5 classic JavaScript interview questions_javascript skills

1: Scope

Copy code The code is as follows:

(function() {
var a = b = 5;
})();
console.log(b);

What will be printed to the console?

Answer

The above code will print 5.

The trick to this problem is that there are two variable declarations, but a is declared using the keyword var. Indicates that it is a local variable of a function. In contrast, b becomes a global variable.

Another trick with this question is that it doesn’t use strict mode (‘use strict’;). If strict mode is enabled, the code will raise a ReferenceError: b is not defined. Remember that strict mode requires explicit specification in order to implement global variable declarations. For example, you should write:

Copy code The code is as follows:

(function() {
'use strict';
var a = window.b = 5;
})();

console.log(b);

2: Create a "native" method

Define a repeatify function for the string object. When passed in an integer n, it returns the result of repeating the string n times. For example:

Copy code The code is as follows:

console.log('hello'.repeatify(3));

should print hellohellohello.

Answer

A possible implementation looks like this:

Copy code The code is as follows:

String.prototype.repeatify = String.prototype.repeatify || function(times) {
var str = '';
for (var i = 0; i < times; i ) {
       str = this;
}
Return str;
};

The current questions test developers’ knowledge about JavaScript inheritance and prototypes. This also verifies that the developer knows how to extend built-in objects (even though this should not be done).

Another important point here is that you need to know how not to override functionality that may already be defined. Test that the function definition did not exist before:

Copy code The code is as follows:

String.prototype.repeatify = String.prototype.repeatify || function(times) {/* code here */};

This technique is especially useful when you are asked to make JavaScript functions compatible.

3: Statement Hoisting

Execute this code and what results will be output.

Copy code The code is as follows:

function test() {
console.log(a);
console.log(foo());
var a = 1;
Function foo() {
Return 2;
}
}

10: test();

Answer

The result of this code is undefined and 2.

The reason is that the declarations of variables and functions are brought forward (moved to the top of the function), but the variables are not assigned any values. So, when printing the variable, it exists in the function (it was declared), but it is still undefined. In other words, the above code is equivalent to the following:

Copy code The code is as follows:

function test() {
var a;
Function foo() {
Return 2;
}

console.log(a);
console.log(foo());

a = 1;
}

test();

4: How this works in JavaScript

What will the following code output? Give your answer.

Copy code The code is as follows:

var fullname = 'John Doe';
var obj = {
Fullname: 'Colin Ihrig',
; prop: {
        fullname: 'Aurelio De Rosa',
        getFullname: function() {
            return this.fullname;
}
}
};

console.log(obj.prop.getFullname());

var test = obj.prop.getFullname;

console.log(test());

Answer

The answer is Aurelio De Rosa and John Doe. The reason is that in a function, the behavior of this depends on how the JavaScript function is called and defined, not just how it is defined.

In the first console.log() call, getFullname() is called as a function of the obj.prop object. Therefore, the context refers to the latter and the function returns the fullname of the object. In contrast, when getFullname() is assigned to the test variable, the context refers to the global object (window). This is because test is implicitly set as a property of the global object. For this reason, the function returns the fullname of the window, which is the value defined on the first line.

5: call() and apply()

Now let's solve the previous problem so that the final console.log() prints Aurelio De Rosa.

Answer

This problem can be changed by forcing the use of call() or apply() to change the function context. Below I'll use call(), but in this case apply() will output the same result:

Copy code The code is as follows:

console.log(test.call(obj.prop));