Javascript ES6 spread operator on undefined [duplicate]

问题

This question already has an answer here:

Spreading undefined in array vs object (1 answer)

Closed 2 years ago.

While developing my react App, I needed to send a conditional prop to a component so I found somewhere a pattern to do so, although it seems really weird to me and I couldn't understand how and why it worked.

If I type:

console.log(...undefined)   // Error 
console.log([...undefined]) // Error
console.log({...undefined}) // Work

When the spread operator is activated on undefined an error is raised, although when the undefined is inside an object, an empty object returned.

I'm quite surprised regarding this behavior, is that really how it supposed to be, can I rely on this and is that a good practice?

回答1:

This behavior is useful for doing something like optional spreading:

function foo(options) {
  const bar = {
    baz: 1, 
    ...(options && options.bar) //options and bar can be undefined
  } 
}

And it gets even better with optional chaining, which is in stage-1:

function foo(options) {
  const bar = {
    baz: 1, 
    ...options?.bar //options and bar can be undefined
  } 
}

a thought: its too bad it doesn't also work for spreading into an array

来源：https://stackoverflow.com/questions/46957194/javascript-es6-spread-operator-on-undefined