解释简单的return关键字

时间:2016-02-07 05:49:43

标签: javascript return-value map-function

所以我想出了一个返回数组的问题的基本解决方案:

 function doubleAll(numbers) {
          var result = []

          numbers.map(function(val, i){
            result.push(val*2);
          })

          return result
    }

然而,官方解决方案只是使用双返回语句,我试图了解操作的顺序以及它是如何工作的。

function doubleAll(numbers) {
      return numbers.map(function double(num) {
        return num * 2
      })
    }

我对幕后发生的事情以及两种做法之间的差异感兴趣。

2 个答案:

答案 0 :(得分:2)

  

map()方法创建一个新数组,其中包含调用a的结果   为此数组中的每个元素提供了函数。

所以在你的例子中:

function doubleAll(numbers) {      
      return numbers.map(function double(num) {
        return num * 2
      }) //returns array 
}

 function doubleAll(numbers) {
          var result = [];
          numbers.map(function(val, i){
            result.push(val*2); //pushing value into result array
          });
          return result;
 }

在第二个示例中,资源被浪费,因为map返回一个数组并且它从未使用过,这仍然可以使用forEach方法实现,而不会浪费资源:

function doubleAll(numbers) {
  var result = [];
  numbers.forEach(function(val, i){
    result.push(val*2); //pushing value into result array
  });
  return result;
}

此填充将增加.map的清晰度:
来自MDN的参考:

// Production steps of ECMA-262, Edition 5, 15.4.4.19
// Reference: http://es5.github.io/#x15.4.4.19
if (!Array.prototype.map) {

  Array.prototype.map = function(callback, thisArg) {

    var T, A, k;

    if (this == null) {
      throw new TypeError(' this is null or not defined');
    }

    // 1. Let O be the result of calling ToObject passing the |this| 
    //    value as the argument.
    var O = Object(this);

    // 2. Let lenValue be the result of calling the Get internal 
    //    method of O with the argument "length".
    // 3. Let len be ToUint32(lenValue).
    var len = O.length >>> 0;

    // 4. If IsCallable(callback) is false, throw a TypeError exception.
    // See: http://es5.github.com/#x9.11
    if (typeof callback !== 'function') {
      throw new TypeError(callback + ' is not a function');
    }

    // 5. If thisArg was supplied, let T be thisArg; else let T be undefined.
    if (arguments.length > 1) {
      T = thisArg;
    }

    // 6. Let A be a new array created as if by the expression new Array(len) 
    //    where Array is the standard built-in constructor with that name and 
    //    len is the value of len.
    A = new Array(len);

    // 7. Let k be 0
    k = 0;

    // 8. Repeat, while k < len
    while (k < len) {

      var kValue, mappedValue;

      // a. Let Pk be ToString(k).
      //   This is implicit for LHS operands of the in operator
      // b. Let kPresent be the result of calling the HasProperty internal 
      //    method of O with argument Pk.
      //   This step can be combined with c
      // c. If kPresent is true, then
      if (k in O) {

        // i. Let kValue be the result of calling the Get internal 
        //    method of O with argument Pk.
        kValue = O[k];

        // ii. Let mappedValue be the result of calling the Call internal 
        //     method of callback with T as the this value and argument 
        //     list containing kValue, k, and O.
        mappedValue = callback.call(T, kValue, k, O);

        // iii. Call the DefineOwnProperty internal method of A with arguments
        // Pk, Property Descriptor
        // { Value: mappedValue,
        //   Writable: true,
        //   Enumerable: true,
        //   Configurable: true },
        // and false.

        // In browsers that support Object.defineProperty, use the following:
        // Object.defineProperty(A, k, {
        //   value: mappedValue,
        //   writable: true,
        //   enumerable: true,
        //   configurable: true
        // });

        // For best browser support, use the following:
        A[k] = mappedValue;
      }
      // d. Increase k by 1.
      k++;
    }

    // 9. return A
    return A;
  };
}

答案 1 :(得分:0)

数组的map方法将对数组的每个成员执行回调并返回结果数组。

function doubleAll(numbers) {      
      return numbers.map(function double(num) {
        return num * 2
      }) //returns an array whose members have each been passed to double(num)
}

在这种情况下,doubleAll()返回numbers.map()的结果,该结果是一个数组,其成员已被回调函数double(num)变异。