Javascript 一次将所有对象调用到函数中，并返回每个对象的特定计算？

如果没有明确的方法，如何调用

var auto

的每个对象

我的意思是不必在函数调用中写入它们，在

speedAverage

函数和

return

中计算平均值

function Car (model, speed) {
    this.model = model;
    this.speed = speed;
}


var speedAverage = function(car1, car2, car3) {
    return (car1.speed + car2.speed + car3.speed)/ veh.length;
};



var auto = [
new Car("BMW", 220),
new Car("Ford", 260),
new Car("DeLorean", 350)
];

console.log("The average between these cars is " + speedAverage(auto[0], auto[1], 
auto[2]));

---

谢谢。

使用此方法，您可以非常轻松地完成此操作。此函数用于调用以数组内容为参数的函数：

speedAverage.apply(null, auto);

---

我认为您希望将数组对象传递给函数：

var speedAverage = function (arr) {
  var sum=0
  for(var i=0; i<arr.length; i++) {
     sum += arr[i];
  }
  return sum/arr.length;
}

var speedAverage=函数（arr）{
var总和=0
对于（var i=0；i向
Car
添加一个静态函数，并在构造函数中推送速度，则不必执行
auto
操作：

var speeds = [];
function Car (model, speed) {
    this.model = model;
    this.speed = speed;
    speeds.push(speed);
}

Car.getAverageSpeed = function() {
  return eval(speeds.join('+'))/speeds.length;
};

new Car("BMW", 220),
new Car("Ford", 260),
new Car("DeLorean", 350)

console.log("The average between these cars is " + Car.getAverageSpeed());

或者，您可以使用以下方法：

var speedAverage = function() {
    return eval(Array.prototype.map.call(arguments, function(s) {
        return s.speed;
    }).join('+'))/arguments.length;
};

var auto = [
    new Car("BMW", 220),
    new Car("Ford", 260),
    new Car("DeLorean", 350)
];

console.log("The average between these cars is " + speedAverage.apply(0, auto));

由于
auto
数组中已有所有数据，请使用reduce函数：

console.log (auto.reduce (
   function (speed, car) { 
     return speed + car.speed; 
   }, 0) / auto.length);

调用
reduce
返回
auto
中所有车辆的速度之和，将其除以车辆数量即可得到平均值

function Car (model, speed) {
    this.model = model;
    this.speed = speed;
}


var speedAverage = function(car1, car2, car3) {
    return (car1.speed + car2.speed + car3.speed)/ veh.length;
};



var auto = [
new Car("BMW", 220),
new Car("Ford", 260),
new Car("DeLorean", 350)
];

console.log("The average between these cars is " + speedAverage(auto[0], auto[1], 
auto[2]));