I'm trying to understand how & and ref correspond. Here's an example where I thought these were equivalent, but one works and the other doesn't:
fn main() {
let t = "
aoeu
aoeu
aoeu
a";
let ls = t.lines();
dbg!(ls.clone().map(|l| &l[..]).collect::<Vec<&str>>().join("\n")); // works
dbg!(ls.clone().map(|ref l| l[..]).collect::<Vec<&str>>().join("\n")); // doesn't work
dbg!(ls.clone().map(|ref l| &l[..]).collect::<Vec<&str>>().join("\n")); // works again!
}
From the docs:
// A `ref` borrow on the left side of an assignment is equivalent to
// an `&` borrow on the right side.
let ref ref_c1 = c;
let ref_c2 = &c;
println!("ref_c1 equals ref_c2: {}", *ref_c1 == *ref_c2);
What would the equivalent to |l| &l[..] be with |ref l|? How does it correspond to the assignment examples in the docs?
Taking a look at the docs page for Lines(The iterator adapter for producing lines from a str), we can see that the item produced by it is:
type Item = &'a str;
Therefore the following happens when trying to do the "doesn't work" version:
dbg!(ls.clone().map(|ref l| l[..]).collect::<Vec<&str>>().join("\n")); # doesn't work
//Can become:
let temp = ls
.clone()
.map(|ref l| l[..])
.collect::<Vec<&str>>()
.join("\n");
println!("{}", temp);
Here we can see a crucial problem. l if of type &&str (Which I will explain below) so indexing into it will create a str, which is unsized and therefore cannot be outside of a pointer of some sort.
Now, onto the real thing you wanted to learn: What a ref pattern does:
When doing pattern matching or destructuring via the let binding, the ref keyword can be used to take references to the fields of a struct/tuple.
What this does is the following:
When we have let ref x = y, we take a reference to y.
When pattern matching on something (Like in your closure arguments you showed) we have a slightly different effect: the value under the reference is moved into the scope and then taken reference to while exposing a way to take the value under the reference. For example:
fn foo(ref x: String) {}
let y: fn(String) = foo;
This works because what is essentially being done is this:
fn foo(x: String) {
let x: &String = &x;
}
So what ref x does is take ownership of x and produce a reference to it.
On the other hand
When we have let &x = y we move a value out of y.
This is the opposite of ref, in that we take ownership of the value under y if we can. For example:
let x = 2;
let y = &x;
let &z = y; //Ok, we're moving a `Copy` type
This is only ok for copy types though as though this isn't exactly the same as let x = *y which would work for owned Boxes.
Related
Consider following code
fn main() {
let s = (&&0,);
let (x,) = s; // &&i32
let (&y,) = s; // &i32
let (&&z,) = s; // i32
let t = &(&0,);
let (x,) = t; // &&i32
let (&y,) = t; // i32
let u = &&(0,);
let (x,) = u; // &i32
let (&y,) = u; // mismatched types expected type `{integer}` found reference `&_`
}
Could someone explain, why & pattern behaves differently in every case? I suppose it is tied somehow to match ergonomics, maybe some coercions come into play? But I can't wrap my head around it.
You are correct, this is due to match ergonomics. The first case should hopefully be self explanatory, but the second and third cases can be a bit counter-intuitive.
In the second case:
(x,) is a non-reference pattern (see the second example in the RFC). The t tuple reference is dereferenced, and x is bound as a ref as it also is a non-reference pattern. Note that t.0 was a reference to begin with, thus resulting in x being a double reference.
(&y,) is also a non-reference pattern. The t tuple is dereferenced again to a (&i32,). However, &y is a reference pattern being matched to a &i32 reference. Hence y is bound with move mode and is an i32.
In the third case:
Using the same reasoning as the second case, u is dereferenced via Deref coercion to an (i32,), and x, a non-reference pattern, is bound in ref mode. Hence x is an &i32.
Again with the same reasoning as the second case, u is dereferenced to an (i32,). The &y reference pattern is then matched to an i32, a non-reference, which causes an error.
What is the purpose of & in the code &i in list? If I remove the &, it produces an error in largest = i, since they have mismatched types (where i is &32 and i is i32). But how does &i convert i into i32?
fn largest(list: &[i32]) -> i32 {
println!("{:?}", list);
let mut largest = list[0];
for &i in list {
if i > largest {
largest = i;
}
}
largest
}
fn main() {
let hey = vec![1, 3, 2, 6, 90, 67, 788, 12, 34, 54, 32];
println!("The largest number is: {}", largest(&hey));
}
Playground
It seems like it is somehow dereferencing, but then why in the below code, is it not working?
fn main() {
let mut hey: i32 = 32;
let x: i32 = 2;
hey = &&x;
}
It says:
4 | hey = &&x;
| ^^^ expected i32, found &&i32
|
= note: expected type `i32`
found type `&&i32`
So normally when you use for i in list, the loop variable i would be of type &i32.
But when instead you use for &i in list, you are not dereferencing anything, but instead you are using pattern matching to explicitly destructure the reference and that will make i just be of type i32.
See the Rust docs about the for-loop loop variable being a pattern and the reference pattern that we are using here. See also the Rust By Example chapter on destructuring pointers.
An other way to solve this, would be to just keep i as it is and then comparing i to a reference to largest, and then dereferencing i before assigning to largest:
fn largest(list: &[i32]) -> i32 {
println!("{:?}", list);
let mut largest = list[0];
for i in list {
if i > &largest {
largest = *i;
}
}
largest
}
fn main() {
let mut hey: i32 = 32;
let x: i32 = 2;
hey = &&x;
}
This simply doesn't work, because here you are assigning hey, which is an i32, to a reference to a reference to an i32. This is quite unrelated to the pattern matching and destructuring in the loop variable case.
This is the effect of destructuring. I won't completely describe that feature here, but in short:
In many syntax contexts (let bindings, for loops, function arguments, ...) , Rust expects a "pattern". This pattern can be a simple variable name, but it can also contain some "destructuring elements", like &. Rust will then bind a value to this pattern. A simple example would be something like this:
let (a, b) = ('x', true);
On the right hand side there is a value of type (char, bool) (a tuple). This value is bound to the left hand pattern ((a, b)). As there is already a "structure" defined in the pattern (specifically, the tuple), that structure is removed and a und b bind to the tuple's elements. Thus, the type of a is char and the type of b is bool.
This works with a couple of structures, including arrays:
let [x] = [true];
Again, on the right side we have a value of type [bool; 1] (an array) and on the left side we have a pattern in the form of an array. The single array element is bound to x, meaning that the type of x is bool and not [bool; 1]!
And unsurprisingly, this also works for references!
let foo = 0u32;
let r = &foo;
let &c = &foo;
Here, foo has the type u32 and consequently, the expression &foo has the type &u32. The type of r is also &u32, as it is a simple let binding. The type of c is u32 however! That is because the "reference was destructured/removed" by the pattern.
A common misunderstanding is that syntax in patterns has exactly the opposite effect of what the same syntax would have in expressions! If you have a variable a of type [T; 1], then the expression [a] has the type [[T; 1]; 1] → it adds stuff. However, if you bind a to the pattern [c], then y has the type T → it removes stuff.
let a = [true]; // type of `a`: `[bool; 1]`
let b = [a]; // type of `b`: `[[bool; 1]; 1]`
let [c] = a; // type of `c`: `bool`
This also explains your question:
It seems like it is somehow dereferencing, but then why in the below code, it is not working?
fn main() {
let mut hey:i32 = 32;
let x:i32 = 2;
hey = &&x;
}
Because you use & on the expression side, where it adds a layer of references.
So finally about your loop: when iterating over a slice (as you do here), the iterator yields reference to the slice's elements. So in the case for i in list {}, i has the type &i32. But the assignment largest = i; requires a i32 on the right hand side. You can achieve this in two ways: either dereference i via the dereference operator * (i.e. largest = *i;) or destructure the reference in the loop pattern (i.e. for &i in list {}).
Related questions:
Iterating over a slice's values instead of references in Rust?
Why is & needed to destructure a list of tuples during iteration?
I'm wondering if someone can help me understand why this program behaves as it does:
fn main() {
let mut x = 456;
{
let mut y = Box::new(&x);
y = Box::new(&mut y);
println!("GOT {}",*y);
}
}
This program compiles under rust 1.35.0 (both 2015 and 2018 editions), and prints
GOT 456
But, I'm confused what's going on here. I'm guessing that this is an example of an auto-dereference. So, in reality, it looks like this:
fn main() {
let mut x = 456;
{
let mut y = Box::new(&x);
y = Box::new(&mut *y);
println!("GOT {}",*y);
}
}
Is that it?
This is a case of deref coercion, but one that is obfuscated by a few other unnecessary parts of the code. The following improvements should be made here:
The mut modifier on variable x is not needed because it is never modified.
The borrow of y in Box::new(&mut y) does not have to be mutable because the variable holds an immutable reference.
The println! implementation also knows to print values behind references, so the explicit * is not needed.
Then, we get the following code:
fn main() {
let x = 456;
{
let mut y = Box::new(&x);
y = Box::new(&y);
println!("GOT {}", y);
}
}
y is a variable of type Box<&i32> which is initially bound to a box created in the outer scope. The subsequent assignment to a new box works because the &y, of type &Box<&i32>, is coerced to &&i32, which can then be put in the box by automatically dereferencing the first borrow. This coercion is required because the variable x can only be assigned values of the same Box<&i32> type.
The lifetime of the reference inside both boxes also ended up being the same, because they refer to the same value in x.
See also:
What is the relation between auto-dereferencing and deref coercion?
I am playing around with Rust references:
fn main() {
let str = String::from("Hallo");
let &x = &str;
}
This produces the following error:
error[E0507]: cannot move out of borrowed content
--> src/main.rs:3:9
|
3 | let &x = &str;
| ^-
| ||
| |hint: to prevent move, use `ref x` or `ref mut x`
| cannot move out of borrowed content
What is going on here?
Adding to wiomoc's answer: depending on what language(s) you've previously known, variable declaration in Rust might be a little different. Whereas in C/C++ you explicitly have to declare that you want a pointer/reference variable:
int *p = &other_int;
In Rust it's enough to just use let, so the above would in Rust be
let p = &other_int;
and when you write
let &s = &string;
It pattern-matches that, so the Rust compiler reads it roughly as "I know I have a reference, and I want to bind whatever it is referring to to the name p". If you're not familiar with pattern-matching, a more obvious example (that works in Rust as well) would be
let point = (23, 42);
let (x, y) = point;
The second line unpacks the right-hand side to match the left-hand side (both are tuples of two values) and binds the variable names on the left to the values at the same position in the structure on the right. In the case above, it's less obvious that it's matching your "structural description".
The result of let &x = &str;, i.e. "I know &str is a reference, please bind whatever it refers to to the variable x" means that you're trying to have x be the same as str, when at that line all you have is a borrowed reference to str. That's why the compiler tells you you can't create an owned value (which x would be, because it's not being created as a reference) from a reference.
You dont need that &at let x
let str = String::from("Hallo");
let x = &str;
Or if you want to declare the type manually
let string = String::from("Hallo");
let x: &str = &string;
How could know the type of a binding if I use auto type deduction when creating a binding? what if the expression on the right side is a borrow(like let x = &5;), will it be value or a borrow? What will happen if I re-assign a borrow or a value?
Just for check, I do can re-assign a borrow if I use let mut x: &mut T = &mut T{}; or let mut x:&T = & T{};, right?
I sense some confusion between binding and assigning:
Binding introduces a new variable, and associates it to a value,
Assigning overwrites a value with another.
This can be illustrated in two simple lines:
let mut x = 5; // Binding
x = 10; // Assigning
A binding may appear in multiple places in Rust:
let statements,
if let/while let conditions,
cases in a match expression,
and even in a for expression, on the left side of in.
Whenever there is a binding, Rust's grammar also allows pattern matching:
in the case of let statements and for expressions, the patterns must be irrefutable,
in the case of if let, while let and match cases, the patterns may fail to match.
Pattern matching means that the type of the variable introduced by the binding differs based on how the binding is made:
let x = &5; // x: &i32
let &y = &5; // y: i32
Assigning always requires using =, the assignment operator.
When assigning, the former value is overwritten, and drop is called on it if it implements Drop.
let mut x = 5;
x = 6;
// Now x == 6, drop was not called because it's a i32.
let mut s = String::from("Hello, World!");
s = String::from("Hello, 神秘德里克!");
// Now s == "Hello, 神秘德里克!", drop was called because it's a String.
The value that is overwritten may be as simple as an integer or float, a more involved struct or enum, or a reference.
let mut r = &5;
r = &6;
// Now r points to 6, drop was not called as it's a reference.
Overwriting a reference does not overwrite the value pointed to by the reference, but the reference itself. The original value still lives on, and will be dropped when it's ready.
To overwrite the pointed to value, one needs to use *, the dereference operator:
let mut x = 5;
let r = &mut x;
*r = 6;
// r still points to x, and now x = 6.
If the type of the dereferenced value requires it, drop will be called:
let mut s = String::from("Hello, World!");
let r = &mut s;
*r = String::from("Hello, 神秘德里克!");
// r still points to s, and now s = "Hello, 神秘德里克!".
I invite you to use to playground to and toy around, you can start from here:
fn main() {
let mut s = String::from("Hello, World!");
{
let r = &mut s;
*r = String::from("Hello, 神秘德里克!");
}
println!("{}", s);
}
Hopefully, things should be a little clearer now, so let's check your samples.
let x = &5;
x is a reference to i32 (&i32). What happens is that the compiler will introduce a temporary in which 5 is stored, and then borrow this temporary.
let mut x: &mut T = T{};
Is impossible. The type of T{} is T not &mut T, so this fails to compile. You could change it to let mut x: &mut T = &mut T{};.
And your last example is similar.