Use for loop on a rust array works correctly:
fn main() {
let v = [1, 2, 3, 4, 5];
for _ in v.into_iter() {}
for _ in v.into_iter() {}
}
But substituting a vec doesn't compile:
fn main() {
let v = vec![1, 2, 3, 4, 5];
for _ in v.into_iter() {}
for _ in v.into_iter() {}
}
The error:
use of moved value: `v`
I understand why this program does not work with vec. But why does it work with array? I was expecting a similar error in the array example, but it gives no error.
As another commenter mentioned, Arrays in Rust implement the Copy trait, and can therefore be passed-by-value multiple times, whereas Vector types must be explicitly clone()d to achieve the same behavior.
In Rust, when a function's parameter is pass-by-value, the compiler defaults to doing a move of a copy on all calls except the last one, on the last call it will transfer ownership of the original, instead of a copy/clone. The compiler will not automatically run clone(), if Copy isn't implemented.
Here is the doc for Copy trait: https://doc.rust-lang.org/std/marker/trait.Copy.html
Array's impl of Copy documentation can be found here: https://doc.rust-lang.org/std/primitive.array.html#impl-Copy-for-%5BT%3B%20N%5D
Here is a great article with more details: https://colinsblog.net/2021-04-16-rust-ownership-comparisons/
Related
I want to call .map() on an array of enums:
enum Foo {
Value(i32),
Nothing,
}
fn main() {
let bar = [1, 2, 3];
let foos = bar.iter().map(|x| Foo::Value(*x)).collect::<[Foo; 3]>();
}
but the compiler complains:
error[E0277]: the trait bound `[Foo; 3]: std::iter::FromIterator<Foo>` is not satisfied
--> src/main.rs:8:51
|
8 | let foos = bar.iter().map(|x| Foo::Value(*x)).collect::<[Foo; 3]>();
| ^^^^^^^ a collection of type `[Foo; 3]` cannot be built from an iterator over elements of type `Foo`
|
= help: the trait `std::iter::FromIterator<Foo>` is not implemented for `[Foo; 3]`
How do I do this?
The issue is actually in collect, not in map.
In order to be able to collect the results of an iteration into a container, this container should implement FromIterator.
[T; n] does not implement FromIterator because it cannot do so generally: to produce a [T; n] you need to provide n elements exactly, however when using FromIterator you make no guarantee about the number of elements that will be fed into your type.
There is also the difficulty that you would not know, without supplementary data, which index of the array you should be feeding now (and whether it's empty or full), etc... this could be addressed by using enumerate after map (essentially feeding the index), but then you would still have the issue of deciding what to do if not enough or too many elements are supplied.
Therefore, not only at the moment one cannot implement FromIterator on a fixed-size array; but even in the future it seems like a long shot.
So, now what to do? There are several possibilities:
inline the transformation at call site: [Value(1), Value(2), Value(3)], possibly with the help of a macro
collect into a different (growable) container, such as Vec<Foo>
...
Update
This can work:
let array: [T; N] = something_iterable.[into_]iter()
.collect::<Vec<T>>()
.try_into()
.unwrap()
In newer version of rust, try_into is included in prelude, so it is not necessary to use std::convert::TryInto. Further, starting from 1.48.0, array support directly convert from Vec type, signature from stdlib source:
fn try_from(mut vec: Vec<T, A>) -> Result<[T; N], Vec<T, A>> {
...
}
Original Answer
as of rustc 1.42.0, if your element impl Copy trait, for simplicity, this just works:
use std::convert::TryInto;
...
let array: [T; N] = something_iterable.[into_]iter()
.collect::<Vec<T>>()
.as_slice()
.try_into()
.unwrap()
collect as_slice try_into + unwrap()
Iterator<T> ------> Vec<T> -------> &[T] ------------------> [T]
But I would just call it a workaround.
You need to include std::convert::TryInto because the try_into method is defined in the TryInto trait.
Below is the signature checked when you call try_into as above, taken from the source. As you can see, that requires your type T implement Copy trait, so theoritically, it will copy all your elements once.
#[stable(feature = "try_from", since = "1.34.0")]
impl<T, const N: usize> TryFrom<&[T]> for [T; N]
where
T: Copy,
[T; N]: LengthAtMost32,
{
type Error = TryFromSliceError;
fn try_from(slice: &[T]) -> Result<[T; N], TryFromSliceError> {
<&Self>::try_from(slice).map(|r| *r)
}
}
While you cannot directly collect into an array for the reasons stated by the other answers, that doesn't mean that you can't collect into a data structure backed by an array, like an ArrayVec:
use arrayvec::ArrayVec; // 0.7.0
use std::array;
enum Foo {
Value(i32),
Nothing,
}
fn main() {
let bar = [1, 2, 3];
let foos: ArrayVec<_, 3> = array::IntoIter::new(bar).map(Foo::Value).collect();
let the_array = foos
.into_inner()
.unwrap_or_else(|_| panic!("Array was not completely filled"));
// use `.expect` instead if your type implements `Debug`
}
Pulling the array out of the ArrayVec returns a Result to deal with the case where there weren't enough items to fill it; the case that was discussed in the other answers.
For your specific problem, Rust 1.55.0 allows you to directly map an array:
enum Foo {
Value(i32),
Nothing,
}
fn main() {
let bar = [1, 2, 3];
let foos = bar.map(Foo::Value);
}
In this case you can use Vec<Foo>:
#[derive(Debug)]
enum Foo {
Value(i32),
Nothing,
}
fn main() {
let bar = [1, 2, 3];
let foos = bar.iter().map(|&x| Foo::Value(x)).collect::<Vec<Foo>>();
println!("{:?}", foos);
}
.collect() builds data structures that can have arbitrary length, because the iterator's item number is not limited in general. (Shepmaster's answer already provides plenty details there).
One possibility to get data into an array from a mapped chain without allocating a Vec or similar is to bring mutable references to the array into the chain. In your example, that'd look like this:
#[derive(Debug, Clone, Copy)]
enum Foo {
Value(i32),
Nothing,
}
fn main() {
let bar = [1, 2, 3];
let mut foos = [Foo::Nothing; 3];
bar.iter().map(|x| Foo::Value(*x))
.zip(foos.iter_mut()).for_each(|(b, df)| *df = b);
}
The .zip() makes the iteration run over both bar and foos in lockstep -- if foos were under-allocated, the higher bars would not be mapped at all, and if it were over-allocated, it'd keep its original initialization values. (Thus also the Clone and Copy, they are needed for the [Nothing; 3] initialization).
You can actually define a Iterator trait extension to do this!
use std::convert::AsMut;
use std::default::Default;
trait CastExt<T, U: Default + AsMut<[T]>>: Sized + Iterator<Item = T> {
fn cast(mut self) -> U {
let mut out: U = U::default();
let arr: &mut [T] = out.as_mut();
for i in 0..arr.len() {
match self.next() {
None => panic!("Array was not filled"),
Some(v) => arr[i] = v,
}
}
assert!(self.next().is_none(), "Array was overfilled");
out
}
}
impl<T, U: Iterator<Item = T>, V: Default + AsMut<[T]>> CastExt<T, V> for U { }
fn main () {
let a: [i32; 8] = (0..8).map(|i| i * 2).cast();
println!("{:?}", a); // -> [0, 2, 4, 6, 8, 10, 12, 14]
}
Here's a playground link.
This isn't possible because arrays do not implement any traits. You can only collect into types which implement the FromIterator trait (see the list at the bottom of its docs).
This is a language limitation, since it's currently impossible to be generic over the length of an array and the length is part of its type. But, even if it were possible, it's very unlikely that FromIterator would be implemented on arrays because it'd have to panic if the number of items yielded wasn't exactly the length of the array.
You may combine arrays map method with Iterator::next.
Example:
fn iter_to_array<Element, const N: usize>(mut iter: impl Iterator<Item = Element>) -> [Element; N] {
// Here I use `()` to make array zero-sized -> no real use in runtime.
// `map` creates new array, which we fill by values of iterator.
let res = [(); N].map(|_| iter.next().unwrap());
// Ensure that iterator finished
assert!(matches!(iter.next(), None));
res
}
I ran into this problem myself — here's a workaround.
You can't use FromIterator, but you can iterate over the contents of a fixed-size object, or, if things are more complicated, indices that slice anything that can be accessed. Either way, mutation is viable.
For example, the problem I had was with an array of type [[usize; 2]; 4]:
fn main() {
// Some input that could come from another function and thus not be mutable
let pairs: [[usize; 2]; 4] = [[0, 0], [0, 1], [1, 1], [1, 0]];
// Copy mutable
let mut foo_pairs = pairs.clone();
for pair in foo_pairs.iter_mut() {
// Do some operation or other on the fixed-size contents of each
pair[0] += 1;
pair[1] -= 1;
}
// Go forth and foo the foo_pairs
}
If this is happening inside a small function, it's okay in my book. Either way, you were going to end up with a transformed value of identical type as the same one, so copying the whole thing first and then mutating is about the same amount of effort as referencing a value in a closure and returning some function of it.
Note that this only works if you plan to compute something that is going to be the same type, up to and including size/length. But that's implied by your use of Rust arrays. (Specifically, you could Value() your Foos or Nothing them as you like, and still be within type parameters for your array.)
On page 327 of Programming Rust you can find the following statement
However, unlike the into_iter() method, which takes the collection by value and consumes it, drain merely borrows a mutable references to the collection, and when the iterator is dropped, it removes any remaining elements from the collection, and leaves it empty.
I'm confused at what it means it says it removes any remaining elements from the collection? I can see with this code when the iterator is dropped the remaining elements from a are still there,
fn main() {
let mut a = vec![1, 2, 3, 4, 5];
{
let b: Vec<i32> = a.drain(0..3).collect();
}
println!("Hello, world! {:?}", a);
}
Perhaps I'm confused at merely the wording. Is there something more to this?
This looks like a bit imprecise wording.
The real meaning of these words is: if you drop the drain iterator without exhausting it, it will drop all the elements used for its creation. As you've asked it to use only the first three elements, it won't empty the entire vector, but rather the first part only; but it will do this even if unused:
fn main() {
let mut a = vec![1, 2, 3, 4, 5];
{
let _ = a.drain(0..3);
}
println!("Hello, world! {:?}", a);
}
Hello, world! [4, 5]
playground
You could understand this in the following way: the "collection" mentioned here is not the initial collection the drain was called on, but rather is "sub-collection", specified by the passed range.
Why does the following code work?
use std::rc::Rc;
fn main () {
let c = vec![1, 2, 3, 4, 5];
let r = Rc::new(c);
println!("{:?}", (**r)[0]);
}
I can understand it working with single deference (println!("{:?}", (*r)[0]);). But not able to understand it working with double-dereference too.
Both, Rc and Vec implements Deref, whichs deref-method is called with the *.
let c = vec![1, 2, 3, 4, 5];
creates a Vec with the given elements with the vec!-macro.
let r = Rc::new(c);
creates a Reference counted Object from the Vector. The Vector is moved into the RC.
println!("{:?}", (**r)[0]);
This one is a bit more tricky: *r dereferences the Rc, so we get the underlying Vector. *rc dereferences the Vector as a slice. slice[0] indexes the first element of the slice, which results in the first element 1. println! finally prints the result.
It might be easier to understand what happens once we build a function prototype around the expression (**r)[0]:
fn foo<T, U>(r: T) -> i32
where
T: Deref<Target=U>,
U: Deref<Target=[i32]>,
{
(**r)[0]
}
Playground
Rc<T>, as is typical for most smart containers in Rust, implements Deref so that it can be used as an ordinary reference to the underlying value. In turn, Vec<T> implements Deref so that it can be used as a slice (Target = [T]). The explicit dereferencing *, when performed twice, applies the two conversions in sequence.
Of course, usually you wouldn't need to do this, because Vec also implements the Index operator.
I have a boxed array of structs and I want to consume this array and insert it into a vector.
My current approach would be to convert the array into a vector, but the corresponding library function does not seem to work the way I expected.
let foo = Box::new([1, 2, 3, 4]);
let bar = foo.into_vec();
The compiler error states
no method named into_vec found for type Box<[_; 4]> in the current scope
I've found specifications here that look like
fn into_vec(self: Box<[T]>) -> Vec<T>
Converts self into a vector without clones or allocation.
... but I am not quite sure how to apply it. Any suggestions?
I think there's more cleaner way to do it. When you initialize foo, add type to it. Playground
fn main() {
let foo: Box<[u32]> = Box::new([1, 2, 3, 4]);
let bar = foo.into_vec();
println!("{:?}", bar);
}
The documentation you link to is for slices, i.e., [T], while what you have is an array of length 4: [T; 4].
You can, however, simply convert those, since an array of length 4 kinda is a slice. This works:
let foo = Box::new([1, 2, 3, 4]);
let bar = (foo as Box<[_]>).into_vec();
Can someone explain why this compiles:
fn main() {
let a = vec![1, 2, 3];
println!("{:?}", a[4]);
}
When running it, I got:
thread '' panicked at 'index out of bounds: the len is 3 but the index is 4', ../src/libcollections/vec.rs:1132
If you would like to access elements of the Vec with index checking, you can use the Vec as a slice and then use its get method. For example, consider the following code.
fn main() {
let a = vec![1, 2, 3];
println!("{:?}", a.get(2));
println!("{:?}", a.get(4));
}
This outputs:
Some(3)
None
In order to understand the issue, you have to think about it in terms of what the compiler sees.
Typically, a compiler never reasons about the value of an expression, only about its type. Thus:
a is of type Vec<i32>
4 is of an unknown integral type
Vec<i32> implements subscripting, so a[4] type checks
Having a compiler reasoning about values is not unknown, and there are various ways to get it.
you can allow evaluation of some expression at compile-time (C++ constexpr for example)
you can encode value into types (C++ non-type template parameters, using Peano's numbers)
you can use dependent typing which bridges the gap between types and values
Rust does not support any of these at this point in time, and while there has been interest for the former two it will certainly not be done before 1.0.
Thus, the values are checked at runtime, and the implementation of Vec correctly bails out (here failing).
Note that the following is a compile time error:
fn main() {
let a = [1, 2, 3];
println!("{:?}", a[4]);
}
error: this operation will panic at runtime
--> src/main.rs:3:22
|
3 | println!("{:?}", a[4]);
| ^^^^ index out of bounds: the length is 3 but the index is 4
|
= note: `#[deny(unconditional_panic)]` on by default
This works because without the vec!, the type is [i32; 3], which does actually carry length information.
With the vec!, it's now of type Vec<i32>, which no longer carries length information. Its length is only known at runtime.
Maybe what you mean is :
fn main() {
let a = vec![1, 2, 3];
println!("{:?}", a[4]);
}
This returns an Option so it will return Some or None. Compare this to:
fn main() {
let a = vec![1, 2, 3];
println!("{:?}", &a[4]);
}
This accesses by reference so it directly accesses the address and causes the panic in your program.