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Tarun Sharma @tkssharma

Javascript Coding Questions 15 min code


Published on March 04, 2020

Answer Questions : Deep concepts

  1. what will be the output of this code
(function () {
    try {
        throw new Error();
    } catch (x) {
        var x = 1, y = 2;
        console.log(x);
    }
    console.log(x);
    console.log(y);
})();
// output 1 undefined 2 
  1. what will be the output of this code
var length = 10;
function fn() {
    console.log(this.length);
}
var obj = {
  length: 5,
  method: function(fn) {
    fn();
    arguments[0]();
  }
};
obj.method(fn, 1);
// output 10, 2
  1. what is the closure with example

A closure is an inner function that has access to the variables in the outer (enclosing) function’s scope chain. The closure has access to variables in three scopes; specifically: (1) variable in its own scope, (2) variables in the enclosing function’s scope, and (3) global variables.

var globalVar = "xyz";

(function outerFunc(outerArg) {
    var outerVar = 'a';
    
    (function innerFunc(innerArg) {
    var innerVar = 'b';
    
    console.log(
        "outerArg = " + outerArg + "\n" +
        "innerArg = " + innerArg + "\n" +
        "outerVar = " + outerVar + "\n" +
        "innerVar = " + innerVar + "\n" +
        "globalVar = " + globalVar);
    
    })(456);
})(123);
  1. what will be the output of this code
for (var i = 0; i < 5; i++) {
	setTimeout(function() { console.log(i); }, i * 1000 );
}

The code sample shown will not display the values 0, 1, 2, 3, and 4 as might be expected; rather, it will display 5, 5, 5, 5, and 5. The reason for this is that each function executed within the loop will be executed after the entire loop has completed and all will therefore reference the last value stored in i, which was 5.

Closures can be used to prevent this problem by creating a unique scope for each iteration, storing each unique value of the variable within its scope, as follows:

for (var i = 0; i < 5; i++) {
    (function(x) {
        setTimeout(function() { console.log(x); }, x * 1000 );
    })(i);
}

This will produce the presumably desired result of logging 0, 1, 2, 3, and 4 to the console.

In an ES2015 context, you can simply use let instead of var in the original code:

for (let i = 0; i < 5; i++) {
	setTimeout(function() { console.log(i); }, i * 1000 );
}
  1. How can you clone an Object
var obj = {a: 1 ,b: 2}
var objclone = Object.assign({},obj);

Now the value of objclone is {a: 1 ,b: 2} but points to a different object than obj. Note the potential pitfall, though: Object.clone() will just do a shallow copy, not a deep copy. This means that nested objects aren’t copied. They still refer to the same nested objects as the original:

let obj = {
    a: 1,
    b: 2,
    c: {
        age: 30
    }
};
var objclone = Object.assign({},obj);
console.log('objclone: ', objclone);
obj.c.age = 45;
console.log('After Change - obj: ', obj);           // 45 - This also changes
console.log('After Change - objclone: ', objclone); // 45

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