JavaScript Trivia Top 37 Flashcards
What will the code below output to the console and why?
(function(){
var a = b = 3;
})();
console. log(“a defined? “ + (typeof a !== ‘undefined’));
console. log(“b defined? “ + (typeof b !== ‘undefined’));
Since both a and b are defined within the enclosing scope of the function, and since the line they are on begins with the var keyword, most JavaScript developers would expect typeof a and typeof b to both be undefined in the above example.
However, that is not the case. The issue here is that most developers incorrectly understand the statement var a = b = 3; to be shorthand for:
var b = 3; var a = b; But in fact, var a = b = 3; is actually shorthand for:
b = 3; var a = b; As a result (if you are not using strict mode), the output of the code snippet would be:
a defined? false b defined? true But how can b be defined outside of the scope of the enclosing function? Well, since the statement var a = b = 3; is shorthand for the statements b = 3; and var a = b;, b ends up being a global variable (since it is not preceded by the var keyword) and is therefore still in scope even outside of the enclosing function.
Note that, in strict mode (i.e., with use strict), the statement var a = b = 3; will generate a runtime error of ReferenceError: b is not defined, thereby avoiding any headfakes/bugs that might othewise result. (Yet another prime example of why you should use use strict as a matter of course in your code!)
What will the code below output to the console and why?
var myObject = {
foo: "bar",
func: function() {
var self = this;
console.log("outer func: this.foo = " + this.foo);
console.log("outer func: self.foo = " + self.foo);
(function() {
console.log("inner func: this.foo = " + this.foo);
console.log("inner func: self.foo = " + self.foo);
}());
}
};
myObject.func();
outer func: this.foo = bar outer func: self.foo = bar inner func: this.foo = undefined inner func: self.foo = bar In the outer function, both this and self refer to myObject and therefore both can properly reference and access foo.
In the inner function, though, this no longer refers to myObject. As a result, this.foo is undefined in the inner function, whereas the reference to the local variable self remains in scope and is accessible there.
What is the significance of, and reason for, wrapping the entire content of a JavaScript source file in a function block?
This is an increasingly common practice, employed by many popular JavaScript libraries (jQuery, Node.js, etc.). This technique creates a closure around the entire contents of the file which, perhaps most importantly, creates a private namespace and thereby helps avoid potential name clashes between different JavaScript modules and libraries.
Another feature of this technique is to allow for an easily referenceable (presumably shorter) alias for a global variable. This is often used, for example, in jQuery plugins. jQuery allows you to disable the $ reference to the jQuery namespace, using jQuery.noConflict(). If this has been done, your code can still use $ employing this closure technique, as follows:
(function($) { /* jQuery plugin code referencing $ */ } )(jQuery);
What is the significance, and what are the benefits, of including ‘use strict’ at the beginning of a JavaScript source file?
The short and most important answer here is that use strict is a way to voluntarily enforce stricter parsing and error handling on your JavaScript code at runtime. Code errors that would otherwise have been ignored or would have failed silently will now generate errors or throw exceptions. In general, it is a good practice.
Some of the key benefits of strict mode include:
>Makes debugging easier. Code errors that would otherwise have been ignored or would have failed silently will now generate errors or throw exceptions, alerting you sooner to problems in your code and directing you more quickly to their source.
>Prevents accidental globals. Without strict mode, assigning a value to an undeclared variable automatically creates a global variable with that name. This is one of the most common errors in JavaScript. In strict mode, attempting to do so throws an error.
>Eliminates this coercion. Without strict mode, a reference to a this value of null or undefined is automatically coerced to the global. This can cause many headfakes and pull-out-your-hair kind of bugs. In strict mode, referencing a a this value of null or undefined throws an error.
>Disallows duplicate parameter values. Strict mode throws an error when it detects a duplicate named argument for a function (e.g., function foo(val1, val2, val1){}), thereby catching what is almost certainly a bug in your code that you might otherwise have wasted lots of time tracking down.
Note: It used to be (in ECMAScript 5) that strict mode would disallow duplicate property names (e.g. var object = {foo: "bar", foo: "baz"};) but as of ECMAScript 2015 this is no longer the case.
>Makes eval() safer. There are some differences in the way eval() behaves in strict mode and in non-strict mode. Most significantly, in strict mode, variables and functions declared inside of an eval() statement are not created in the containing scope (they are created in the containing scope in non-strict mode, which can also be a common source of problems).
>Throws error on invalid usage of delete. The delete operator (used to remove properties from objects) cannot be used on non-configurable properties of the object. Non-strict code will fail silently when an attempt is made to delete a non-configurable property, whereas strict mode will throw an error in such a case.
Consider the two functions below. Will they both return the same thing? Why or why not?
function foo1()
{
return {
bar: "hello"
};
}
function foo2()
{
return
{
bar: "hello"
};
}
foo1 returns:
Object {bar: "hello"}
foo2 returns:
undefined
Not only is this surprising, but what makes this particularly gnarly is that foo2() returns undefined without any error being thrown.
The reason for this has to do with the fact that semicolons are technically optional in JavaScript (although omitting them is generally really bad form). As a result, when the line containing the return statement (with nothing else on the line) is encountered in foo2(), a semicolon is automatically inserted immediately after the return statement.
No error is thrown since the remainder of the code is perfectly valid, even though it doesn’t ever get invoked or do anything (it is simply an unused code block that defines a property bar which is equal to the string “hello”).
This behavior also argues for following the convention of placing an opening curly brace at the end of a line in JavaScript, rather than on the beginning of a new line. As shown here, this becomes more than just a stylistic preference in JavaScript.
What will the code below output? Explain your answer.
console. log(0.1 + 0.2);
console. log(0.1 + 0.2 == 0.3);
An educated answer to this question would simply be: “You can’t be sure. it might print out 0.3 and true, or it might not. Numbers in JavaScript are all treated with floating point precision, and as such, may not always yield the expected results.”
The example provided above is classic case that demonstrates this issue. Surprisingly, it will print out:
0.30000000000000004
false
A typical solution is to compare the absolute difference between two numbers with the special constant Number.EPSILON:
function areTheNumbersAlmostEqual(num1, num2) {
return Math.abs( num1 - num2 ) < Number.EPSILON;
}
console.log(areTheNumbersAlmostEqual(0.1 + 0.2, 0.3));
In what order will the numbers 1-4 be logged to the console when the code below is executed? Why?
(function() {
console.log(1);
setTimeout(function(){console.log(2)}, 1000);
setTimeout(function(){console.log(3)}, 0);
console.log(4);
})();
1
4
3
2
Let’s first explain the parts of this that are presumably more obvious:
1 and 4 are displayed first since they are logged by simple calls to console.log() without any delay
2 is displayed after 3 because 2 is being logged after a delay of 1000 msecs (i.e., 1 second) whereas 3 is being logged after a delay of 0 msecs.
OK, fine. But if 3 is being logged after a delay of 0 msecs, doesn’t that mean that it is being logged right away? And, if so, shouldn’t it be logged before 4, since 4 is being logged by a later line of code?
The answer has to do with properly understanding JavaScript events and timing.
The browser has an event loop which checks the event queue and processes pending events. For example, if an event happens in the background (e.g., a script onload event) while the browser is busy (e.g., processing an onclick), the event gets appended to the queue. When the onclick handler is complete, the queue is checked and the event is then handled (e.g., the onload script is executed).
Similarly, setTimeout() also puts execution of its referenced function into the event queue if the browser is busy.
When a value of zero is passed as the second argument to setTimeout(), it attempts to execute the specified function “as soon as possible”. Specifically, execution of the function is placed on the event queue to occur on the next timer tick. Note, though, that this is not immediate; the function is not executed until the next tick. That’s why in the above example, the call to console.log(4) occurs before the call to console.log(3) (since the call to console.log(3) is invoked via setTimeout, so it is slightly delayed).
Write a simple function (less than 160 characters) that returns a boolean indicating whether or not a string is a palindrome.
The following one line function will return true if str is a palindrome; otherwise, it returns false.
function isPalindrome(str) {
str = str.replace(/\W/g, '').toLowerCase();
return (str == str.split('').reverse().join(''));
}
For example:
console. log(isPalindrome(“level”)); // logs ‘true’
console. log(isPalindrome(“levels”)); // logs ‘false’
console. log(isPalindrome(“A car, a man, a maraca”)); // logs ‘true’
Write a sum method which will work properly when invoked using either syntax below.
console. log(sum(2,3)); // Outputs 5
console. log(sum(2)(3)); // Outputs 5
METHOD 1
function sum(x) {
if (arguments.length == 2) {
return arguments[0] + arguments[1];
} else {
return function(y) { return x + y; };
}
}
In JavaScript, functions provide access to an arguments object which provides access to the actual arguments passed to a function. This enables us to use the length property to determine at runtime the number of arguments passed to the function.
If two arguments are passed, we simply add them together and return.
Otherwise, we assume it was called in the form sum(2)(3), so we return an anonymous function that adds together the argument passed to sum() (in this case 2) and the argument passed to the anonymous function (in this case 3).
METHOD 2
function sum(x, y) {
if (y !== undefined) {
return x + y;
} else {
return function(y) { return x + y; };
}
}
When a function is invoked, JavaScript does not require the number of arguments to match the number of arguments in the function definition. If the number of arguments passed exceeds the number of arguments in the function definition, the excess arguments will simply be ignored. On the other hand, if the number of arguments passed is less than the number of arguments in the function definition, the missing arguments will have a value of undefined when referenced within the function. So, in the above example, by simply checking if the 2nd argument is undefined, we can determine which way the function was invoked and proceed accordingly.
Consider the following code snippet:
for (var i = 0; i < 5; i++) {
var btn = document.createElement('button');
btn.appendChild(document.createTextNode('Button ' + i));
btn.addEventListener('click', function(){ console.log(i); });
document.body.appendChild(btn);
}
(a) What gets logged to the console when the user clicks on “Button 4” and why?
(b) Provide one or more alternate implementations that will work as expected.
(a) No matter what button the user clicks the number 5 will always be logged to the console. This is because, at the point that the onclick method is invoked (for any of the buttons), the for loop has already completed and the variable i already has a value of 5. (Bonus points for the interviewee if they know enough to talk about how execution contexts, variable objects, activation objects, and the internal “scope” property contribute to the closure behavior.)
(b) The key to making this work is to capture the value of i at each pass through the for loop by passing it into a newly created function object. Here are four possible ways to accomplish this:
for (var i = 0; i < 5; i++) {
var btn = document.createElement('button');
btn.appendChild(document.createTextNode('Button ' + i));
btn.addEventListener('click', (function(i) {
return function() { console.log(i); };
})(i));
document.body.appendChild(btn);
}
Alternatively, you could wrap the entire call to btn.addEventListener in the new anonymous function:
for (var i = 0; i < 5; i++) {
var btn = document.createElement('button');
btn.appendChild(document.createTextNode('Button ' + i));
(function (i) {
btn.addEventListener('click', function() { console.log(i); });
})(i);
document.body.appendChild(btn);
}
Or, we could replace the for loop with a call to the array object’s native forEach method:
['a', 'b', 'c', 'd', 'e'].forEach(function (value, i) {
var btn = document.createElement('button');
btn.appendChild(document.createTextNode('Button ' + i));
btn.addEventListener('click', function() { console.log(i); });
document.body.appendChild(btn);
});
Lastly, the simplest solution, if you’re in an ES6/ES2015 context, is to use let i instead of var i:
for (let i = 0; i < 5; i++) {
var btn = document.createElement('button');
btn.appendChild(document.createTextNode('Button ' + i));
btn.addEventListener('click', function(){ console.log(i); });
document.body.appendChild(btn);
}
Assuming d is an “empty” object in scope, say:
var d = {};
…what is accomplished using the following code?
[ ‘zebra’, ‘horse’ ].forEach(function(k) {
d[k] = undefined;
});
The snippet of code shown above sets two properties on the object d. Ideally, any lookup performed on a JavaScript object with an unset key evaluates to undefined. But running this code marks those properties as “own properties” of the object.
This is a useful strategy for ensuring that an object has a given set of properties. Passing this object to Object.keys will return an array with those set keys as well (even if their values are undefined).
What will the code below output to the console and why?
var arr1 = “john”.split(‘’);
var arr2 = arr1.reverse();
var arr3 = “jones”.split(‘’);
arr2.push(arr3);
console.log(“array 1: length=” + arr1.length + “ last=” + arr1.slice(-1));
console.log(“array 2: length=” + arr2.length + “ last=” + arr2.slice(-1));
“array 1: length=5 last=j,o,n,e,s”
“array 2: length=5 last=j,o,n,e,s”
arr1 and arr2 are the same (i.e. [‘n’,’h’,’o’,’j’, [‘j’,’o’,’n’,’e’,’s’] ]) after the above code is executed for the following reasons:
Calling an array object’s reverse() method doesn’t only return the array in reverse order, it also reverses the order of the array itself (i.e., in this case, arr1).
The reverse() method returns a reference to the array itself (i.e., in this case, arr1). As a result, arr2 is simply a reference to (rather than a copy of) arr1. Therefore, when anything is done to arr2 (i.e., when we invoke arr2.push(arr3);), arr1 will be affected as well since arr1 and arr2 are simply references to the same object.
And a couple of side points here that can sometimes trip someone up in answering this question:
Passing an array to the push() method of another array pushes that entire array as a single element onto the end of the array. As a result, the statement arr2.push(arr3); adds arr3 in its entirety as a single element to the end of arr2 (i.e., it does not concatenate the two arrays, that’s what the concat() method is for).
Like Python, JavaScript honors negative subscripts in calls to array methods like slice() as a way of referencing elements at the end of the array; e.g., a subscript of -1 indicates the last element in the array, and so on.
What will the code below output to the console and why ?
console. log(1 + “2” + “2”);
console. log(1 + +”2” + “2”);
console. log(1 + -“1” + “2”);
console. log(+”1” + “1” + “2”);
console. log( “A” - “B” + “2”);
console. log( “A” - “B” + 2);
"122" "32" "02" "112" "NaN2" NaN
The fundamental issue here is that JavaScript (ECMAScript) is a loosely typed language and it performs automatic type conversion on values to accommodate the operation being performed. Let’s see how this plays out with each of the above examples.
Example 1: 1 + “2” + “2” Outputs: “122” Explanation: The first operation to be performed in 1 + “2”. Since one of the operands (“2”) is a string, JavaScript assumes it needs to perform string concatenation and therefore converts the type of 1 to “1”, 1 + “2” yields “12”. Then, “12” + “2” yields “122”.
Example 2: 1 + +”2” + “2” Outputs: “32” Explanation: Based on order of operations, the first operation to be performed is +”2” (the extra + before the first “2” is treated as a unary operator). Thus, JavaScript converts the type of “2” to numeric and then applies the unary + sign to it (i.e., treats it as a positive number). As a result, the next operation is now 1 + 2 which of course yields 3. But then, we have an operation between a number and a string (i.e., 3 and “2”), so once again JavaScript converts the type of the numeric value to a string and performs string concatenation, yielding “32”.
Example 3: 1 + -“1” + “2” Outputs: “02” Explanation: The explanation here is identical to the prior example, except the unary operator is - rather than +. So “1” becomes 1, which then becomes -1 when the - is applied, which is then added to 1 yielding 0, which is then converted to a string and concatenated with the final “2” operand, yielding “02”.
Example 4: +”1” + “1” + “2” Outputs: “112” Explanation: Although the first “1” operand is typecast to a numeric value based on the unary + operator that precedes it, it is then immediately converted back to a string when it is concatenated with the second “1” operand, which is then concatenated with the final “2” operand, yielding the string “112”.
Example 5: “A” - “B” + “2” Outputs: “NaN2” Explanation: Since the - operator can not be applied to strings, and since neither “A” nor “B” can be converted to numeric values, “A” - “B” yields NaN which is then concatenated with the string “2” to yield “NaN2”.
Example 6: “A” - “B” + 2 Outputs: NaN Explanation: As exlained in the previous example, “A” - “B” yields NaN. But any operator applied to NaN with any other numeric operand will still yield NaN.
The following recursive code will cause a stack overflow if the array list is too large. How can you fix this and still retain the recursive pattern?
var list = readHugeList();
var nextListItem = function() {
var item = list.pop();
if (item) {
// process the list item...
nextListItem();
}
};
The potential stack overflow can be avoided by modifying the nextListItem function as follows:
var list = readHugeList();
var nextListItem = function() {
var item = list.pop();
if (item) {
// process the list item...
setTimeout( nextListItem, 0);
}
};
The stack overflow is eliminated because the event loop handles the recursion, not the call stack. When nextListItem runs, if item is not null, the timeout function (nextListItem) is pushed to the event queue and the function exits, thereby leaving the call stack clear. When the event queue runs its timed-out event, the next item is processed and a timer is set to again invoke nextListItem. Accordingly, the method is processed from start to finish without a direct recursive call, so the call stack remains clear, regardless of the number of iterations.
What is a “closure” in JavaScript? Provide an 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.
Here is an example:
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);
In the above example, variables from innerFunc, outerFunc, and the global namespace are all in scope in the innerFunc. The above code will therefore produce the following output:
outerArg = 123 innerArg = 456 outerVar = a innerVar = b globalVar = xyz
