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。

而不是 50 个参数，他是一个只有 4 个参数的例子：

// 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，但它似乎没有提供一种方法来更改命名变量本身的值。

是否可以运行通过一个函数修改所有命名变量（在函数签名中命名），然后在以后使用这些修改后的参数（使用相同的变量名）之前修改它们中的每一个？

Answer 1

你是这个意思？

将迭代参数转换为数组并映射它

Array.from(arguments) 或 [...arguments] 将在此处工作

这实际上并没有像 TJ

指出的那样修改实际参数数组

// 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 2

您说过要修改 "named variables" 的值，所以我假设您指的是形式参数（show、me 等）

The arguments object can access all the arguments passed into a function, but it doesn't seem to offer a way to change the values of the named-variables themselves.

确实如此，但仅限于宽松模式，而非严格模式：

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。严格模式只更新arguments[x]，不更新形参； link 回到它被删除。（值得注意的是，严格模式是模块和 class 构造中的默认模式。）

实例：

// 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，但不会修改形参的值。例如，您可以使用休息参数：

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

实例：

"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"

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}: `;
}

实例：

"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 3

arguments 对象的属性实际上是 setter。如果您在参数上重新分配属性，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 中几乎没有其他东西表现出这种极其不直观的行为。考虑改变你的功能。

Answer 4

可以使用javascript的参数对象！！ 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"

更改通过函数签名命名的所有变量的值

Change Values of All Variables Named via Function Signature

javascript

ecmascript-next