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1. var obj ={a:1}; (function (obj) {obj = {a:2};})       (obj); Q. does the value of obj change?

var obj = {a:1};

(function (obj) {

obj = {A:2};

}) (obj);

// Ask obj will the value change?

The external obj does not change.
Because the input parameter of obj in the anonymous function is equal to the creation of a local variable, obj, which points to a new object. If changed to (function () {obj = {a:2};})  (obj); It's going to change.


2. var obj = {a:1, func:function () {(function () {a=2;} (); }} ; does the value of a in Obj.func ()//obj change? in the anonymous function. This What's that pointing to? ?

var obj = {

A:1,

Func:function () {

(function () {

a=2;

}();

}

} ;

Obj.func ();

//obj in a will the value change? ? in the anonymous function. This What's that pointing to? ?

The A in obj does not change.  The This in the anonymous function points to the Global object window. This is equivalent to adding a property named A to window.
To change the value of a in obj, you should:

(function () {

THIS.A = 2

}). Call (this);

or as defined in obj

Func:function () {

var = this;

(function () {

self.a=2;

})();}

3. to implement functions within the function every 5 seconds to call themselves this function, after the first stop, how to do ?

(function () {

var index = 0;

function fn () {

if (Index < 100) {

index++;

SetTimeout (function () {

FN ();

},5000);

}

}

FN ();

})();

4. Click on a ul five li elements, respectively pop their serial number, how to do?
Method 1:

var oLi = document.getelementsbytagname (' li ');

for (var i=0; i<olis.length; i++) {

Olis[i].onclick = (function (j) {

Returnfunction () {

Alert (j);

}

}) (i);

}

Method 2:

var oLi = document.getelementsbytagname (' li ');

for (var i=0; i<oli.length; i++) {

(function (j) {

oli[j].onclick= function () {

Alert (j);

};

}) (i);

}

Method 3:

var oLi = document.getelementsbytagname (' li ');

for (var i=0; i<oli.length; i++) {

Oli[i].index = i;

Oli[i].onclick =function () {

alert (This.index);

}

}

5. JS Implementation of array de-weight how to implement ?
Method 1. Create a new temporary array to hold the existing elements in the array

var a = new Array (1,2,2,2,2,5,3,2,9,5,6,3);

Array.prototype.unique1 = function () {

var n = []; a new temporary array

for (var i=0;i<this.length; i++) {

        // If you put the current array I has been saved in a temporary array , so skip it.

if (N.indexof (this[i]) = =-1) {

N.push (This[i]);

}

}

return n;

}

Console.log (A.unique1 ());

Method 2. using a hash table to store existing elements

Array.prototype.unique2 = function () {

var hash = {},

n = []; Hash as a hash table , n as a temporary array

for (var i=0;i<this.length; i++) {

If (!hash[this[i]]) {// if There is no current item in the hash table

Hash[this[i]] = true; deposit hash Table

N.push (This[i]); the current element is push into the temporary array

}

}

return n;

}

Method 3. use IndexOf to determine if the position of the first occurrence of an array element is the current position

Array.prototype.unique3 = function () {

var n = [This[0]];

For (var i=1;i<this.length; i++)// start traversing from the second item

{

        // if the current array element appears in the array for the first time, the position is not I

        // description is a repeating element

if (This.indexof (this[i]) = = i) {

N.push (This[i]);

}

}

return n;

}

Method 4. First sort and then go to heavy

Array.prototype.unique4 = function () {

This.sort (function (A, b) {return a B;});

var n = [This[0]];

for (var i=1;i<this.length; i++) {

if (This[i]!=this[i-1]) {

N.push (This[i]);

}

}

return n;

}

The first and third methods both use indexof (), and the execution mechanism of the function iterates over the array
The second method uses a hash table, which is the fastest.
The third method also has a sort of complexity calculation.
Then we did a test, randomly generating 1 million 0-1000 array results as follows:

The third method is always nearly twice times the second method, and the fourth method is related to the range of the array,
If it is an array of 0-100

And if it is 0-10000, method four look at the effect is good

And the second method is always the best, but it's a space-changing time
The entire code is as follows

var a = [];

for (var i=0; i<1000000; i++) {

A.push (Math.ceil (Math.random () *10000));

}

Array.prototype.unique1 = function () {

var n = []; a new temporary array

for (var i=0;i<this.length; i++) {

        // If you put the current array I has been saved in a temporary array , so skip it.

if (N.indexof (this[i]) = =-1) {

N.push (This[i]);

}

}

return n;

}

Array.prototype.unique2 = function () {

var hash = {},

n = []; Hash as a hash table , n as a temporary array

for (var i=0;i<this.length; i++) {

If (!hash[this[i]]) {// if There is no current item in the hash table

Hash[this[i]] = true; deposit hash Table

N.push (This[i]); the current element is push into the temporary array

}

}

return n;

}

Array.prototype.unique3 = function () {

var n = [This[0]];

For (var i=1;i<this.length; i++)// start traversing from the second item

{

        // if the current array element appears in the array for the first time, the position is not I

        // description is a repeating element

if (This.indexof (this[i]) = = i) {

N.push (This[i]);

}

}

return n;

}

Array.prototype.unique4 = function () {

This.sort (function (A, b) {return a B;});

var n = [This[0]];

for (var i=1;i<this.length; i++) {

if (This[i]!=this[i-1]) {

N.push (This[i]);

}

}

return n;

}

var begin1 = new Date ();

A.unique1 ();

var end1 = new Date ();

var begin2 = new Date ();

A.unique2 ();

var end2 = new Date ();

var begin3 = new Date ();

A.unique3 ();

var end3 = new Date ();

var begin4 = new Date ();

A.unique4 ();

var end4 = new Date ();

console.log (" method One Execution time :" + (end1-begin1));

console.log (" method Two execution time :" + (end2-begin2));

console.log (" method Three execution time :" + (end3-begin3));

console.log (" method four execution time :" + (END4-BEGIN4));

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Web Front End Questions finishing (program)