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

对于函数重载,可以这样做。

function addCSS(el, prop, val) {
  return {
    2: function() {
      // when two arguments are set
      // now prop is an oject
      for (var i in prop) {
          el.style[i] = prop[i];
      }
    },
    3: function() {
      // when three arguments are set
      el.style[prop] = val;
    }
    }[arguments.length]();
}
// usage
var el = document.getElementById("demo");
addCSS(el, "color", "blue");
addCSS(el, {
    "backgroundColor": "black",
  "padding": "10px"
});

其他回答

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.

在JS中没有实际的重载,无论如何我们仍然可以用几种方式模拟方法重载:

方法# 1: 使用对象

function test(x,options){
  if("a" in options)doSomething();
  else if("b" in options)doSomethingElse();
}
test("ok",{a:1});
test("ok",{b:"string"});

方法# 2: 使用rest (spread)参数

function test(x,...p){
 if(p[2])console.log("3 params passed"); //or if(typeof p[2]=="string")
else if (p[1])console.log("2 params passed");
else console.log("1 param passed");
}

方法# 3: 使用未定义的

function test(x, y, z){
 if(typeof(z)=="undefined")doSomething();
}

方法# 4: 类型检查

function test(x){
 if(typeof(x)=="string")console.log("a string passed")
 else ...
}

你可以使用John Resig的addMethod。使用此方法,您可以根据参数计数“重载”方法。

// addMethod - By John Resig (MIT Licensed)
function addMethod(object, name, fn){
    var old = object[ name ];
    object[ name ] = function(){
        if ( fn.length == arguments.length )
            return fn.apply( this, arguments );
        else if ( typeof old == 'function' )
            return old.apply( this, arguments );
    };
}

我还创建了该方法的替代方法,使用缓存来保存函数的变体。这里描述了不同之处

// addMethod - By Stavros Ioannidis
function addMethod(obj, name, fn) {
  obj[name] = obj[name] || function() {
    // get the cached method with arguments.length arguments
    var method = obj[name].cache[arguments.length];

    // if method exists call it 
    if ( !! method)
      return method.apply(this, arguments);
    else throw new Error("Wrong number of arguments");
  };

  // initialize obj[name].cache
  obj[name].cache = obj[name].cache || {};

  // Check if a method with the same number of arguments exists  
  if ( !! obj[name].cache[fn.length])
    throw new Error("Cannot define multiple '" + name +
      "' methods with the same number of arguments!");

  // cache the method with fn.length arguments
  obj[name].cache[fn.length] = function() {
    return fn.apply(this, arguments);
  };
}

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

看看这个。它很酷。 http://ejohn.org/blog/javascript-method-overloading/ 技巧Javascript,让你做这样的调用:

var users = new Users();
users.find(); // Finds all
users.find("John"); // Finds users by name
users.find("John", "Resig"); // Finds users by first and last name