Defining and Instantiating Structs
Structs are similar to tuples, which were discussed in “The Tuple Type” section. Like tuples, the pieces of a struct can be different types. Unlike with tuples, you’ll name each piece of data so it’s clear what the values mean. As a result of these names, structs are more flexible than tuples: you don’t have to rely on the order of the data to specify or access the values of an instance.
To define a struct, we enter the keyword struct and name the entire struct. A
struct’s name should describe the significance of the pieces of data being
grouped together. Then, inside curly brackets, we define the names and types of
the pieces of data, which we call fields. For example, Listing 5-1 shows a
struct that stores information about a user account.
struct User { active: bool, username: String, email: String, sign_in_count: u64, } fn main() {}
Listing 5-1: A User struct definition
To use a struct after we’ve defined it, we create an instance of that struct
by specifying concrete values for each of the fields. We create an instance by
stating the name of the struct and then add curly brackets containing key: value pairs, where the keys are the names of the fields and the values are the
data we want to store in those fields. We don’t have to specify the fields in
the same order in which we declared them in the struct. In other words, the
struct definition is like a general template for the type, and instances fill
in that template with particular data to create values of the type. For
example, we can declare a particular user as shown in Listing 5-2.
struct User { active: bool, username: String, email: String, sign_in_count: u64, } fn main() { let user1 = User { email: String::from("someone@example.com"), username: String::from("someusername123"), active: true, sign_in_count: 1, }; }
Listing 5-2: Creating an instance of the User
struct
To get a specific value from a struct, we can use dot notation. If we wanted
just this user’s email address, we could use user1.email wherever we wanted
to use this value. If the instance is mutable, we can change a value by using
the dot notation and assigning into a particular field. Listing 5-3 shows how
to change the value in the email field of a mutable User instance.
struct User { active: bool, username: String, email: String, sign_in_count: u64, } fn main() { let mut user1 = User { email: String::from("someone@example.com"), username: String::from("someusername123"), active: true, sign_in_count: 1, }; user1.email = String::from("anotheremail@example.com"); }
Listing 5-3: Changing the value in the email field of a
User instance
Note that the entire instance must be mutable; Rust doesn’t allow us to mark only certain fields as mutable. As with any expression, we can construct a new instance of the struct as the last expression in the function body to implicitly return that new instance.
Listing 5-4 shows a build_user function that returns a User instance with
the given email and username. The active field gets the value of true, and
the sign_in_count gets a value of 1.
struct User { active: bool, username: String, email: String, sign_in_count: u64, } fn build_user(email: String, username: String) -> User { User { email: email, username: username, active: true, sign_in_count: 1, } } fn main() { let user1 = build_user( String::from("someone@example.com"), String::from("someusername123"), ); }
Listing 5-4: A build_user function that takes an email
and username and returns a User instance
It makes sense to name the function parameters with the same name as the struct
fields, but having to repeat the email and username field names and
variables is a bit tedious. If the struct had more fields, repeating each name
would get even more annoying. Luckily, there’s a convenient shorthand!
Using the Field Init Shorthand when Variables and Fields Have the Same Name
Because the parameter names and the struct field names are exactly the same in
Listing 5-4, we can use the field init shorthand syntax to rewrite
build_user so that it behaves exactly the same but doesn’t have the
repetition of email and username, as shown in Listing 5-5.
struct User { active: bool, username: String, email: String, sign_in_count: u64, } fn build_user(email: String, username: String) -> User { User { email, username, active: true, sign_in_count: 1, } } fn main() { let user1 = build_user( String::from("someone@example.com"), String::from("someusername123"), ); }
Listing 5-5: A build_user function that uses field init
shorthand because the email and username parameters have the same name as
struct fields
Here, we’re creating a new instance of the User struct, which has a field
named email. We want to set the email field’s value to the value in the
email parameter of the build_user function. Because the email field and
the email parameter have the same name, we only need to write email rather
than email: email.
Creating Instances From Other Instances With Struct Update Syntax
It’s often useful to create a new instance of a struct that uses most of an old instance’s values but changes some. You can do this using struct update syntax.
First, Listing 5-6 shows how we create a new User instance in user2 without
the update syntax. We set a new value for email but otherwise use the same
values from user1 that we created in Listing 5-2.
struct User { active: bool, username: String, email: String, sign_in_count: u64, } fn main() { let user1 = User { email: String::from("someone@example.com"), username: String::from("someusername123"), active: true, sign_in_count: 1, }; let user2 = User { active: user1.active, username: user1.username, email: String::from("another@example.com"), sign_in_count: user1.sign_in_count, }; }
Listing 5-6: Creating a new User instance using one of
the values from user1
Using struct update syntax, we can achieve the same effect with less code, as
shown in Listing 5-7. The syntax .. specifies that the remaining fields not
explicitly set should have the same value as the fields in the given instance.
struct User { active: bool, username: String, email: String, sign_in_count: u64, } fn main() { let user1 = User { email: String::from("someone@example.com"), username: String::from("someusername123"), active: true, sign_in_count: 1, }; let user2 = User { email: String::from("another@example.com"), ..user1 }; }
Listing 5-7: Using struct update syntax to set a new
email value for a User instance but use the rest of the values from
user1
The code in Listing 5-7 also creates an instance in user2 that has a
different value for email but has the same values for the username,
active, and sign_in_count fields from user1. The ..user1 must come last
to specify that any remaining fields should get their values from the
corresponding fields in user1, but we can choose to specify values for as
many fields as we want in any order, regardless of the order of the fields in
the struct’s definition.
Note that the struct update syntax is like assignment with = because it moves
the data, just as we saw in the “Ways Variables and Data Interact: Move”
section. In this example, we can no longer use user1
after creating user2 because the String in the username field of user1
was moved into user2. If we had given user2 new String values for both
email and username, and thus only used the active and sign_in_count
values from user1, then user1 would still be valid after creating user2.
The types of active and sign_in_count are types that implement the Copy
trait, so the behavior we discussed in the “Stack-Only Data: Copy”
section would apply.
Using Tuple Structs without Named Fields to Create Different Types
You can also define structs that look similar to tuples, called tuple structs. Tuple structs have the added meaning the struct name provides but don’t have names associated with their fields; rather, they just have the types of the fields. Tuple structs are useful when you want to give the whole tuple a name and make the tuple be a different type from other tuples, and naming each field as in a regular struct would be verbose or redundant.
To define a tuple struct, start with the struct keyword and the struct name
followed by the types in the tuple. For example, here are definitions and
usages of two tuple structs named Color and Point:
fn main() { struct Color(i32, i32, i32); struct Point(i32, i32, i32); let black = Color(0, 0, 0); let origin = Point(0, 0, 0); }
Note that the black and origin values are different types, because they’re
instances of different tuple structs. Each struct you define is its own type,
even though the fields within the struct have the same types. For example, a
function that takes a parameter of type Color cannot take a Point as an
argument, even though both types are made up of three i32 values. Otherwise,
tuple struct instances behave like tuples: you can destructure them into their
individual pieces, you can use a . followed by the index to access an
individual value, and so on.
Unit-Like Structs Without Any Fields
You can also define structs that don’t have any fields! These are called
unit-like structs because they behave similarly to (), the unit type that
we mentioned in “The Tuple Type” section. Unit-like
structs can be useful in situations in which you need to implement a trait on
some type but don’t have any data that you want to store in the type itself.
We’ll discuss traits in Chapter 10. Here’s an example of declaring and
instantiating a unit struct named AlwaysEqual:
fn main() { struct AlwaysEqual; let subject = AlwaysEqual; }
To define AlwaysEqual, we use the struct keyword, the name we want, then a
semicolon. No need for curly brackets or parentheses! Then we can get an
instance of AlwaysEqual in the subject variable in a similar way: using the
name we defined, without any curly brackets or parentheses. Imagine we’ll be
implementing behavior for this type that every instance is always equal to
every instance of every other type, perhaps to have a known result for testing
purposes. We wouldn’t need any data to implement that behavior! You’ll see in
Chapter 10 how to define traits and implement them on any type, including
unit-like structs.
Ownership of Struct Data
In the
Userstruct definition in Listing 5-1, we used the ownedStringtype rather than the&strstring slice type. This is a deliberate choice because we want instances of this struct to own all of its data and for that data to be valid for as long as the entire struct is valid.It’s possible for structs to store references to data owned by something else, but to do so requires the use of lifetimes, a Rust feature that we’ll discuss in Chapter 10. Lifetimes ensure that the data referenced by a struct is valid for as long as the struct is. Let’s say you try to store a reference in a struct without specifying lifetimes, like this, which won’t work:
Filename: src/main.rs
struct User { username: &str, email: &str, sign_in_count: u64, active: bool, } fn main() { let user1 = User { email: "someone@example.com", username: "someusername123", active: true, sign_in_count: 1, }; }The compiler will complain that it needs lifetime specifiers:
$ cargo run Compiling structs v0.1.0 (file:///projects/structs) error[E0106]: missing lifetime specifier --> src/main.rs:2:15 | 2 | username: &str, | ^ expected named lifetime parameter | help: consider introducing a named lifetime parameter | 1 | struct User<'a> { 2 | username: &'a str, | error[E0106]: missing lifetime specifier --> src/main.rs:3:12 | 3 | email: &str, | ^ expected named lifetime parameter | help: consider introducing a named lifetime parameter | 1 | struct User<'a> { 2 | username: &str, 3 | email: &'a str, | error: aborting due to 2 previous errors For more information about this error, try `rustc --explain E0106`. error: could not compile `structs` To learn more, run the command again with --verbose.In Chapter 10, we’ll discuss how to fix these errors so you can store references in structs, but for now, we’ll fix errors like these using owned types like
Stringinstead of references like&str.