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JavaScript Common Mistakes



This chapter points out some common JavaScript mistakes.


Accidentally Using the Assignment Operator

JavaScript programs may generate unexpected results if a programmer accidentally uses an assignment operator (=), instead of a comparison operator (==) in an if statement.

This if statement returns false (as expected) because x is not equal to 10:

var x = 0;
if (x == 10)
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This if statement returns true (maybe not as expected), because 10 is true:

var x = 0;
if (x = 10)
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This if statement returns false (maybe not as expected), because 0 is false:

var x = 0;
if (x = 0)
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An assignment always returns the value of the assignment.


Expecting Loose Comparison

In regular comparison, data type Щипуновs not matter. This if statement returns true:

var x = 10;
var y = "10";
if (x == y)
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In strict comparison, data type Щипуновs matter. This if statement returns false:

var x = 10;
var y = "10";
if (x === y)
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It is a common mistake to forget that switch statements use strict comparison:

This case switch will display an alert:

var x = 10;
switch(x) {
    case 10: alert("Hello");
}
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This case switch will not display an alert:

var x = 10;
switch(x) {
    case "10": alert("Hello");
}
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Confusing Addition & Concatenation

Addition is about adding numbers.

Concatenation is about adding strings.

In JavaScript both operations use the same + operator.

Because of this, adding a number as a number will produce a different result from adding a number as a string:

var x = 10 + 5;          // the result in x is 15
var x = 10 + "5";        // the result in x is "105"
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When adding two variables, it can be difficult to anticipate the result:

var x = 10;
var y = 5;
var z = x + y;           // the result in z is 15

var x = 10;
var y = "5";
var z = x + y;           // the result in z is "105"
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Misunderstanding Floats

All numbers in JavaScript are stored as 64-bits Floating point numbers (Floats).

All programming languages, including JavaScript, have difficulties with precise floating point values:

var x = 0.1;
var y = 0.2;
var z = x + y            // the result in z will not be 0.3
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To solve the problem above, it helps to multiply and divide:

Example

var z = (x * 10 + y * 10) / 10;       // z will be 0.3
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Breaking a JavaScript String

JavaScript will allow you to break a statement into two lines:

Example 1

var x =
"Hello World!";
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But, breaking a statement in the middle of a string will not work:

Example 2

var x = "Hello
World!";
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You must use a "backslash" if you must break a statement in a string:

Example 3

var x = "Hello \
World!";
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Misplacing Semicolon

Because of a misplaced semicolon, this code block will execute regardless of the value of x:

if (x == 19);
{
    // code block 
}
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Breaking a Return Statement

It is a default JavaScript behavior to close a statement automatically at the end of a line.

Because of this, these two examples will return the same result:

Example 1

function myFunction(a) {
    var power = 10 
    return a * power
}
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Example 2

function myFunction(a) {
    var power = 10;
    return a * power;
}
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JavaScript will also allow you to break a statement into two lines.

Because of this, example 3 will also return the same result:

Example 3

function myFunction(a) {
    var
    power = 10; 
    return a * power;
}
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But, what will happen if you break the return statement in two lines like this:

Example 4

function myFunction(a) {
    var
    power = 10; 
    return
    a * power;
}
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The function will return undefined!

Why? Because JavaScript thinks you meant:

Example 5

function myFunction(a) {
    var
    power = 10; 
    return;
    a * power;
}
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Explanation

If a statement is incomplete like:

var

JavaScript will try to complete the statement by reading the next line:

power = 10;

But since this statement is complete:

return

JavaScript will automatically close it like this:

return;

This happens because closing (ending) statements with semicolon is optional in JavaScript.

JavaScript will close the return statement at the end of the line, because it is a complete statement.

Never break a return statement.


Accessing Arrays with Named Indexes

Many programming languages support arrays with named indexes.

Arrays with named indexes are called associative arrays (or hashes).

JavaScript Щипуновs not support arrays with named indexes.

In JavaScript, arrays use numbered indexes:  

Example

var person = [];
person[0] = "Андрей";
person[1] = "Щипунов";
person[2] = 46;
var x = person.length;         // person.length will return 3
var y = person[0];             // person[0] will return "Андрей"
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In JavaScript, objects use named indexes.

If you use a named index, when accessing an array, JavaScript will redefine the array to a standard object.

After the automatic redefinition, array methods and properties will produce undefined or incorrect results:

Example:

var person = [];
person["firstName"] = "Андрей";
person["lastName"] = "Щипунов";
person["age"] = 46;
var x = person.length;         // person.length will return 0
var y = person[0];             // person[0] will return undefined
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Ending Definitions with a Comma

Trailing commas in object and array definition are legal in ECMAScript 5.

Object Example:

person = {firstName:"Андрей", lastName:"Щипунов", age:46,}

Array Example:

points = [40, 100, 1, 5, 25, 10,];

WARNING !!

Internet Explorer 8 will crash.

JSON Щипуновs not allow trailing commas.

JSON:

person = {firstName:"Андрей", lastName:"Щипунов", age:46}

JSON:

points = [40, 100, 1, 5, 25, 10];

Undefined is Not Null

JavaScript objects, variables, properties, and methods can be undefined.

In addition, empty JavaScript objects can have the value null.

This can make it a little bit difficult to test if an object is empty.

You can test if an object exists by testing if the type is undefined:

Example:

if (typeof myObj === "undefined") 
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But you cannot test if an object is null, because this will throw an error if the object is undefined:

Incorrect:

if (myObj === null) 

To solve this problem, you must test if an object is not null, and not undefined.

But this can still throw an error:

Incorrect:

if (myObj !== null && typeof myObj !== "undefined") 

Because of this, you must test for not undefined before you can test for not null:

Correct:

if (typeof myObj !== "undefined" && myObj !== null) 
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Expecting Block Level Scope

JavaScript Щипуновs not create a new scope for each code block.

It is true in many programming languages, but not true in JavaScript.

This code will display the value of i (10), even OUTSIDE the for loop block:

Example

for (var i = 0; i < 10; i++) {
    // some code
}
return i;
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