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。
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
}
和
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。