具有默认值分配的ES6对象解构 [英] ES6 object destructuring with a default value assignment

问题描述

考虑以下代码:

const log = ({a,b=a}) => console.log(a,b);
log({a:'a'})

为变量 b 分配了值 a .将其编译到es5时确实可以使用，但是我不确定这是否是有效的es6语法.

The variable b is assigned the value a. This does work when transpiling it to es5, but I am not entirely sure if this is valid es6 syntax.

我可以在变形对象中进行这种默认值分配吗?

Am I allowed to do this kind of default value assignment in a destructured object?

推荐答案

const log = ({a,b=a}) => console.log(a,b);
log('a')

在语法上有效，但在语义上无效，因为您正试图解构一个字符串原语，该字符串原语被装箱到临时的 String 对象包装器中，并试图同时获取 a 和 b 属性，它们始终是 undefined ，因为包装对象是临时的，并且仅出于操作触发装箱本身的目的而创建.

is syntactically valid but semantically invalid, since you are trying to destructure a string primitive, which gets boxed into a temporary String object wrapper, and tries to get both a and b properties, which are always undefined since the wrapper object is temporary and only created for the sake of the operation triggering the boxing itself.

因此，您在调用它时具有 undefined，undefined .

Thus you have undefined, undefined when you call it.

具有默认值的解构操作在语义上可能是有效的在您的情况下，像这样的通话:

The destructuring operation with defaults could be semantically valid in your case with a call like this:

const log = ({a,b=a}) => console.log(a,b);
log({a: 'a'}) // a,a

UPD:

但是请注意，提供默认值的顺序很重要，因此这不会工作

But beware that order of supplying default value matters, so this won't work

const log = ({a=b,b}) => console.log(a,b);
log({a: 'a'}) // error

因为解构发生在参数对象初始化之后，并且从左到右进行求值，所以 b 尚未解构，并且在我们尝试对 a 进行解构时知道如果未定义 a .

because destructuring happens after argument object initialization and is evaluated from left to right, thus b is not yet destructured and known by the time we destructure a to try to reference it in case a is undefined.

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