JavaScript函数带有默认参数和一个多重词

发布于 2025-02-07 06:12:36 字数 730 浏览 1 评论 0原文

假设我有这样的函数声明：

function manyParams(a, b=1, ...n) { ... }

现在很明显，我可以以一种可以提供a，b的方式调用此函数在数组n下。

示例：

function manyParams(a, b=1, ...n) {
    console.log(JSON.stringify(a));
    console.log(JSON.stringify(b));
    console.log(n.length);
}

如果您使用以下参数调用此函数nyparams（1，1，“ John”，“ stacy”，“ Mike”，“ Bob”，“ Carly”）;，您将获得此输出，这是有道理的：

1
1
5

但是，如果我想在b中使用默认值，在仍提供许多参数的同时，我如何指定“ john”是开始n？我认为我可以做这样的事情：nyparams（a = 1，n =“ john”，“ stacy”，“ mike”，“ bob”，“ carly”）;，但这不是'工作。运行这些参数为我提供以下输出：

1
"John"
4

原文

Let's say I have a function declaration like this:

function manyParams(a, b=1, ...n) { ... }

Now clearly I can call this function in such a way that I can provide a, b, and then any number of additional arguments under an array n.

Example:

function manyParams(a, b=1, ...n) {
    console.log(JSON.stringify(a));
    console.log(JSON.stringify(b));
    console.log(n.length);
}

If you call this function with the following arguments manyParams(1, 1, "John", "Stacy", "Mike", "Bob", "carly");, you get this output, which makes sense:

1
1
5

However, what if I want to use the default value in b, while still providing many arguments, how do I specify that "John" is the start of n? I would think I could do something like this: manyParams(a=1, n="John", "Stacy", "Mike", "Bob", "carly"); but that doesn't work.
Running those arguments gives me the following output: