Problem understanding this code

[Rajesh Khan]

Hi i m currently going over an open source JS library and I am trying to comprehend it. I ran into something like this I deleted a lot of stuff in this code but i wanted to know what the extend method does , I believe its suppose to copy the properties of second into the first but whats after that ?? the extend function is not missing anything else. What does it do after copying the properties. IT neither returns the value and neither are the objects passed to this function global.

var qq=qq || {}; //What does this mean
qq.extend = function(first, second)
{   for (var prop in second)
    {
        first[prop] = second[prop];
    }
};  


qq.FileBasic = function ab() 
{
    this._options = 
    {
        propA: 'valA',
        propB: 'valB',
        propC:  function(id, fileName){},              
    };
    qq.extend(this._options, {PropD:"ValD",PropE:"ValE"});
};

[Eric Pascarello]

    var qq=qq || {}; //If qq is already defined use it OR create a new object 
    qq.extend = function(first, second)  //Copies one object into another object
    {   for (var prop in second)  //Looks through all of the properties in the second object
        {  
            first[prop] = second[prop];  //stores the prop into the first object.
        }  
        //There is no return since objects are passed by reference, not value. So when we add items to first they will remain there.
    };    

Since objects are passed by reference, you can not just store the values, you need to loop through them to make a copy.

var a = { "apple" : "fruit", "anteater" : "animal" };
var b = { "ball" : "toy", "baseball" : "toy" };

console.log('a["foo"]:\t' + a["foo"] ); //should write undefined to console
console.log('b["foo"]:\t' + b["foo"] ); //should write undefined to console

var items1 = a;
items1["foo"] = "bar";

console.log('a["foo"]:\t' +  a["foo"] ); //should write bar to console
console.log('items1["foo"]:\t' +  items1["foo"] ); //should write bar to console

var items2 = {};
for(var prop in b) {
    items2 [prop] = b[prop]
}
items2["foo"] = "bar";

console.log('b["foo"]:\t' + b["foo"] ); //should write undefined to console
console.log('items2["foo"]:\t' +  items2["foo"] ); //should write bar to console

Eric

[Rajesh Khan]

Eric Pascarello wrote:

    var qq=qq || {}; //If qq is already defined use it OR create a new object 
    qq.extend = function(first, second)  //Copies one object into another object
    {   for (var prop in second)  //Looks through all of the properties in the second object
        {  
            first[prop] = second[prop];  //stores the prop into the first object.
        }  
        //There is no return since objects are passed by reference, not value. So when we add items to first they will remain there.
    };    

Since objects are passed by reference, you can not just store the values, you need to loop through them to make a copy.

var a = { "apple" : "fruit", "anteater" : "animal" };
var b = { "ball" : "toy", "baseball" : "toy" };

console.log('a["foo"]:\t' + a["foo"] ); //should write undefined to console
console.log('b["foo"]:\t' + b["foo"] ); //should write undefined to console

var items1 = a;
items1["foo"] = "bar";

console.log('a["foo"]:\t' +  a["foo"] ); //should write bar to console
console.log('items1["foo"]:\t' +  items1["foo"] ); //should write bar to console

var items2 = {};
for(var prop in b) {
    items2 [prop] = b[prop]
}
items2["foo"] = "bar";

console.log('b["foo"]:\t' + b["foo"] ); //should write undefined to console
console.log('items2["foo"]:\t' +  items2["foo"] ); //should write bar to console

Eric

Thanks Eric for clarifying that out. It helped a lot.