在Javascript中伪造函数重载的最好方法是什么?

我知道在Javascript中不可能像在其他语言中那样重载函数。 如果我需要一个函数有两个使用foo(x)和foo(x,y,z),这是最好的/首选的方式:

一开始就用不同的名字 使用可选参数,如y = y || 'default' 使用参数数量 检查参数类型 还是如何?


当前回答

最好的方法取决于函数和参数。你的每个选择在不同的情况下都是一个好主意。我通常按照以下顺序尝试这些方法,直到其中一个有效为止:

Using optional arguments like y = y || 'default'. This is convenient if you can do it, but it may not always work practically, e.g. when 0/null/undefined would be a valid argument. Using number of arguments. Similar to the last option but may work when #1 doesn't work. Checking types of arguments. This can work in some cases where the number of arguments is the same. If you can't reliably determine the types, you may need to use different names. Using different names in the first place. You may need to do this if the other options won't work, aren't practical, or for consistency with other related functions.

其他回答

对于你的用例,这是我用ES6解决它的方法(因为它已经是2017年底了):

const foo = (x, y, z) => {
  if (y && z) {
    // Do your foo(x, y, z); functionality
    return output;
  }
  // Do your foo(x); functionality
  return output;
}

显然,您可以将其用于任何数量的参数,并相应地更改条件语句。

我不确定最佳实践,但我是这样做的:

/*
 * Object Constructor
 */
var foo = function(x) {
    this.x = x;
};

/*
 * Object Protoype
 */
foo.prototype = {
    /*
     * f is the name that is going to be used to call the various overloaded versions
     */
    f: function() {

        /*
         * Save 'this' in order to use it inside the overloaded functions
         * because there 'this' has a different meaning.
         */   
        var that = this;  

        /* 
         * Define three overloaded functions
         */
        var f1 = function(arg1) {
            console.log("f1 called with " + arg1);
            return arg1 + that.x;
        }

        var f2 = function(arg1, arg2) {
             console.log("f2 called with " + arg1 + " and " + arg2);
             return arg1 + arg2 + that.x;
        }

        var f3 = function(arg1) {
             console.log("f3 called with [" + arg1[0] + ", " + arg1[1] + "]");
             return arg1[0] + arg1[1];
        }

        /*
         * Use the arguments array-like object to decide which function to execute when calling f(...)
         */
        if (arguments.length === 1 && !Array.isArray(arguments[0])) {
            return f1(arguments[0]);
        } else if (arguments.length === 2) {
            return f2(arguments[0], arguments[1]);
        } else if (arguments.length === 1 && Array.isArray(arguments[0])) {
            return f3(arguments[0]);
        }
    } 
}

/* 
 * Instantiate an object
 */
var obj = new foo("z");

/*
 * Call the overloaded functions using f(...)
 */
console.log(obj.f("x"));         // executes f1, returns "xz"
console.log(obj.f("x", "y"));    // executes f2, returns "xyz"
console.log(obj.f(["x", "y"]));  // executes f3, returns "xy"

虽然默认参数没有重载,但它可能解决开发人员在这方面面临的一些问题。输入是严格由顺序决定的,你不能重新排序,因为你希望在经典重载:

function transformer(
    firstNumber = 1,
    secondNumber = new Date().getFullYear(),
    transform = function multiply(firstNumber, secondNumber) {
        return firstNumber * secondNumber;
    }
) {
    return transform(firstNumber, secondNumber);
}

console.info(transformer());
console.info(transformer(8));
console.info(transformer(2, 6));
console.info(transformer(undefined, 65));

function add(firstNumber, secondNumber) {
    return firstNumber + secondNumber;
}
console.info(transformer(undefined, undefined, add));
console.info(transformer(3, undefined, add));

结果(2020年):

2020
16160
12
65
2021
2023

更多信息:https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Functions/Default_parameters

不是每个人都知道可以在函数签名中直接进行解构赋值。

得益于此,您可以轻松地定义非常灵活的方法签名,恕我直言,这比Java方法重载更优越。

例子:

const myFunction = (({a, b, c}) => {
    console.log(a, b, c);
});

myFunction({a: 1, b: 2});
myFunction({a: 1, b: 2, c: 3});

您甚至不需要考虑参数的顺序,并且调用语句和目标方法签名之间具有命名一致性。

你也可以指定默认值:

const myFunction = (({a = 1, b = 2, c} = {}) => {
    console.log(a, b, c);
});

有两种方法可以更好地解决这个问题:

如果您希望保留很大的灵活性,则传递一个字典(关联数组) 将一个对象作为参数,并使用基于原型的继承来增加灵活性。