ES6 (ii) extension of deconstruction Assignment + string

Explain the previous deconstruction assignment

① "..." in the assignment of the deconstruction.

Let [a,... b]= [1];     
b   //[]     ... Represents the variable B to match all remaining elements returns an array that is not matched to return []  
Note: "... b" can only be put in the last

② The data type must be the same on both sides of an equal-value solution assignment

That

let [] = [] or let {} = {}

However: The set structure also allows the use of arrays for the assignment of destructors

New Set ([1,2,3,4]) a  //1b  //2

Tip: If you're not sure if the structure can be deconstructed, determine if it has a iterator interface. The judging method is as follows:

function* fibs () {  var x= 0;   var y= 1;    while (true) {    yield x;     = [Y, x+ y];  }} var [A, B, C, D, E, f] = fibs (); F//  5//   fibs is a generator function, Native has a iterator interface. The deconstruction assignment gets the value from this interface in turn. 

③ Deconstruction assignment applies not only to let/const but also to Var directives

④ refactoring assignment allows default values to be added

// 4

(The matching pattern of the deconstructed assignment is = = =) Note: The default value does not take effect if the deconstruction assignment is not strictly equal to undefined

let [a = 4]=//4

let [b = 4]=[null]b//null

⑤, I've made my own mistakes.

let [x = y, y = 1] = [];     // error  because x=y when Y has not been declared    will not be promoted!!! Remember
var [a = b, b = 1] = [];      a===undefined  b = = 1

Deconstruction assignment of an object

① objects are not ordered in the same way as arrays, so the object's deconstruction assignment is matched by the ' key '.

var {Foo:baz} = {foo: ' aaa ', bar: ' BBB '//  ' aaa '      foo is a matching pattern  baz is a variable

② object and array nesting if you look at the example, it's almost there.

var obj = {  p: [    ' Hello ',    ' World ' }  }; var {p: [x, {y}]} = obj;      //p is a match pattern  [x, {y}] is a value   then look at each item in the array

// "Hello"
//

③ returns an undefined object when it does not exist as an array.

Deconstruction assignment of a string

const [A, B, C, D, e] = ' Hello ';  A  //  hB  //  eC  //  L

let {length : len} = ‘hello‘;len // 5 匹配的是字符串的length属性

Dry Goods: Use

(1) value of the swap variable　

[A,b]=[b,a]

(2) Returning multiple values from a function

function example () {  return [1, 2, 3];} var [A, B, c] = example ();

(3) Definition of function parameters

function f ([x, Y, z]) {...} f ([1, 2, 3]);

(4) Extracting JSON data

var =  {jsondata,  "OK",  data: [867, 5309 = Jsondata;

(5) Default value for function parameters

function f ([x = 1, y = 2, z]) {...}

(6) Traverse the map structure

var New Map (); Map.set (' first ', ' Hello '); Map.set (' second ', ' World ') ;  for (Let [key, value] of the map)  {+ "is" + value);} // First is Hello // second is world

//get key name for  (let [key" of Map) {//... }//get key value for  (let [ ,value] of Map{//...              /span>          

(7) Specify method of input module

const {Sourcemapconsumer, Sourcenode} = require ("Source-map");

　　

ES6 (ii) extension of deconstruction Assignment + string