rust-book-cn/nostarch/chapter4.md
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# Ownership
Rusts 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 well 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 isnt 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, were going to learn through examples, focusing on a very
common data structure: strings.
## Variable binding scope
Lets take a step back and look at the very basics again. Now that were past
basic syntax, we wont include all of the `fn main() {` stuff in examples, so
if youre 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:
```rust
let s = "hello";
```
This variable binding refers to a string literal. Its valid from the point at
which its declared, until the end of the current _scope_. That is:
```rust
{ // s is not valid here, its 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} // this scope is now over, and s is no longer valid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. Lets build
on top of this understanding by introducing a new type: `String`.
## Strings
String literals are convenient, but they arent the only way that you use strings.
For one thing, theyre 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:
```rust
let s = String::from("hello");
```
We havent 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()`.
Well discuss this syntax more in the “Method Syntax” and “Modules” chapters.
This kind of string can be mutated:
```rust
let mut s = String::from("hello");
s.push_str(", world!");
```
## Memory and allocation
So, whats 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 cant 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 were
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, dont need
to think about it. Without GC, its the programmers 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, thats a bug too. We need to pair exactly one
`allocate()` with exactly one `free()`.
Rust takes a different path. Remember our example? Heres a version with
`String`:
```rust
{
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,
> Rusts take on this concept has a number of differences, and so we dont tend
> to use the same term. If youre 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!
Lets go over the first one of those right now.
## Move
What would you expect this code to do?
```rust
let x = 5;
let y = x;
```
You might say “Make a copy of `5`.” Thatd be correct! We now have two
bindings, `x` and `y`, and both equal `5`.
Now lets look at `String`. What would you expect this code to do?
```rust
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`. Whats that mean? Well,
lets take a look at what `String` looks like under the covers:
<img alt="string" src="img/foo1.png" class="center" />
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 dont worry about it too much
if it doesnt make sense, and just ignore the capacity.
> Weve 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 dont 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:
<img alt="s1 and s2" src="img/foo2.png" class="center" />
_Not_ this:
<img alt="s1 and s2 to two places" src="img/foo4.png" class="center" />
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.
Theres 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! Thats bad.
So whats 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. Thats
also not good, and it doesnt solve our efficiency concerns either.
Lets 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, doesnt 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:
```rust,ignore
let s1 = String::from("hello");
let s2 = s1;
println!("{}", s1);
```
Youll get an error like this:
```text
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:
<img alt="s1 and s2 to the same place" src="img/foo3.png" class="center" />
With only `s2` valid, when it goes out of scope, it will free the memory, and were done!
## Ownership Rules
This leads us to the Ownership Rules:
> 1. Each value in Rust has a variable binding thats called its 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, theres a design choice thats 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? Theres a common method for that: `clone()`. Heres an example
of `clone()` in action:
```rust
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:
<img alt="s1 and s2 to two places" src="img/foo4.png" class="center" />
When you see a call to `clone()`, you know that some arbitrary code is being
executed, which may be expensive. Its a visual indicator that something
different is going on here.
## Copy
Theres one last wrinkle that we havent talked about yet. This code works:
```rust
let x = 5;
let y = x;
println!("{}", x);
```
But why? We dont have a call to `clone()`. Why didnt `x` get moved into `y`?
For types that do not have any kind of complex storage requirements, like
integers, typing `clone()` is busy work. Theres no reason we would ever want
to prevent `x` from being valid here, as theres no situation in which its
incorrect. In other words, theres no difference between deep and shallow
copying here, so calling `clone()` wouldnt 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 havent 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 cant get it wrong: the compiler will throw an error if you try to use
a type that moves incorrectly, as we saw above.
Heres some types that youve 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:
```rust
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. Heres
the same example, but with some annotations showing where things go into and
out of scope:
```rust
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 its 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:
```rust
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:
```rust
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
}
```
Its the same pattern, every time: assigning something moves it, and when an
owner goes out of scope, if it hasnt 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? Its quite annoying that anything I pass in
also needs passed back. Look at this function:
```rust
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. Its a lot of work for a pattern that should be common.
Luckily for us, Rust has such a feature, and its what the next section is about.
# References and Borrowing
At the end of the last section, we had some example Rust that wasnt very
good. Here it is again:
```rust
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:
```rust
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, youll 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. Heres a diagram:
DIAGRAM GOES HERE of a &String pointing at a String, with (ptr, len, capacity)
Lets take a closer look at the function call here:
```rust
# 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:
Lets add some explanatory annotations:
```rust
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.
```
Its the same process as before, except that because we dont have ownership,
we dont 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 wont need to return them to give back ownership.
Theres another word for what references do, and thats borrowing. Just like
with real life, if I own something, you can borrow it from me. When youre 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 doesnt work:
```rust,ignore
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
}
```
Heres the error:
```text
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. Were not allowed
to modify something we have a reference to.
## Mutable references
We can fix this bug! Just a small tweak:
```rust
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:
```rust,ignore
let mut s = String::from("hello");
let r1 = &mut s;
let r2 = &mut s;
```
Heres the error:
```text
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 youd like.
As always, we can use `{}`s to create a new scope, allowing for multiple mutable
references. Just not _simultaneous_ ones:
```rust
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:
```rust,ignore
let mut s = String::from("hello");
let r1 = &s; // no problem
let r2 = &s; // no problem
let r3 = &mut s; // BIG PROBLEM
```
Heres the error:
```text
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 dont expect the values to suddenly change out
from under them! Multiple immutable references are okay, however.
## Dangling references
In languages with pointers, its 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.
Lets try to create a dangling reference:
```rust,ignore
fn main() {
let reference_to_nothing = dangle();
}
fn dangle() -> &String {
let s = String::from("hello");
&s
}
```
Heres the error:
```text
error: missing lifetime specifier [E0106]
fn dangle() -> &String {
^~~~~~~
help: this functions 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 havent learned about yet,
lifetimes. The message does contain the key to why this code is a problem,
though:
```text
this functions return type contains a borrowed value, but there is no value
for it to be borrowed from
```
Lets examine exactly what happens with `dangle()`:
```rust,ignore
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`! Thats
no good. Rust wont let us do this.
The correct code here is to return the `String` directly:
```rust
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
Heres a recap of what weve 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, weve talked about types that have ownership, like `String`, and ones
that dont, 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.
Heres a small programming problem: write a function which takes a string,
and returns the first word you find. If we dont find a space in the string,
then the whole string is a word, so the whole thing should be returned.
Lets think about the signature of this function:
```rust,ignore
fn first_word(s: &String) -> ?
```
This function, `first_word`, takes a `&String` as an argument. We dont want
ownership, so this is fine. But what should we return? We dont really have a
way to talk about _part_ of a string. We could return the index of the end of
the word, though. Lets try that:
```rust
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()
}
```
Lets break that down a bit:
```rust
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,
// were 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 its 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 didnt find a space, so this whole thing must be a
// word. So return the length.
s.len()
}
```
This works, but theres a problem. Were returning a `usize` on its own, but
its only a meaningful number in the context of the `&String` itself. In other
words, because its a separate value from the `String`, theres no guarantee
that it will still be valid in the future. Consider this:
```rust
# 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! Theres no more word here.
}
```
This is bad! Its even worse if we wanted to write a `second_word()`
function. Its signature would have to look like this:
```rust,ignore
fn second_word(s: &String) -> (usize, usize) {
```
Now were 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:
```rust
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`,
its 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 Rusts `..` syntax, if you want to start at zero, you can drop the zero.
In other words, these are equal:
```rust
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:
```rust
let s = String::from("hello");
let len = s.len();
let slice = &s[1..len];
let slice = &s[1..];
```
With this in mind, lets re-write `first_word()` to return a slice:
```rust
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()`:
```rust,ignore
fn second_word(s: &String) -> &str {
```
Same deal. We now have a straightforward API, thats 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:
```rust,ignore
# 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!
}
```
Heres the error:
```text
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 its 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.
```rust
let s = "Hello, world!";
```
The type of `s` here is `&str`: Its 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 `String`s leads us to
one more improvement on `first_word()`, and thats its signature:
```rust,ignore
fn first_word(s: &String) -> &str {
```
A more experienced Rustacean would write this one instead:
```rust,ignore
fn first_word(s: &str) -> &str {
```
Why is this? Well, we arent trying to modify `s` at all. And we can take
a string slice thats the full length of a `String`, so we havent lost
the ability to talk about full `String`s. And additionally, we can take
string slices of string literals too, so this function is more useful, but
with no loss of functionality:
```rust
# 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 theres a more
general slice type, too. Consider arrays:
```rust
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:
```rust
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. Youll use this kind
of slice for all sorts of other collections. Well discuss these other slices
in detail when we talk about vectors, in Chapter 9.1.