更改通过函数签名命名的所有变量的值

Question

假设我有一个有 50 个参数的函数，我需要修改在函数签名中创建的每个命名变量的值。

他只有 4 个参数，而不是 50 个参数：

// Each of these strings are padded with intentional and unnecessary whitespace:
let show = "  show ";
let me = " me  ";
let the = " the ";
let bunny = "  bunny   ";

function showMeTheBunny(show, me, the, bunny)
{
	// I want to trim each argument, without having to do this:
	show = show.trim();
	me = me.trim();
	the = the.trim();
	bunny = bunny.trim();
	// The above lines, within this function,
	// are the ones I want to replace with a loop (if possible)

	return `${show} ${me} ${the} ${bunny}: ????`; 
}
console.log(showMeTheBunny(show, me, the, bunny)); // output: "show me the bunny: ????"

arguments 对象可以访问传递给函数的所有参数，但它似乎没有提供更改命名变量本身值的方法。

是否可以通过一个修改每个变量的函数来运行所有命名变量（在函数签名中命名），然后再使用这些修改后的参数（使用相同的变量名）？

Answer 1

您说过要修改“命名变量”的值，所以我假设您的意思是形式参数（show、me 等）

arguments 对象可以访问传递给函数的所有参数，但它似乎没有提供一种方法来更改命名变量本身的值。

确实如此，但仅限于松散模式，而不是严格模式：

function showMeTheBunny(show, me, the, bunny)
{
    for (let n = 0; n < arguments.length; ++n)
    {
        arguments[n] = arguments[n].trim();
    }
    return `${show} ${me} ${the} ${bunny}: ?`;
}

其中，arguments[0] = arguments[0].trim() 更新show 形式参数的值，arguments[1] = arguments[1].trim() 更新me 等。但仅限于松散模式。在严格模式下，只会更新arguments[x]，不会更新形参；返回它的链接被删除。 （值得注意的是，严格模式是模块和class 构造中的默认设置。）

现场示例：

// Each of these strings are padded with intentional and unnecessary whitespace:
let show = "  show ";
let me = " me  ";
let the = " the ";
let bunny = "  bunny   ";

function showMeTheBunny(show, me, the, bunny)
{
    for (let n = 0; n < arguments.length; ++n)
    {
        arguments[n] = arguments[n].trim();
    }
	return `${show} ${me} ${the} ${bunny}: ?`;
}
console.log(showMeTheBunny(show, me, the, bunny)); // output: "show me the bunny"

还有其他方法，但它们不会修改形参的值。例如，您可以使用 rest 参数：

function showMeTheBunny(...rest)
{
    rest = rest.map(entry => entry.trim());
    const [show, me, the, bunny] = rest;
    return `${show} ${me} ${the} ${bunny}: ?`;
}

现场示例：

"use strict";
// Each of these strings are padded with intentional and unnecessary whitespace:
let show = "  show ";
let me = " me  ";
let the = " the ";
let bunny = "  bunny   ";

function showMeTheBunny(...rest)
{
    rest = rest.map(entry => entry.trim());
    const [show, me, the, bunny] = rest;
	return `${show} ${me} ${the} ${bunny}: ?`;
}
console.log(showMeTheBunny(show, me, the, bunny)); // output: "show me the bunny"

这在严格模式下有效。

另一种选择是接受具有参数属性的对象，然后（再次）使用解构来获取单个变量：

function showMeTheBunny(args)
{
    for (const [name, value] of Object.entries(args)) {
        args[name] = value.trim();
    }
    const {show, me, the, bunny} = args;
    return `${show} ${me} ${the} ${bunny}: ?`;
}

现场示例：

"use strict";
// Each of these strings are padded with intentional and unnecessary whitespace:
let show = "  show ";
let me = " me  ";
let the = " the ";
let bunny = "  bunny   ";

function showMeTheBunny(args)
{
    for (const [name, value] of Object.entries(args)) {
        args[name] = value.trim();
    }
    const {show, me, the, bunny} = args;
	return `${show} ${me} ${the} ${bunny}: ?`;
}
console.log(showMeTheBunny({show, me, the, bunny})); // output: "show me the bunny"

这也适用于严格模式。

Answer 2

arguments 对象的属性实际上是 setter。如果您重新分配参数的属性，相应的变量名称也会更改。因此，您可以遍历 arguments 并重新分配它们：

// Each of these strings are padded with intentional and unnecessary whitespace:
let show = "  show ";
let me = " me  ";
let the = " the ";
let bunny = "  bunny   ";

function showMeTheBunny(show, me, the, bunny)
{
  [...arguments].forEach((arg, i) => {
    arguments[i] = arg.trim();
  });
	return `${show} ${me} ${the} bunny`; 
}
console.log(showMeTheBunny(show, me, the, bunny)); // output: "show me the bunny"

但这真的很奇怪。 Javascript 中几乎没有其他任何东西表现出这种极其不直观的行为。考虑改变你的功能。

Answer 3

你是说这个吗？

将迭代参数转换为数组并映射

Array.from(arguments) 或 [...arguments] 将在这里工作

这实际上并没有像 TJ 所指出的那样修改实际的参数数组

// Each of these strings are padded with intentional and unnecessary whitespace:
let show = "  show ";
let me = " me  ";
let the = " the ";
let bunny = "  bunny   ";

function showMeTheBunny(show, me, the, bunny) {
  return Array.from(arguments).map(el => el.trim()).join(" ")+": ?";
}
console.log(showMeTheBunny(show, me, the, bunny)); // output: "show me the bunny"

Answer 4

你可以使用javascript的arguments对象！！ https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Functions/arguments

// Each of these strings are padded with intentional and unnecessary whitespace:
let show = "  show ";
let me = " me  ";
let the = " the ";
let bunny = "  bunny   ";

function showMeTheBunny(show, me, the, bunny)
{
    var retString = "";
	for (var i = 0; i < arguments.length; i++) {
       retString += arguments[i].trim() + " ";
    }

	return retString.trim(); 
}
console.log(showMeTheBunny(show, me, the, bunny)); // output: "show me the bunny"