Destructure对象参数，还要引用参数作为对象吗？ [英] Destructure object parameter, but also have reference to the parameter as an object?

问题描述

使用ES6，你可以在函数参数中构造对象：

With ES6 you can destructure objects in function arguments:

({name, value}) => { console.log(name, value) }

等效的ES5将是：

function(params) { console.log(params.name, params.value) }

但如果我想引用 params 对象和嵌套属性<$ c $，该怎么办？ c>值和名称？这是我得到的最接近的，但缺点是它不能用于箭头函数，因为它们无法访问参数对象：

But what if I would like a reference both to the params object and the nested properties value and name? This is the closest I got, but the drawback is that it won't work with arrow functions because they do not have access to the arguments object:

function({name, value}) {
  const params = arguments[0]
  console.log(params, name, value)
}

推荐答案

arguments 在箭头函数中不可用，这会影响在ES6中如何一致地处理函数参数。

arguments isn't available in arrow functions, and this affects how function parameters should be consistently treated in ES6.

如果原始参数是使用时，它应该在函数内部进行解构：

If original parameter is used, it should be destructured inside function:

(param) => {
  const {name, value} = param;
  // ...
}

如果有多个参数可以参考参数协商发生（例如，类似于 arguments.length ），应该使用rest参数：

If several arguments are expected and some argument negotiation takes place (for example, similarly to arguments.length), rest parameter should be used:

(...args) => {
  const [param] = args;
  // ...
}

Boilerplate变量解构有其好处;这种方式更容易在断点处调试 param 或 args ，即使它们当前没有被使用。

Boilerplate variable destructuring has its benefits; it's easier this way to debug param or args at breakpoint even if they aren't currently in use.

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