value}){
    console.log(value, "in Class")
  }
}

let calRef = new cal;
calRef.ref({
  value: "could walk"
})


// didn't get the constructor value
try{
  calRef.ref()
}catch(e){
  console.log('error')
}

Answer 1

这记录在API guide ...

Note that if you use an arrow function with a computed property, this won’t be the component’s instance, but you can still access the instance as the function’s first argument

new Vue({
  el: "#app",
  data: {
    value: "text",
  },
  computed: {
    all (vm) {
      console.log('from all, is vm equal to this?', vm === this)
      return vm.value
    },
    arrowed: vm => vm.value.toUpperCase(),
    destructuredArrow: ({ value }) => value.toUpperCase()
  }
});

<script src="https://cdn.jsdelivr.net/npm/vue/dist/vue.min.js"></script>

<div id="app">
  <pre>value = {{ value }}</pre>
  <pre>all = {{ all }}</pre>
  <pre>arrowed = {{ arrowed }}</pre>
  <pre>destructuredArrow = {{ destructuredArrow }}</pre>
</div>

刚刚发现了一个方便的额外提示...同样的模式也适用于组件的 data函数

export default {
  props: { someProp: String },
  data: vm => ({ localCopy: vm.someProp })
}