What is a potential pitfall with using typeof bar === "object" to determine if bar is an object? How can this pitfall be avoided?

Although typeof bar === "object" is a reliable way of checking if bar is an object, the surprising gotcha in JavaScript is that null is also considered an object!

Therefore, the following code will, to the surprise of most developers, log true (not false) to the console:

var bar = null;
console.log(typeof bar === "object");  // logs true!

As long as one is aware of this, the problem can easily be avoided by also checking if bar is null:

console.log((bar !== null) && (typeof bar === "object"));  // logs false

To be entirely thorough in our answer, there are two other things worth noting:

First, the above solution will return false if bar is a function. In most cases, this is the desired behavior, but in situations where you want to also return true for functions, you could amend the above solution to be:

console.log((bar !== null) && ((typeof bar === "object") 
|| (typeof bar === "function")));

Second, the above solution will return true if bar is an array (e.g., if var bar = [];). In most cases, this is the desired behavior, since arrays are indeed objects, but in situations where you want to also false for arrays, you could amend the above solution to be:

console.log((bar !== null) && (typeof bar === "object") 
&& (toString.call(bar) !== "[object Array]"));

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();

The above code will output the following to the console:

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, 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 property names or parameter values. Strict mode throws an error when it detects a duplicate named property in an object (e.g., var object = {foo: "bar", foo: "baz"};) or 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.


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”
};
}

Surprisingly, these two functions will not return the same thing. Rather:

console.log("foo1 returns:");
console.log(foo1());
console.log("foo2 returns:");
console.log(foo2());

will yield:

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 is NaN? What is its type? How can you reliably test if a value is equal to NaN?

The NaN property represents a value that is “not a number”. This special value results from an operation that could not be performed either because one of the operands was non-numeric (e.g., "abc" / 4), or because the result of the operation is non-numeric (e.g., an attempt to divide by zero).

While this seems straightforward enough, there are a couple of somewhat surprising characteristics of NaN that can result in hair-pulling bugs if one is not aware of them.

For one thing, although NaN means “not a number”, its type is, believe it or not, Number:

console.log(typeof NaN === "number"); // logs "true"

A semi-reliable way to test whether a number is equal to NaN is with the built-in function isNaN()


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

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);
})();

The values will be logged in the following order:

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 80 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

There are (at least) two ways to do this:

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.


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