rust-book-cn/nostarch/chapter4.md

Ownership

Rust’s central feature is called ‘ownership’. It is a feature that is straightforward to explain, but has deep implications for the rest of the language.

Rust is committed to both safety and speed. One of the key tools for balancing between them is “zero-cost abstractions”: the various abstractions in Rust do not pose a global performance penalty. The ownership system is a prime example of a zero-cost abstraction. All of the analysis we’ll talk about in this guide is done at compile time. You do not pay any run-time cost for any of these features.

However, this system does have a certain cost: learning curve. Many new Rustaceans experience something we like to call ‘fighting with the borrow checker’, where the Rust compiler refuses to compile a program that the author thinks is valid. This can happen because the programmer isn’t used to thinking carefully about ownership, or is thinking about it differently from the way that Rust does. You probably will experience something similar at first. There is good news, however: more experienced Rust developers report that once they work with the rules of the ownership system for a period of time, they fight the borrow checker less and less. Keep at it!

This chapter will give you a foundation for understanding the rest of the language. To do so, we’re going to learn through examples, focusing on a very common data structure: strings.

Variable binding scope

Let’s take a step back and look at the very basics again. Now that we’re past basic syntax, we won’t include all of the fn main() { stuff in examples, so if you’re following along, you will have to put them inside of a main() function. This lets our examples be a bit more concise, letting us focus on the actual details, rather than boilerplate.

Anyway, here it is:

let s = "hello";

This variable binding refers to a string literal. It’s valid from the point at which it’s declared, until the end of the current scope. That is:

{                      // s is not valid here, it’s not yet in scope
    let s = "hello";   // s is valid from this point forward

    // do stuff with s
}                      // this scope is now over, and s is no longer valid

In other words, there are two important points in time here:

• When s comes ‘into scope’, it is valid.
• It remains so until it ‘goes out of scope’.

At this point, things are similar to other programming languages. Let’s build on top of this understanding by introducing a new type: String.

Strings

String literals are convenient, but they aren’t the only way that you use strings. For one thing, they’re immutable. For another, not every string is literal: what about taking user input and storing it in a string?

For this, Rust has a second string type, String. You can create a String from a string literal using the from function:

let s = String::from("hello");

We haven’t seen the double colon (::) syntax yet. It is a kind of scope operator, allowing us to namespace this particular from() function under the String type itself, rather than using some sort of name like string_from(). We’ll discuss this syntax more in the “Method Syntax” and “Modules” chapters.

This kind of string can be mutated:

let mut s = String::from("hello");

s.push_str(", world!");

Memory and allocation

So, what’s the difference here? Why can String be mutated, but literals cannot? The difference comes down to how these two types deal with memory.

In the case of a string literal, because we know the contents of the string at compile time, we can hard-code the text of the string directly into the final executable. This means that string literals are quite fast and efficient. But these properties only come from its immutability; we can’t put an arbitrary-sized blob of memory into the binary for each string!

With String, to support a mutable, growable string, we need to allocate an unknown amount of memory to hold the contents. This means two things:

1. The memory must be requested from the operating system at runtime.
2. We need a way of giving this memory back to the operating system when we’re done with our String.

That first part is done by us: when we call String::from(), its implementation requests the memory it needs. This is pretty much universal in programming languages.

The second case, however, is different. In languages with a garbage collector (‘GC’), the GC handles that second case, and we, as the programmer, don’t need to think about it. Without GC, it’s the programmer’s responsibility to identify when memory is no longer being used, and explicitly return it, just as it was requested. Doing this correctly has historically been a difficult problem. If we forget, we will waste memory. If we do it too early, we will have an invalid variable. If we do it twice, that’s a bug too. We need to pair exactly one allocate() with exactly one free().

Rust takes a different path. Remember our example? Here’s a version with String:

{
    let s = String::from("hello"); // s is valid from this point forward

    // do stuff with s
}                                  // this scope is now over, and s is no longer valid

We have a natural point at which we can return the memory our String needs back to the operating system: when it goes out of scope! When a variable goes out of scope, a special function is called. This function is called drop(), and it is where the author of String can put the code to return the memory.

Aside: This pattern is sometimes called “Resource Aquisition Is Initialization” in C++, or “RAII” for short. While they are very similar, Rust’s take on this concept has a number of differences, and so we don’t tend to use the same term. If you’re familliar with this idea, keep in mind that it is roughly similar in Rust, but not identical.

This pattern has a profound impact on the way that Rust code is written. It may seem obvious right now, but things can get tricky in more advanced situations! Let’s go over the first one of those right now.

Move

What would you expect this code to do?

let x = 5;
let y = x;

You might say “Make a copy of 5.” That’d be correct! We now have two bindings, x and y, and both equal 5.

Now let’s look at String. What would you expect this code to do?

let s1 = String::from("hello");
let s2 = s1;

You might say “copy the String!” This is both correct and incorrect at the same time. It does a shallow copy of the String. What’s that mean? Well, let’s take a look at what String looks like under the covers:

A String is made up of three parts: a pointer to the memory that holds the contents of the string, a length, and a capacity. The length is how much memory the String is currently using. The capacity is the total amount of memory the String has gotten from the operating system. The difference between length and capacity matters, but not in this context, so don’t worry about it too much if it doesn’t make sense, and just ignore the capacity.

We’ve talked about two kinds of composite types: arrays and tuples. String is a third type: a struct, which we will cover the details of in the next chapter of the book. For now, thinking about String as a tuple is close enough.

When we assign s1 to s2, the String itself is copied. But not all kinds of copying are the same. Many people draw distinctions between ‘shallow copying’ and ‘deep copying’. We don’t use these terms in Rust. We instead say that something is ‘moved’ or ‘cloned’. Assignment in Rust causes a ‘move’. In other words, it looks like this:

Not this:

When moving, Rust makes a copy of the data structure itself, the contents of s1 are copied, but if s1 contains a reference, like it does in this case, Rust will not copy the things that those references refer to.

There’s a problem here! Both data pointers are pointing to the same place. Why is this a problem? Well, when s2 goes out of scope, it will free the memory that data points to. And then s1 goes out of scope, and it will also try to free the memory that data points to! That’s bad.

So what’s the solution? Here, we stand at a crossroads. There are a few options. One would be to declare that assignment will also copy out any data. This works, but is inefficient: what if our String contained a novel? Also, it only works for memory. What if, instead of a String, we had a TcpConnection? Opening and closing a network connection is very similar to allocating and freeing memory. The solution that we could use there is to allow the programmer to hook into the assignment, similar to drop(), and write code fix things up. That would work, but now, an = can run arbitrary code. That’s also not good, and it doesn’t solve our efficiency concerns either.

Let’s take a step back: the root of the problem is that s1 and s2 both think that they have control of the memory, and therefore needs to free it. Instead of trying to copy the allocated memory, we could say that s1 is no longer valid, and therefore, doesn’t need to free anything. This is in fact the choice that Rust makes. Check it out what happens when you try to use s1 after s2 is created:

let s1 = String::from("hello");
let s2 = s1;

println!("{}", s1);

You’ll get an error like this:

5:22 error: use of moved value: `s1` [E0382]
println!("{}", s1);
               ^~
5:24 note: in this expansion of println! (defined in <std macros>)
3:11 note: `s1` moved here because it has type `collections::string::String`, which is moved by default
 let s2 = s1;
     ^~

We say that s1 was moved into s2. When a value moves, its data is copied, but the original variable binding is no longer usable. That solves our problem:

With only s2 valid, when it goes out of scope, it will free the memory, and we’re done!

Ownership Rules

This leads us to the Ownership Rules:

1. Each value in Rust has a variable binding that’s called it’s ‘owner’.
2. There can only be one owner at a time.
3. When the owner goes out of scope, the value will be drop()ped.

Furthermore, there’s a design choice that’s implied by this: Rust will never automatically create ‘deep’ copies of your data. Any automatic copying must be inexpensive.

Clone

But what if we do want to deeply copy the String’s data, and not just the String itself? There’s a common method for that: clone(). Here’s an example of clone() in action:

let s1 = String::from("hello");
let s2 = s1.clone();

println!("{}", s1);

This will work just fine. Remember our diagram from before? In this case, it is doing this:

When you see a call to clone(), you know that some arbitrary code is being executed, which may be expensive. It’s a visual indicator that something different is going on here.

Copy

There’s one last wrinkle that we haven’t talked about yet. This code works:

let x = 5;
let y = x;

println!("{}", x);

But why? We don’t have a call to clone(). Why didn’t x get moved into y?

For types that do not have any kind of complex storage requirements, like integers, typing clone() is busy work. There’s no reason we would ever want to prevent x from being valid here, as there’s no situation in which it’s incorrect. In other words, there’s no difference between deep and shallow copying here, so calling clone() wouldn’t do anything differently from the usual shallow copying.

Rust has a special annotation that you can place on types, called Copy. If a type is Copy, an older binding is still usable after assignment. Integers are an example of such a type; most of the primitive types are Copy.

While we haven’t talked about how to mark a type as Copy yet, you might ask yourself “what happens if we made String Copy?” The answer is, you cannot. Remember drop()? Rust will not let you make something Copy if it has implemented drop(). If you need to do something special when the value goes out of scope, being Copy will be an error.

So what types are Copy? You can check the documentation for the given type to be sure, but as a rule of thumb, any group of simple scalar values can be Copy, but nothing that requires allocation or is some form of resource is Copy. And you can’t get it wrong: the compiler will throw an error if you try to use a type that moves incorrectly, as we saw above.

Here’s some types that you’ve seen so far that are Copy:

• All of the integer types, like u32.
• The booleans, true and false.
• All of the floating point types, like f64.
• Tuples, but only if they contain types which are also Copy. (i32, i32) is Copy, but (i32, String) is not!

Ownership and functions

Passing a value to a function has similar semantics as assigning it:

fn main() {
    let s = String::from("hello");

    takes_ownership(s);

    let x = 5;

    makes_copy(x);
}

fn takes_ownership(some_string: String) {
    println!("{}", some_string);
}

fn makes_copy(some_integer: i32) {
    println!("{}", some_integer);
}

Passing a binding to a function will move or copy, just like assignment. Here’s the same example, but with some annotations showing where things go into and out of scope:

fn main() {
    let s = String::from("hello");  // s goes into scope.

    takes_ownership(s);             // s moves into the function...
                                    // ... and so is no longer valid here.
    let x = 5;                      // x goes into scope.

    makes_copy(x);                  // x would move into the function,
                                    // but i32 is Copy, so it’s okay to still
                                    // use x afterward.

} // Here, x goes out of scope, then s. But since s was moved, nothing special
  // happens.

fn takes_ownership(some_string: String) { // some_string comes into scope.
    println!("{}", some_string);
} // Here, some_string goes out of scope and `drop()` is called. The backing
  // memory is freed.

fn makes_copy(some_integer: i32) { // some_integer comes into scope.
    println!("{}", some_integer);
} // Here, some_integer goes out of scope. Nothing special happens.

Remember: If we tried to use s after the call to takes_ownership(), Rust would throw a compile-time error! These static checks protect us from mistakes.

Returning values can also transfer ownership:

fn main() {
    let s1 = gives_ownership();

    let s2 = String::from("hello");

    let s3 = takes_and_gives_back(s2);
}

fn gives_ownership() -> String {
    let some_string = String::from("hello");

    some_string
}

fn takes_and_gives_back(a_string: String) -> String {

    a_string
}

With simililar annotations:

fn main() {
    let s1 = gives_ownership();         // gives_ownership moves its return
                                        // value into s1.

    let s2 = String::from("hello");     // s2 comes into scope

    let s3 = takes_and_gives_back(s2);  // s2 is moved into
                                        // takes_and_gives_back, which also
                                        // moves its return value into s3.
} // Here, s3 goes out of scope, and is dropped. s2 goes out of scope, but was
  // moved, so nothing happens. s1 goes out of scope, and is dropped.

fn gives_ownership() -> String {             // gives_ownership will move its
                                             // return value into the function
                                             // that calls it.

    let some_string = String::from("hello"); // some_string comes into scope.

    some_string                              // some_string is returned, and
                                             // moves out to the calling
                                             // function.
}

// takes_and_gives_back will both take a String and return one
fn takes_and_gives_back(a_string: String) -> String { // a_string comes into scope

    a_string  // a_string is returned, and moves out to the calling function
}

It’s the same pattern, every time: assigning something moves it, and when an owner goes out of scope, if it hasn’t been moved, it will drop().

This might seem a bit tedious, and it is. What if I want to let a function use a value, but not take ownership? It’s quite annoying that anything I pass in also needs passed back. Look at this function:

fn main() {
    let s1 = String::from("hello");

    let (s2, len) = calculate_length(s1);

    println!("The length of '{}' is {}.", s2, len);
}

fn calculate_length(s: String) -> (String, usize) {
    let length = s.len(); // len() returns the length of a String.

    (s, length)
}

This is too much ceremony: we have to use a tuple to give back the String as well as the length. It’s a lot of work for a pattern that should be common.

Luckily for us, Rust has such a feature, and it’s what the next section is about.

References and Borrowing

At the end of the last section, we had some example Rust that wasn’t very good. Here it is again:

fn main() {
    let s1 = String::from("hello");

    let (s2, len) = calculate_length(s1);

    println!("The length of '{}' is {}.", s2, len);
}

fn calculate_length(s: String) -> (String, usize) {
    let length = s.len(); // len() returns the length of a String.

    (s, length)
}

The issue here is that we have to return the String back to the calling function so that it could still use it.

There is a better way. It looks like this:

fn main() {
    let s1 = String::from("hello");

    let len = calculate_length(&s1);

    println!("The length of '{}' is {}.", s1, len);
}

fn calculate_length(s: &String) -> usize {
    let length = s.len();

    length
}

First, you’ll notice all of the tuple stuff is gone. Next, that we pass &s1 into calculate_lengths(). And in its definition, we take &String rather than String.

These &s are called ‘references’, and they allow you to refer to some value without taking ownership of it. Here’s a diagram:

DIAGRAM GOES HERE of a &String pointing at a String, with (ptr, len, capacity)

Let’s take a closer look at the function call here:

# fn calculate_length(s: &String) -> usize {
#     let length = s.len();
# 
#     length
# }
let s1 = String::from("hello");

let len = calculate_length(&s1);

The &s1 syntax lets us create a reference from s1. This reference refers to the value of s1, but does not own it. Because it does not own it, the value it points to will not be dropped when the reference goes out of scope.

Likewise, the signature of the function uses & to indicate that it takes a reference as an argument:

Let’s add some explanatory annotations:

fn calculate_length(s: &String) -> usize { // s is a reference to a String
    let length = s.len();

    length
} // Here, s goes out of scope. But since it does not have ownership of what
  // it refers to, nothing happens.

It’s the same process as before, except that because we don’t have ownership, we don’t drop what a reference points to when the reference goes out of scope. This lets us write functions which take references as arguments instead of the values themselves, so that we won’t need to return them to give back ownership.

There’s another word for what references do, and that’s ‘borrowing’. Just like with real life, if I own something, you can borrow it from me. When you’re done, you have to give it back.

Speaking of which, what if you try to modify something you borrow from me? Try this code out. Spoiler alert: it doesn’t work:

fn main() {
    let s = String::from("hello");

    change(&s);
}

fn change(some_string: &String) {
    some_string.push_str(", world");  // push_str() appends a literal to a String
}

Here’s the error:

8:16 error: cannot borrow immutable borrowed content `*some_string` as mutable
 some_string.push_str(", world");  // push_str() appends a literal to a String
 ^~~~~~~~~~~

Just like bindings are immutable by default, so are references. We’re not allowed to modify something we have a reference to.

Mutable references

We can fix this bug! Just a small tweak:

fn main() {
    let mut s = String::from("hello");

    change(&mut s);
}

fn change(some_string: &mut String) {
    some_string.push_str(", world");  // push_str() appends a literal to a String
}

First, we had to change s to be mut. Then, we had to create a mutable reference with &mut s and accept a mutable reference with some_string: &mut String.

Mutable references have one big restriction, though. This code fails:

let mut s = String::from("hello");

let r1 = &mut s;
let r2 = &mut s;

Here’s the error:

5:20 error: cannot borrow `s` as mutable more than once at a time [E0499]
    let r2 = &mut s;
                  ^
4:20 note: previous borrow of `s` occurs here; the mutable borrow prevents
           subsequent moves, borrows, or modification of `s` until the borrow
           ends
    let r1 = &mut s;
                  ^
7:2 note: previous borrow ends here
fn main() {

}
^

The error is what it says on the tin: you cannot borrow something more than once at a time in a mutable fashion. This restriction allows for mutation, but in a very controlled fashion. It is something that new Rustaceans struggle with, because most languages let you mutate whenever you’d like.

As always, we can use {}s to create a new scope, allowing for multiple mutable references. Just not simultaneous ones:

let mut s = String::from("hello");

{
    let r1 = &mut s;

} // r1 goes out of scope here, so we can make a new reference with no problems.

let r2 = &mut s;

There is a simlar rule for combining the two kinds of references. This code errors:

let mut s = String::from("hello");

let r1 = &s; // no problem
let r2 = &s; // no problem
let r3 = &mut s; // BIG PROBLEM

Here’s the error:

19: 6:20 error: cannot borrow `s` as mutable because it is also borrowed as
                immutable [E0502]
    let r3 = &mut s; // BIG PROBLEM
                  ^
15: 4:16 note: previous borrow of `s` occurs here; the immutable borrow
               prevents subsequent moves or mutable borrows of `s` until the
               borrow ends
    let r1 = &s; // no problem
              ^
8:2 note: previous borrow ends here
fn main() {

}
^

Whew! We also cannot have a mutable reference while we have an immutable one. Users of an immutable reference don’t expect the values to suddenly change out from under them! Multiple immutable references are okay, however.

Dangling references

In languages with pointers, it’s easy to create a “dangling pointer” by freeing some memory while keeping around a pointer to that memory. In Rust, by contrast, the compiler guarantees that references will never be dangling: if we have a reference to something, the compiler will ensure that it will not go out of scope before the reference does.

Let’s try to create a dangling reference:

fn main() {
    let reference_to_nothing = dangle();
}

fn dangle() -> &String {
    let s = String::from("hello");

    &s
}

Here’s the error:

error: missing lifetime specifier [E0106]
fn dangle() -> &String {
               ^~~~~~~
help: this function’s return type contains a borrowed value, but there is no
      value for it to be borrowed from
help: consider giving it a ‘static lifetime

This error message refers to a feature we haven’t learned about yet, ‘lifetimes’. The message does contain the key to why this code is a problem, though:

this function’s return type contains a borrowed value, but there is no value
for it to be borrowed from

Let’s examine exactly what happens with dangle():

fn dangle() -> &String { // dangle returns a reference to a String

    let s = String::from("hello"); // s is a new String

    &s // we return a reference to the String, s
} // Here, s goes out of scope, and is dropped. Its memory goes away.
  // Danger!

Because s is created inside of dangle(), when the code of dangle() is finished, it will be deallocated. But we tried to return a reference to it. That means this reference would be pointing to an invalid String! That’s no good. Rust won’t let us do this.

The correct code here is to return the String directly:

fn no_dangle() -> String {
    let s = String::from("hello");

    s
}

This works, no problem. Ownership is moved out, nothing is deallocated.

The Rules of References

Here’s a recap of what we’ve talked about. The Rules of References:

1. At any given time, you may have either, but not both of:
   1. One mutable reference.
   2. Any number of immutable references .
2. References must always be valid.

While these rules are not complicated on their own, they can be tricky when applied to real code.

Slices

So far, we’ve talked about types that have ownership, like String, and ones that don’t, like &String. There is a second kind of type which does not have ownership: slices. Slices let you reference a contiguous sequence of elements in a collection, rather than the whole collection itself.

Here’s a small programming problem: write a function which takes a string, and returns the first word you find. If we don’t find a space in the string, then the whole string is a word, so the whole thing should be returned.

Let’s think about the signature of this function:

fn first_word(s: &String) -> ?

This function, first_word, takes a &String as an argument. We don’t want ownership, so this is fine. But what should we return? We don’t really have a way to talk about part of a string. We could return the index of the end of the word, though. Let’s try that:

fn first_word(s: &String) -> usize {
    let bytes = s.as_bytes();

    for (i, &byte) in bytes.iter().enumerate() {
        if byte == 32 {
            return i;
        }
    }

    s.len()
}

Let’s break that down a bit:

fn first_word(s: &String) -> usize {

    // Since we need to go through the String element by element, and
    // check if a value is a space, we will convert our String to an
    // array of bytes, using the `.as_bytes()` method.
    let bytes = s.as_bytes();

    // We discussed using the iter() method with for in Chapter 3.7. Here,
    // we’re adding another method: enumerate(). While iter() returns each
    // element, enumerate() modifies the result of iter(), and returns a
    // tuple instead. The first element of the tuple is the index, and the
    // second element is a reference to the element itself. This is a bit
    // nicer than calculating the index ourselves.
    //
    // Since it’s a tuple, we can use patterns, just like elsewhere in Rust.
    // So we match against the tuple with i for the index, and &byte for
    // the byte itself.
    for (i, &byte) in bytes.iter().enumerate() {

        // 32 is the value of a space in UTF-8
        if byte == 32 {

            // We found a space! Return this position.
            return i;
        }
    }

    // If we got here, we didn’t find a space, so this whole thing must be a
    // word. So return the length.
    s.len()
}

This works, but there’s a problem. We’re returning a usize on its own, but it’s only a meaningful number in the context of the &String itself. In other words, because it’s a separate value from the String, there’s no guarantee that it will still be valid in the future. Consider this:

# fn first_word(s: &String) -> usize {
#     let bytes = s.as_bytes();
# 
#     for (i, &byte) in bytes.iter().enumerate() {
#         if byte == 32 {
#             return i;
#         }
#     }
# 
#     s.len()
# }

fn main() {
    let mut s = String::from("hello world");

    let word = first_word(&s);

    s.clear(); // This empties the String, making it equal to "".

    // word is now totally invalid! There’s no more word here.
}

This is bad! It’s even worse if we wanted to write a second_word() function. Its signature would have to look like this:

fn second_word(s: &String) -> (usize, usize) {

Now we’re tracking both a start and and ending index. Even more chances for things to go wrong. We now have three unrelated variable bindings floating around which need to be kept in sync.

Luckily, Rust has a solution to this probem: string slices.

String slices

A string slice looks like this:

let s = String::from("hello world");

let hello = &s[0..5];
let world = &s[5..9];

This looks just like taking a reference to the whole String, but with the extra [0..5] bit. Instead of being a reference to the entire String, it’s a reference to an internal position in the String, but it also keeps track of the number of elements that it refers to as well. In other words, it looks like this:

DIAGRAM GOES HERE of s, hello, and world

With Rust’s .. syntax, if you want to start at zero, you can drop the zero. In other words, these are equal:

let s = String::from("hello");

let slice = &s[0..2];
let slice = &s[..2];

By the same token, if you want to go to the maximum value, which for slices is the last element, you can drop the trailing number. In other words, these are equal:

let s = String::from("hello");

let len = s.len();

let slice = &s[1..len];
let slice = &s[1..];

With this in mind, let’s re-write first_word() to return a slice:

fn first_word(s: &String) -> &str {
    let bytes = s.as_bytes();

    for (i, &byte) in bytes.iter().enumerate() {
        if byte == 32 {
            return &s[0..i];
        }
    }

    &s[..]
}

Now, we have a single value, the &str. It contains both elements that we care about: a reference to the starting point, and the number of elements. This would also work for a second_word():

fn second_word(s: &String) -> &str {

Same deal. We now have a straightforward API, that’s much harder to mess up.

But what about our error condition from before? Slices also fix that. Using the slice version of first_word() will throw an error:

# fn first_word(s: &String) -> &str {
#     let bytes = s.as_bytes();
# 
#     for (i, &byte) in bytes.iter().enumerate() {
#         if byte == 32 {
#             return &s[0..i];
#         }
#     }
# 
#     &s[..]
# }
fn main() {
    let mut s = String::from("hello world");

    let word = first_word(&s);

    s.clear(); // Error!
}

Here’s the error:

17:6 error: cannot borrow `s` as mutable because it is also borrowed as
            immutable [E0502]
    s.clear(); // Error!
    ^
15:29 note: previous borrow of `s` occurs here; the immutable borrow prevents
            subsequent moves or mutable borrows of `s` until the borrow ends
    let word = first_word(&s);
                           ^
18:2 note: previous borrow ends here
fn main() {

}
^

Remember the borrowing rules? If we have an immutable reference to something, we cannot also take a mutable reference. Since clear() needs to truncate the String, it tries to take a mutable reference, which fails. Not only has Rust made our API easier to use, but it’s also eliminated an entire class of errors at compile time!

String literals are slices

Remember how we talked about string literals being stored inside of the binary itself? Now that we know about slices, we can now properly understand string literals.

let s = "Hello, world!";

The type of s here is &str: It’s a slice, pointing to that specific point of the binary. This is also why string literals are immutable; &str is an immutable reference.

String slices as arguments

Knowing that you can take slices of both literals and Strings leads us to one more improvement on first_word(), and that’s its signature:

fn first_word(s: &String) -> &str {

A more experienced Rustacean would write this one instead:

fn first_word(s: &str) -> &str {

Why is this? Well, we aren’t trying to modify s at all. And we can take a string slice that’s the full length of a String, so we haven’t lost the ability to talk about full Strings. And additionally, we can take string slices of string literals too, so this function is more useful, but with no loss of functionality:

# fn first_word(s: &str) -> &str {
#     let bytes = s.as_bytes();
# 
#     for (i, &byte) in bytes.iter().enumerate() {
#         if byte == 32 {
#             return &s[0..i];
#         }
#     }
# 
#     &s[..]
# }
fn main() {
    let s = String::from("hello world");
    let word = first_word(&s[..]);

    let s = "hello world";
    let word = first_word(&s[..]);

    let word = first_word(s); // since literals are &strs, this works too!
}

Other slices

String slices, as you might imagine, are specific to strings. But there’s a more general slice type, too. Consider arrays:

let a = [1, 2, 3, 4, 5];

Just like we may want to refer to a part of a string, we may want to refer to part of an array:

let a = [1, 2, 3, 4, 5];

let slice = &a[1..3];

This slice has the type &[i32]. It works the exact same way as string slices do, with a reference to the first element, and a length. You’ll use this kind of slice for all sorts of other collections. We’ll discuss these other slices in detail when we talk about vectors, in Chapter 9.1.