展开运算符 vs array.concat() [英] spread operator vs array.concat()

问题描述

扩展运算符和array.concat()

let parts = ['four', 'five'];
let numbers = ['one', 'two', 'three'];
console.log([...numbers, ...parts]);

Array.concat() 函数

let parts = ['four', 'five'];
let numbers = ['one', 'two', 'three'];
console.log(numbers.concat(parts));

两个结果是一样的.那么，我们想在什么样的场景中使用它们呢?哪一种最适合性能?

Both results are same. So, what kind of scenarios we want to use them? And which one is best for performance?

推荐答案

好吧 console.log(['one', 'two', 'three', 'four', 'five']) 也有相同的结果，那么为什么在这里使用任何一个呢?:P

Well console.log(['one', 'two', 'three', 'four', 'five']) has the same result as well, so why use either here? :P

通常，当您有两个(或更多)来自任意来源的数组时，您将使用 concat，并且如果附加元素始终属于数组之前是已知的.因此，如果您的代码中有一个带有 concat 的数组文字，请使用扩展语法，否则只需使用 concat :

In general you would use concat when you have two (or more) arrays from arbitrary sources, and you would use the spread syntax in the array literal if the additional elements that are always part of the array are known before. So if you would have an array literal with concat in your code, just go for spread syntax, and just use concat otherwise:

[...a, ...b] // bad :-(
a.concat(b) // good :-)

[x, y].concat(a) // bad :-(
[x, y, ...a]    // good :-)

在处理非数组值时，这两种选择的行为也大不相同.

Also the two alternatives behave quite differently when dealing with non-array values.

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