Problem 1: Requirement # Requirements What is the value of a[5] in the array dec
ID: 3714426 • Letter: P
Question
Problem 1:
Requirement # Requirements
What is the value of a[5] in the array declared below:
var a = [2, 4, 6, 'Joe', 10, [1, 2], {lastName: 'Smith'}];
Problem 2:
Requirement # Requirements
var myQuote="He who laughs, lasts.";
Give the value of each of the following expressions:
a. myQuote.toUpperCase()
b. myQuote.indexOf("a")
c. myQuote.indexOf("@", 1)
d. myQuote.indexOf("who")
e. myQuote.indexOf("a", 10)
f. myQuote.lastIndexOf("a")
g. myQuote.lastIndexOf("a", 7)
h. myQuote.lastIndexOf("a", 10)
i. myQuote.charAt(7)
j. myQuote.charAt(myQuote.length)
k. myQuote.charAt(myQuote.length - 3)
Problem 3:
Requirement # Requirements
Explain the below code. Include the final output to console.log().
Tip: Higher Order Function (Callback).
function Thing(name) {
this.name = name;
}
Thing.prototype.doSomething = function(callback, salutation) {
callback.call(this, salutation);
}
function Afunction(salutation) {
console.log(salutation + " " + this.name);
}
var t = new Thing('John Smith');
t.doSomething(Afunction, 'Hello');
Problem 4:
Requirement # Requirements
What are the advantages of using Objects in JavaScript. Min 50 words
Submit Answer
Problem 5:
Requirement # Requirements
Create a an Object constructor function called MedRecord with the properties firstName, lastName, medRecord, medID, insuranceInfo.
Define a MedRecord prototype method called changeInsurance() that accepts the name of a patient new insurance information and update object insuranceInfo property.
Problem 6:
Requirement # Requirements
Put the below data in a JSON formatted array where each individual is its own object. Then write code that will print each person's first letter of their first name and last name (example: msmith)
firstName: Mike lastName: Smith
firstName: Anna lastName: House
firstName: Mark lastName: McDonald
Explanation / Answer
1:
Index starts with 0.
So, 5th index is the 6th element that is 1,2
Problem 2:
a. myQuote.toUpperCase() = HE WHO LAUGHS, LASTS.
Give the string in upper case.
b. myQuote.indexOf("a") = 8
The very first index when a is found in the given string.
c. myQuote.indexOf("@", 1) = -1
Not found in the string
d. myQuote.indexOf("who") = 3
The first time "Who" occured in the string is at index 3
e. myQuote.indexOf("a", 10) = 16
This gives the index of first occurence of 'a' in the string starting from 10th index
f. myQuote.lastIndexOf("a") = 16
The very last time a occured in the string is 16.
g. myQuote.lastIndexOf("a", 7) = -1
There is no existance of a till 7th index.
h. myQuote.lastIndexOf("a", 10) = 8
Checking in the string till the 10th index and gives the index of the last occurence of the character
i. myQuote.charAt(7) = l
7th index is the eighth character.
j. myQuote.charAt(myQuote.length) = Nothing
because the indices range from 0 to myQuote.length-1
k. myQuote.charAt(myQuote.length - 3) = t
3rd character from ending