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7 tips to handle undefined in JavaScript
When I started to learn JavaScript about eight years ago, a bizarre situation for me was the existence of both undefined and null that represent empty values. What is the clear difference between them? They both seem to define empty values, and moreover the comparison null == undefined evaluates to true.
Most of the modern languages like Ruby, Python or Java have a single null value (nilor null), which seems a reasonable approach.
In case of JavaScript, the interpreter returns undefined when accessing a variable or object property that is not yet initialized. For example:
let company;
company;
let person = { name: 'John Smith' };
person.age;
On the other side null represents a missing object reference. JavaScript by itself does not set variables or object properties to null.
Some native methods like String.prototype.match() can return null to denote a missing object. Take a look at the sample:
let array = null;
array;
let movie = { name: 'Starship Troopers', musicBy: null };
movie.musicBy;
'abc'.match(/[0-9]/);
Because JavaScript is very permissive, developers have the temptation to access uninitialized values. I'm guilty of such bad practice too.
Often such risky actions generate undefined related errors, ending the script lightning fast. Related common error messages are:
TypeError: 'undefined' is not a functionTypeError: Cannot read property '<prop-name>' of undefined- and alike type errors.
The JavaScript developer can understand the irony of this joke:
function undefined() {
}
To reduce the risk of such errors, you have to understand the cases when undefined is generated. And more important suppress its appearance and spread within your application, increasing the code durability.
Let's detail the exploration of undefined and its effect on code safety.
1. What is undefined
JavaScript has 6 primitive types:
- Boolean:
true or false - Number:
1, 6.7, 0xFF - String:
"Gorilla and banana" - Symbol:
Symbol("name") (starting ES2015) - Null:
null - Undefined:
undefined.
And a separated object type: {name: "Dmitri"}, ["apple", "orange"].
From these 6 primitive types, undefined is a special value with its own type Undefined. According to ECMAScript specification:
Undefined value primitive value is used when a variable has not been assigned a value.
The standard clearly defines that you will receive an undefined value when accessing uninitialized variables, non existing object properties, non existing array elements and alike. For instance:
let number;
number;
let movie = { name: 'Interstellar' };
movie.year;
let movies = ['Interstellar', 'Alexander'];
movies[3];
As the above example demonstrates, accessing:
- an uninitialized variable
number - a non-existing object property
movie.year - or a non-existing array element
movies[3]
are evaluated to undefined.
The ECMAScript specification defines the type of undefined value:
Undefined type is a type whose sole value is the undefined value.
In this sense, typeof operator returns 'undefined' string for an undefined value:
typeof undefined === 'undefined';
Of course typeof works nicely to verify whether a variable contains an undefinedvalue:
let nothing;
typeof nothing === 'undefined';
2. Common scenarios that create undefined
2.1 Uninitialized variable
A declared variable that is not yet assigned with a value (uninitialized) is by default undefined.
Plain and simple:
let myVariable;
myVariable;
myVariable is declared and not yet assigned with a value. Accessing the variable evaluates to undefined.
An efficient approach to solve the troubles of uninitialized variables is whenever possible assign an initial value. The less the variable exists in an uninitialized state, the better. Ideally you would assign a value right away after declaration const myVariable = 'Initial value', but this is not always possible.
Tip 1: Favor const, otherwise use let, but say goodbye to var
In my opinion, one of the best features of ECMAScript 2015 is the new way to declare variables using const and let. It is a big step forward that these declarations are block scoped (contrary to older function scoped var) and exist in a temporal dead zone until the declaration line.
When the variable receives a value once and forever, I recommend to use a constdeclaration. It creates an immutable binding.
One of the nice features of const is that you have to assign an initial value to the variable const myVariable = 'initial'. The variable is not exposed to the uninitialized state and to access undefined is simply not possible.
Let's check the function that verifies whether a word is a palindrome:
function isPalindrome(word) {
const length = word.length;
const half = Math.floor(length / 2);
for (let index = 0; index < half; index++) {
if (word[index] !== word[length - index - 1]) {
return false;
}
}
return true;
}
isPalindrome('madam');
isPalindrome('hello');
length and half variables are assigned with a value once. Seems reasonable to declare them as const, since these variables are not going to change.
If you need to rebind the variable (i.e. assign multiple times), apply a let declaration. Whenever possible assign an initial value to it right away, e.g. let index = 0.
What about the old school var? In terms of ES2015, my suggestion is stop using it at all.
var declaration problem is the variable hoisting in the entire function scope. You can declare a var variable somewhere at the end of the function scope, but still it can accessed before declaration: and you'll get an undefined.
function bigFunction() {
myVariable;
var myVariable = 'Initial value';
myVariable;
}
bigFunction();
myVariable is accessible and contains undefined even before the declaration line: var myVariable = 'Initial value'.
Contrary, a let (including const) variable cannot be accessed before the declaration line. It happens because the variable is in a temporal dead zone before the declaration. And that's nice, because you have less chances to access an undefined.
The above example updated with let (instead of var) throws a ReferenceError, because the variable in the temporal dead zone is not accessible.
function bigFunction() {
myVariable;
let myVariable = 'Initial value';
myVariable;
}
bigFunction();
Encouraging the usage of const for immutable bindings or let otherwise ensures a practice that exposes you much less to uninitialized variables.
Tip 2: Increase cohesion
Cohesion characterizes the degree to which the elements of a module (namespace, class, method, block of code) belong together. The measurement of the cohesion is usually described as high cohesion or low cohesion.
High cohesion is preferable because it suggests to design the elements of the module to focus solely on a single task. It makes the module:
- Focused and understandable: easier to understand what the module does
- Maintainable and easier to refactor: the change in the module affects fewer modules
- Reusable: being focusing on a single task, it makes the module easier to reuse
- Testable: you would easier test a module that's focused on a single task
High cohesion accompanied with loose coupling is the characteristic of a well designed system.
A code block by itself might be considered a small module. To profit from the benefits of high cohesion, you need to keep the variables as close as possible to the code block that uses them.
For instance, if a variable solely exists to form the logic of a block scope, then declare and allow the variable to live only within that block (using const or let declarations). Do not expose this variable to the outer block scope, since the outer block shouldn't care about this variable.
One classic example of unnecessary extended life of variables is the usage of for cycle inside a function:
function someFunc(array) {
var index, item, length = array.length;
for (index = 0; index < length; index++) {
item = array[index];
}
return 'some result';
}
index, item and length variables are declared at the beginning of function body. However they are used only near the end. So what is the problem with such approach?
All the way between the declaration at the top and the usage in for statement the variables index, item are uninitialized and exposed to undefined. They have an unreasonably long lifecycle in the entire function scope.
A better approach is to move these variables as close as possible to their usage place:
function someFunc(array) {
const length = array.length;
for (let index = 0; index < length; index++) {
const item = array[index];
}
return 'some result';
}
index and item variables exist only in the block scope of for statement. They don't have any meaning outside of for.
length variable is declared close to the source of its usage too.
Why is the modified version better than the initial one? Let's see:
- The variables are not exposed to uninitialized state, thus you have no risk of accessing
undefined - Moving the variables as close as possible to their usage place increases the code readability
- High cohesive chunks of code are easier to refactor and extract into separated functions when necessary
2.2 Accessing non-existing property
When accessing a non-existing object property, JavaScript returns undefined.
Let's demonstrate that in an example:
let favoriteMovie = {
title: 'Blade Runner'
};
favoriteMovie.actors;
favoriteMovie is an object with a single property title. Accessing a non-existing property actors using a property accessor favoriteMovie.actors is evaluated to undefined.
By itself accessing a non-existing property does not throw an error. The real problem appears when trying to get data from a non-existing property value. This is the most common undefined related trap, reflected in the well known error message TypeError: Cannot read property <prop> of undefined.
Let's slightly modify the previous code snippet to illustrate a TypeError throw:
let favoriteMovie = {
title: 'Blade Runner'
};
favoriteMovie.actors[0];
favoriteMovie does not have the property actors, so favoriteMovie.actors evaluates to undefined.
As result, accessing the first item of an undefined value using the expression favoriteMovie.actors[0] throws a TypeError.
The permissive nature of JavaScript that allows to access non-existing properties is a source of confusion: the property may be set, or may be not. The ideal way to bypass this problem is to restrict the object to have always defined the properties that it holds.
Unfortunately you often don't have control over the objects that you work with. Such objects may have different set of properties in diverse scenarios. So you have to handle all these scenarios manually.
Let's implement a function append(array, toAppend) that adds at the beginning and/or at the end of an array new elements. toAppend parameter accepts an object with properties:
first: element inserted at the beginning of arraylast: element inserted at the end of array.
The function returns a new array instance, without altering the original array (i.e. it's a pure function).
The first version of append(), a bit naive, may look like this:
function append(array, toAppend) {
const arrayCopy = array.slice();
if (toAppend.first) {
arrayCopy.unshift(toAppend.first);
}
if (toAppend.last) {
arrayCopy.push(toAppend.last);
}
return arrayCopy;
}
append([2, 3, 4], { first: 1, last: 5 });
append(['Hello'], { last: 'World' });
append([8, 16], { first: 4 });
Because toAppend object can omit first or last properties, it is obligatory to verify whether these properties exist in toAppend.
A property accessor evaluates to undefined if the property does not exist. The first temptation to check weather first or last properties are present is to verify them against undefined. Let's do the verification in conditionals if(toAppend.first){} and if(toAppend.last){}...
Not so fast. There is a serious drawback in this approach. undefined, as well as false, null, 0, NaN and '' are falsy values.
In the current implementation of append(), the function doesn't allow to insert falsy elements:
append([10], { first: 0, last: false });
0 and false are falsy. Because if(toAppend.first){} and if(toAppend.last){} actually compare against falsy, these elements are not inserted into the array. The function returns the initial array [10] without modifications.
The tips that follow explain how to correctly check the property existence.
Tip 3: Check the property existence
Fortunately, JavaScript offers a bunch of ways to determine if the object has a specific property:
obj.prop !== undefined: compare against undefined directlytypeof obj.prop !== 'undefined': verify the property value typeobj.hasOwnProperty('prop'): verify whether the object has an own property'prop' in obj: verify whether the object has an own or inherited property
My recommendation is to use in operator. It has a short and sweet syntax. in operator presence suggests a clear intent of checking whether an object has a specific property, without accessing the actual property value.
obj.hasOwnProperty('prop') is a nice solution too. It's slightly longer than in operator and verifies only in object's own properties.
The 2 ways that involve comparing with undefined might work... But it seems to me that obj.prop !== undefined and typeof obj.prop !== 'undefined' look verbose and weird, and expose to a suspicions path of dealing directly with undefined.
Let's improve append(array, toAppend) function using in operator:
function append(array, toAppend) {
const arrayCopy = array.slice();
if ('first' in toAppend) {
arrayCopy.unshift(toAppend.first);
}
if ('last' in toAppend) {
arrayCopy.push(toAppend.last);
}
return arrayCopy;
}
append([2, 3, 4], { first: 1, last: 5 });
append([10], { first: 0, last: false });
'first' in toAppend (and 'last' in toAppend) is true whether the corresponding property exists, false otherwise.
The usage of in operator fixes the problem with inserting falsy elements 0 and false. Now, adding these elements at the beginning and at the end of [10] produces the expected result [0, 10, false].
Tip 4: Destructuring to access object properties
When accessing an object property, sometimes it's necessary to indicate a default value if the property does not exist.
You might use in accompanied with ternary operator to accomplish that:
const object = { };
const prop = 'prop' in object ? object.prop : 'default';
prop;
The usage of ternary operator syntax becomes daunting when the number of properties to check increases. For each property you have to create a new line of code to handle the defaults, increasing an ugly wall of similar looking ternary operators.
In order to use a more elegant approach, let's get familiar with a great ES2015 feature called object destructuring.
Object destructuring allows inline extraction of object property values directly into variables, and setting a default value if the property does not exist. A convenient syntax to avoid dealing directly with undefined.
Indeed, the property extraction now looks short and meaningful:
const object = { };
const { prop = 'default' } = object;
prop;
To see things in action, let's define an useful function that wraps a string in quotes.
quote(subject, config) accepts the first argument as the string to be wrapped. The second argument config is an object with the properties:
char: the quote char, e.g. ' (single quote) or " (double quote). Defaults to ".skipIfQuoted: the boolean value to skip quoting if the string is already quoted. Defaults to true.
Applying the benefits of the object destructuring, let's implement quote():
function quote(str, config) {
const { char = '"', skipIfQuoted = true } = config;
const length = str.length;
if (skipIfQuoted
&& str[0] === char
&& str[length - 1] === char) {
return str;
}
return char + str + char;
}
quote('Hello World', { char: '*' });
quote('"Welcome"', { skipIfQuoted: true });
const { char = '"', skipIfQuoted = true } = config destructuring assignment in one line extracts the properties char and skipIfQuoted from config object.
If some properties are not available in config object, the destructuring assignment sets the default values: '"' for char and false for skipIfQuoted.
Fortunately, the function still has room for improvements.
Let's move the destructuring assignment right into the parameters section. And set a default value (an empty object { }) for config parameter, to skip the second argument when default settings are enough.
function quote(str, { char = '"', skipIfQuoted = true } = {}) {
const length = str.length;
if (skipIfQuoted
&& str[0] === char
&& str[length - 1] === char) {
return str;
}
return char + str + char;
}
quote('Hello World', { char: '*' });
quote('Sunny day');
Notice that a destructuring assignment replaces the config parameter in function's signature. I like that: quote() becomes one line shorter.
= {} on the right side of destructuring assignment ensures that an empty object is used if the second argument is not specified at all quote('Sunny day').
Object destructuring is a powerful feature that handles efficiently the extraction of properties from objects. I like the possibility to specify a default value to be returned when the accessed property doesn't exist. As result, you avoid undefined and the problem related to handling it.
Tip 5: Fill the object with default properties
If there is no need to create variables for every property like the destructuring assignment does, the object that misses some properties can be filled with default values.
The ES2015 Object.assign(target, source1, source2, ...) copies the values of all enumerable own properties from one or more source objects into the target object. The function returns the target object.
For instance, you need to access the properties of unsafeOptions object, which not always contains its full set of properties.
To avoid undefined when accessing a non-existing property from unsafeOptions, let's make some adjustments:
- Define an object
defaults that holds the default property values - Call
Object.assign({ }, defaults, unsafeOptions) to build a new object options. The new object receives all properties from unsafeOptions, but the missing ones are taken from defaults.
const unsafeOptions = {
fontSize: 18
};
const defaults = {
fontSize: 16,
color: 'black'
};
const options = Object.assign({}, defaults, unsafeOptions);
options.fontSize;
options.color;
unsafeOptions contains only fontSize property. defaults object defines the default values for properties fontSize and color.
Object.assign() takes the first argument as a target object {}. The target object receives the value of fontSize property from unsafeOptions source object. And the value of color property from defaults source object, because unsafeOptions doesn't contain color.
The order in which the source objects are enumerated does matter: later source object properties overwrite earlier ones.
You are now safe to access any property of options object, including options.colorthat wasn't available in unsafeOptions initially.
Fortunately exists an easier and lighter way to fill the object with default properties. I recommend to use a new JavaScript feature (now at stage 3) that allows to spread properties in object initializers.
Instead of Object.assign() invocation, use the object spread syntax to copy into target object all own and enumberable properties from source objects:
const unsafeOptions = {
fontSize: 18
};
const defaults = {
fontSize: 16,
color: 'black'
};
const options = {
...defaults,
...unsafeOptions
};
options.fontSize;
options.color;
The object initializer spreads properties from defaults and unsafeOptions source objects. The order in which the source objects are specified is important: later source object properties overwrite earlier ones.
Filling an incomplete object with default property values is an efficient strategy to make your code safe and durable. No matter the situation, the object always contains the full set of properties: and undefined cannot be generated.
2.3 Function parameters
The function parameters implicitly default to undefined.
Normally a function that is defined with a specific number of parameters should be invoked with the same number of arguments. In such case the parameters get the values you expect:
function multiply(a, b) {
a;
b;
return a * b;
}
multiply(5, 3);
The invocation multiply(5, 3) makes the parameters a and b receive the corresponding 5 and 3 values. The multiplication is calculated as expected: 5 * 3 = 15.
What does happen when you omit an argument on invocation? The parameter inside the function becomes undefined.
Let's slightly modify the previous example by calling the function with just one argument:
function multiply(a, b) {
a;
b;
return a * b;
}
multiply(5);
function multiply(a, b) { } is defined with two parameters a and b.
The invocation multiply(5) is performed with a single argument: as result aparameter is 5, but b parameter is undefined.
Tip 6: Use default parameter value
Sometimes a function does not require the full set of arguments on invocation. You can simply set defaults for parameters that don't have a value.
Recalling the previous example, let's make an improvement. If b parameter is undefined, it gets assigned with a default value of 2:
function multiply(a, b) {
if (b === undefined) {
b = 2;
}
a;
b;
return a * b;
}
multiply(5);
The function is invoked with a single argument multiply(5). Initially a parameter is 2and b is undefined.
The conditional statement verifies whether b is undefined. If it happens, b = 2assignment sets a default value.
While the provided way to assign default values works, I don't recommend comparing directly against undefined. It's verbose and looks like a hack.
A better approach is to use the ES2015 default parameters feature. It's short, expressive and no direct comparisons with undefined.
Modifying the previous example with a default parameter for b indeed looks great:
function multiply(a, b = 2) {
a;
b;
return a * b;
}
multiply(5);
multiply(5, undefined);
b = 2 in the function signature makes sure that if b is undefined, the parameter is defaulted to 2.
ES2015 default parameters feature is intuitive and expressive. Always use it to set default values for optional parameters.
2.4 Function return value
Implicitly, without return statement, a JavaScript function returns undefined.
In JavaScript a function that doesn't have any return statements implicitly returns an undefined:
function square(x) {
const res = x * x;
}
square(2);
square() function does not return any computation results. The function invocation result is undefined.
The same situation happens when return statement is present, but without an expression nearby:
function square(x) {
const res = x * x;
return;
}
square(2);
return; statement is executed, but it doesn't return any expression. The invocation result is also undefined.
Of course, indicating near return the expression to be returned works as expected:
function square(x) {
const res = x * x;
return res;
}
square(2);
Now the function invocation is evaluated to 4, which is 2 squared.
Tip 7: Don't trust the automatic semicolon insertion
The following list of statements in JavaScript must end with semicolons (;):
- empty statement
let, const, var, import, export declarations- expression statement
debugger statementcontinue statement, break statementthrow statementreturn statement
If you use one of the above statements, be sure to indicate a semicolon at the end:
function getNum() {
let num = 1;
return num;
}
getNum();
At the end of both let declaration and return statement an obligatory semicolon is written.
What happens when you don't want to indicate these semicolons? For instance to reduce the size of the source file.
In such situation ECMAScript provides an Automatic Semicolon Insertion (ASI) mechanism, which inserts for you the missing semicolons.
Being helped by ASI, you can remove the semicolons from the previous example:
function getNum() {
let num = 1
return num
}
getNum()
The above text is a valid JavaScript code. The missing semicolons are automatically inserted for you.
At first sight, it looks pretty promising. ASI mechanism lets you skip the unnecessary semicolons. You can make the JavaScript code smaller and easier to read.
There is one small, but annoying trap created by ASI. When a newline stands between return and the returned expression return \n expression, ASI automatically inserts a semicolon before the newline return; \n expression.
What does mean inside a function to have return; statement? The function returns undefined. If you don't know in details the mechanism of ASI, the unexpectedly returned undefined is misleading.
For instance, let's study the returned value of getPrimeNumbers() invocation:
function getPrimeNumbers() {
return
[ 2, 3, 5, 7, 11, 13, 17 ]
}
getPrimeNumbers()
Between return statement and the array literal expression exists a new line. JavaScript automatically inserts a semicolon after return, interpreting the code as follows:
function getPrimeNumbers() {
return;
[ 2, 3, 5, 7, 11, 13, 17 ];
}
getPrimeNumbers();
The statement return; makes the function getPrimeNumbers() to return undefinedinstead of the expected array.
The problem is solved by removing the newline between return and array literal:
function getPrimeNumbers() {
return [
2, 3, 5, 7, 11, 13, 17
];
}
getPrimeNumbers();
My recommendation is to study how exactly Automatic Semicolon Insertion works to avoid such situations.
Of course, never put a newline between return and the returned expression.
2.5 void operator
void expression evaluates the expression and returns undefined no matter the result of evaluation.
void 1;
void (false);
void {name: 'John Smith'};
void Math.min(1, 3);
One use case of void operator is to suppress expression evaluation to undefined, relying on some side-effect of the evaluation.
3. undefined in arrays
You get undefined when accessing an array element with an out of bounds index.
const colors = ['blue', 'white', 'red'];
colors[5];
colors[-1];
colors array has 3 elements, thus valid indexes are 0, 1 and 2.
Because there are no array elements at indexes 5 and -1, the accessors colors[5] andcolors[-1] are undefined.
In JavaScript you might encounter so called sparse arrays. Theses are arrays that have gaps, i.e. at some indexes no elements are defined.
When a gap (aka empty slot) is accessed inside a sparse array, you also get an undefined.
The following example generates sparse arrays and tries to access their empty slots:
const sparse1 = new Array(3);
sparse1;
sparse1[0];
sparse1[1];
const sparse2 = ['white', ,'blue']
sparse2;
sparse2[1];
sparse1 is created corresponding by invoking an Array constructor with a numeric first argument. It has 3 empty slots.
sparse2 is created with an array literal with the missing second element.
In any of these sparse arrays accessing an empty slot evaluates to undefined.
When working with arrays, to escape catching undefined, be sure to use valid array indexes and avoid at all creating sparse arrays.
4. Difference between undefined and null
A reasonable question appears: what is the main difference between undefined and null? Both special values imply an empty state.
The main difference is that undefined represents a value of a variable that wasn't yet initialized, while null represents an intentional absence of an object.
Let's explore the difference in some examples.
The variable number is defined, however is not assigned with an initial value:
let number;
number;
number variable is undefined, which clearly indicates an uninitialized variable.
The same uninitialized concept happens when a non-existing object property is accessed:
const obj = { firstName: 'Dmitri' };
obj.lastName;
Because lastName property does not exist in obj, JavaScript correctly evaluates obj.lastName to undefined.
In other cases you know that a variable expects to hold an object or a function to return an object. But for some reason you can't instantiate the object. In such case null is a meaningful indicator of a missing object.
For example, clone() is a function that clones a plain JavaScript object. The function is expected to return an object:
function clone(obj) {
if (typeof obj === 'object' && obj !== null) {
return Object.assign({}, obj);
}
return null;
}
clone({name: 'John'});
clone(15);
clone(null);
However clone() might be invoked with a non-object argument: 15 or null (or generally a primitive value, null or undefined). In such case the function cannot create a clone, so it returns null - the indicator of a missing object.
typeof operator makes the distinction between the two values:
typeof undefined;
typeof null;
The strict quality operator === correctly differentiates undefined from null:
let nothing = undefined;
let missingObject = null;
nothing === missingObject;
5. Conclusion
The existence of undefined is a consequence of JavaScript's permissive nature that allows the usage of:
- uninitialized variables
- non-existing object properties or methods
- out of bounds indexes to access array elements
- the invocation result of a function that returns nothing
Mostly comparing directly against undefined is a bad practice, because you probably rely on a permitted but discouraged practice mentioned above.
An efficient strategy is to reduce at minimum the appearance of undefined keyword in your code. In the meantime, always remember about its potential appearance in a surprising way, and prevent that by applying beneficial habits such as:
- reduce the usage of uninitialized variables
- make the variables lifecycle short and close to the source of their usage
- whenever possible assign an initial value to variables
- favor
const, otherwise use let - use default values for insignificant function parameters
- verify the properties existence or fill the unsafe objects with default properties
- avoid the usage of sparse arrays
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