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8.7 JS 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)
Try it yourself
This if statement returns true (maybe not as expected), because 10 is true:
var x = 0;
if (x = 10)
Try it yourself
This if statement returns true (maybe not as expected), because 10 is true:
var x = 0;
if (x = 10)
Try it yourself
This if statement returns false (maybe not as expected), because 0 is false:
var x = 0;
if (x = 0)
Try it yourself
An assignment always returns the value of the assignment.
Expecting Loose Comparison
In regular comparison, data type does not matter. This if statement returns true:
var x = 10;
var y = "10";
if (x == y)
Try it yourself
In strict comparison, data type does matter. This if statement returns false:
var x = 10;
var y = "10";
if (x === y)
Try it yourself
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");
}
Try it yourself
This case switch will not display an alert
var x = 10;
switch(x) {
case "10": alert("Hello");
}
Try it yourself
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"
Try it yourself
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"
Try it yourself
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
Try it yourself
To solve the problem above, it helps to multiply and divide:
var z = (x * 10 + y * 10) / 10; // z will be 0.3
Breaking a JavaScript String
JavaScript will allow you to break a statement into two lines:
var x =
"Hello World!";
But, breaking a statement in the middle of a string will not work:
var x = "Hello
World!";
You must use a "backslash" if you must break a statement in a string:
var x = "Hello \
World!";
Misplacing Semicolon
Because of a misplaced semicolon, this code block will execute regardless of the value of x:
if (x == 19);
{
// code block
}
Try it yourself
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:
function myFunction(a) {
var power = 10
return a * power
}
function myFunction(a) {
var power = 10;
return a * power;
}
JavaScript will also allow you to break a statement into two lines.
Because of this, example 3c will also return the same result:
function myFunction(a) {
var
power = 10;
return a * power;
}
But, what will happen if you break the return statement in two lines like this:
function myFunction(a) {
var
power = 10;
return
a * power;
}
The function will return undefined!
Why? Because JavaScript thought you meant:
function myFunction(a) {
var
power = 10;
return;
a * power;
}
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 does not support arrays with named indexes.
In JavaScript, arrays use numbered indexes :
var person = [];
person[0] = "John";
person[1] = "Doe";
person[2] = 46;
var x = person.length; // person.length will return 3
var y = person[0]; // person[0] will return "John"
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:
var person = [];
person["firstName"] = "John";
person["lastName"] = "Doe";
person["age"] = 46;
var x = person.length; // person.length will return 0
var y = person[0]; // person[0] will return undefined
Ending Definitions with a Comma
Trailing commas in object and array definition are legal in ECMAScript 5.
Object Example:
person = {firstName:"John", lastName:"Doe", age:46,}
Array Example:
points = [40, 100, 1, 5, 25, 10,];
WARNING !!
Internet Explorer 8 will crash.
JSON does not allow trailing commas.
JSON:
person = {"firstName":"John", "lastName":"Doe", "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 :
if (typeof myObj === "undefined")
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)
Try it yourself
JavaScript icons used in the buttons provided by ICONS8.COM . Smartphone icons created by Freepik - Flaticon
Example files created in this module:
JS Classes
JS Class Methods part 1
JS Class Methods part 2
JS Modules
Create Object from JSON String
JS Debuggers The console.log() Method
JS Debuggers The debugger Keyword
JS Style Guide - Line Length
JS Literal Constructors
JS Variables
JS Variables - Automatic Type Conversions
JS Variables - NaN
JS Comparisons
JS Functions - Parameter Defaults
End Your Switches with Defaults
JS String Objects cannot be safely compared
JS Objects cannot be compared
JS Common JavaScript Mistakes
JS Breaking a JavaScript Statement part 1
JS Breaking a JavaScript Statement part 2
JS Breaking a JavaScript Statement part 3
JS Breaking a Return Statement part a
JS Breaking a Return Statement part b
JS Breaking a Return Statement part c
JS Breaking a Return Statement part d
JS Breaking a Return Statement part 3f
JS Accessing Arrays with Named Indexes part 1
JS Accessing Arrays with Named Indexes part 2
JS Objects - Undefined is Not Null
JS Performance - Reduce DOM Access