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:
let x = 0;if (x == 10)
Try it Yourself »
This if statement returns true (maybe not
as expected), because 10 is
true:
let x = 0;if (x = 10)
Try it Yourself »
This if statement returns false (maybe not
as expected), because 0 is
false:
let 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:
let x = 10;
let y = "10";if (x == y)
Try it Yourself »
In strict comparison, data type does matter. This if statement returns false:
let x = 10;
let 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:
let x = 10;
switch(x) { case 10: alert("Hello");
}
Try it Yourself »
This case switch will not display an alert:
let 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:
let x = 10;
x = 10 + 5; //
Now x is 15
let y = 10;
y += "5";
// Now y is "105"
Try it Yourself »
When adding two variables, it can be difficult to anticipate the result:
let x = 10;
let y = 5;
let z = x + y; // Now z is 15
let x = 10;
let y = "5";
let z = x + y; // Now 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:
let x = 0.1;
let y = 0.2;let 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:
Example
let z = (x * 10 + y * 10) / 10; // z will be 0.3
Try it Yourself »
Breaking a JavaScript String
JavaScript will allow you to break a statement into two lines:
Example 1
let x ="Hello World!";
Try it Yourself »
But, breaking a statement in the middle of a string will not work:
Example 2
let x = "HelloWorld!";
Try it Yourself »
You must use a "backslash" if you must break a statement in a string:
Example 3
let x = "Hello \World!";
Try it Yourself »
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:
Example 1
function myFunction(a) {
let power = 10 return a * power}
Try it Yourself »
Example 2
function myFunction(a) {
let power = 10; return a * power;}
Try it Yourself »
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) {
let power = 10; return a * power;}
Try it Yourself »
But, what will happen if you break the return statement in two lines like
this:
Example 4
function myFunction(a) {
let power = 10; return a * power;}
Try it Yourself »
The function will return undefined!
Why? Because JavaScript thought you meant:
Example 5
function myFunction(a) {
let power = 10; return; a * power;}
Try it Yourself »
Explanation
If a statement is incomplete like:
let
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:
Example
const person = [];
person[0] = "John";
person[1] = "Doe";
person[2] = 46;person.length;
// person.length will return 3person[0];
// person[0] will return "John"
Try it Yourself »
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:
const person = [];
person["firstName"] = "John";
person["lastName"] = "Doe";
person["age"] = 46;person.length; // person.length will
return 0person[0];
// person[0] will return undefined
Try it Yourself »
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:
Example:
if (typeof myObj === "undefined")
Try it Yourself »
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 »
★
+1
Reference: https://www.w3schools.com/js/js_mistakes.asp