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 does 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 does 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"
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
if (z == 0.3)
// this if test will fail
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To solve the problem above, it helps to multiply and divide:
Breaking a JavaScript String
JavaScript will allow you to break a statement into two lines:
But, breaking a statement in the middle of a string will not work:
You must use a "backslash" if you must break a statement in a string:
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:
JavaScript will also allow you to break a statement into two lines.
Because of this, example 3 will also return the same result:
But, what will happen if you break the return statement in two lines like this:
The function will return undefined!
Why? Because JavaScript thinks you meant:
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:
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
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"
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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"] = "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
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Ending an Array Definition with a Comma
Incorrect:
points = [40, 100, 1, 5, 25, 10,];
Some JSON and JavaScript engines will fail, or behave unexpectedly.
Correct:
points = [40, 100, 1, 5, 25, 10];Ending an Object Definition with a Comma
Incorrect:
person = {firstName:"John", lastName:"Doe", age:46,}
Some JSON and JavaScript engines will fail, or behave unexpectedly.
Correct:
person = {firstName:"John", lastName:"Doe", age:46}
Undefined is Not Null
With JavaScript, null is for objects, undefined is for variables, properties, and methods.
To be null, an object has to be defined, otherwise it will be undefined.
If you want to test if an object exists, this will throw an error if the object is undefined:
Incorrect:
if (myObj !== null && typeof myObj !== "undefined")
Because of this, you must test typeof() first:
Correct:
if (typeof myObj !== "undefined" && myObj !== null)
Expecting Block Level Scope
JavaScript does not create a new scope for each code block.
It is true in many programming languages, but not true in JavaScript.
It is a common mistake, among new JavaScript developers, to believe that this code returns undefined: