I have a struct
struct Foo {
foo1: String,
foo2: String,
foo3: String,
foo4: String,
// ...
}
I would like to create an instance of Foo from a vector.
let x = vec!["a".to_string(), "b".to_string(), "c".to_string(), "d".to_string()];
match x.as_slice() {
&[ref a, ref b, ref c, ref d] => {
let foo = Foo {
foo1: a.to_string(),
foo2: b.to_string(),
foo3: c.to_string(),
foo4: d.to_string(),
};
},
_ => unreachable!(),
}
Do I have to copy the strings? Is there any better way to destructure the vector into a, b, c, d as well as transferring the ownership?
Actually, I don't mind x is completely destroyed after the destructuring. So I hope there is a pattern match for vectors apart from slices as well. For now it seems we can only destructure slices.
Do I have to copy the strings?
Not if you are willing to give up destructuring. I'm a big fan of itertools:
use itertools::Itertools; // 0.8.2
fn main() {
let x = vec![
"a".to_string(),
"b".to_string(),
"c".to_string(),
"d".to_string(),
];
if let Some((foo1, foo2, foo3, foo4)) = x.into_iter().tuples().next() {
let foo = Foo {
foo1,
foo2,
foo3,
foo4,
};
}
}
This transfers ownership of the vector (and thus the members) to an iterator, then the tuples adapter chunks up the values into a tuple. We take the first one of those and construct the value.
You could also use drain if you didn't want to give up ownership of the entire vector:
if let Some((foo1, foo2, foo3, foo4)) = x.drain(..4).tuples().next() {
Is there any better way to destructure the vector into a, b, c, d as well as transferring the ownership?
No, there is no mechanism to take ownership of a part of a Vec without creating another Vec (or another type that has the same limits) except for an iterator.
Destructuring slices isn't stable, and you can't move out of a slice because it's just a borrow — if you moved out, what would the Vec's destructor do?
Mutating the vector is the way to go here:
let mut x = vec!["a".to_string(), "b".to_string(), "c".to_string(), "d".to_string()];
let foo = Foo {
foo4: x.pop().unwrap(),
foo3: x.pop().unwrap(),
foo2: x.pop().unwrap(),
foo1: x.pop().unwrap(),
};
println!("{:?}", foo);
playground
If you're willing to use nightly this is a one liner with #![feature(box_patterns)]:
#![feature(box_patterns)]
let x = vec!["a".to_string(), "b".to_string(), "c".to_string(), "d".to_string()];
let Ok(box [a, b, c, d]) = <Box<[String; 4]>>::try_from(x) else { unreachable!() };
Note on stability:
Do be aware that these box patterns will likely be replaced with a more general "deref" patterns in the future and thus most likely will never be stabilized as-is.
Whether those will allow movement of ownership is unknown, due to there not existing any kind of DerefInto trait, only Deref and DerefMut which borrow.
I expect, however, that the box_patterns feature won't be removed until there exists another way to do this. (Or at least, until another feature replaces it that can provide the same feature), so you can (mostly) safely use this feature and not be locked out in the future. In any case you can use one of the other answers to re-implement this should that happen.
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.)
still new to Rust and I am learning about the mutability rules of the language. I am trying to understand if structures, objects, arrays are deeply immutable.
So I have this struct
struct Foo {
bar: String,
}
And I would like to see what the rules for deep mutability are. When creating the struct, the struct must be defined as mut in order for it to have properties changed on it.
let mut f = Foo{
bar: String::from("hello")
};
f.bar = String::from("foo"); // Must be mut to change a property
println!("{:?}", f.bar);
However when I define an immutable struct and move it into a mutable vector, the struct becomes mutable.
let f = Foo{
bar: String::from("hello")
};
let mut strings: Vec<Foo> = vec![f];
strings[0].bar = String::from("foo");
println!("{:?}", strings[0].bar);
I am guessing that f is being copied into the mutable vector (and dropped from it's scope?) and the copy is being reassigned as mutable but I am not really sure.
Can I get some insight into what is going on?
mut is not a property of the value. It is a property of the binding (I am specifically not talking about mutable references, i.e. &mut, only let mut).
let v means "I will not attempt to change the value v contains through v, either directly or indirectly".
But once it's no longer within v, i.e. it was moved, it's fine to change it.
The following code, thus, compiles successfully:
let v = Foo { bar: String::new() };
let mut mutable_v = v;
mutable_v.bar = "abc".to_owned();
From the moment we moved the value out of v, the mutability of v is not relevant anymore.
And in the same manner, from the moment we moved the value out of f and into the vec strings, the mutability of f doesn't matter anymore, but the mutability of strings - because we don't change the value through f but through strings.
I am trying to flatten a vector of Enum in Rust, but I am having some issues:
enum Foo {
A(i32),
B(i32, i32),
}
fn main() {
let vf = vec![Foo::A(1), Foo::A(2), Foo::B(3, 4)];
let vi: Vec<i32> = vf
.iter()
.map(|f| match f {
Foo::A(i) => [i].into_iter(),
Foo::B(i, j) => [i, j].into_iter(),
})
.collect(); // this does not compile
// I want vi = [1, 2, 3, 4]. vf must still be valid
}
I could just use a regular for loop and insert elements into an existing vector, but that would be no fun. I'd like to know if there is a more idiomatic Rust way of doing it.
Here's a way to do it that produces an iterator (rather than necessarily a vector, as the fold() based solution does).
use std::iter::once;
enum Foo {
A(i32),
B(i32, i32),
}
fn main() {
let vf = vec![Foo::A(1), Foo::A(2), Foo::B(3, 4)];
let vi: Vec<i32> = vf
.iter()
.flat_map(|f| {
match f {
&Foo::A(i) => once(i).chain(None),
&Foo::B(i, j) => once(i).chain(Some(j)),
}
})
.collect();
dbg!(vi);
}
This does essentially the same thing that you were attempting, but in a way which will succeed. Here are the parts I changed, in the order they appear in the code:
I used .flat_map() instead of .map(). flat_map accepts a function which returns an iterator and produces the elements of that iterator ("flattening") whereas .map() would have just given the iterator.
I used & in the match patterns. This is because, since you are using .iter() on the vector (which is appropriate for your requirement “vf must still be valid”), you have references to enums, and pattern matching on a reference to an enum will normally give you references to its elements, but we almost certainly want to handle the i32s by value instead. There are several other things I could have done, such as using the * dereference operator on the values instead, but this is concise and tidy.
You tried to .into_iter() an array. Unfortunately, in current Rust this does not do what you want and you can't actually return that iterator, for somewhat awkward reasons (which will be fixed in an upcoming Rust version). And then, if it did mean what you wanted, then you'd get an error because the two match arms have unequal types — one is an iterator over [i32; 1] and the other is an iterator over [i32; 2].
Instead, you need to build two possible iterators which are clearly of the same type. There are lots of ways to do this, and the way I picked was to use Iterator::chain to combine once(i), an iterator that returns the single element i, with an Option<i32> (which implements IntoIterator) that contains the second element j if it exists.
Notice that in the first match arm I wrote the seemingly useless expression .chain(None); this is so that the two arms have the same type. Another way to write the same thing, which is arguably clearer since it doesn't duplicate code that has to be identical, is:
let (i, opt_j) = match f {
&Foo::A(i) => (i, None),
&Foo::B(i, j) => (i, Some(j)),
};
once(i).chain(opt_j)
In either case, the iterator's type is std::iter::Chain<std::iter::Once<i32>, std::option::IntoIter<i32>> — you don't need to know this exactly, just notice that there must be a type which handles both the A(i) and the B(i, j) cases.
First of all, you need to change the i32 references to owned values by e.g. dereferencing them. Then you can circumvent proxying through inlined arrays by using fold():
enum Foo {
A(i32),
B(i32, i32),
}
fn main() {
let vf = vec![Foo::A(1), Foo::A(2), Foo::B(3, 4)];
let vi: Vec<i32> = vf
.iter()
.fold(Vec::new(), |mut acc, f| {
match f {
Foo::A(i) => acc.push(*i),
Foo::B(i, j) => {
acc.push(*i);
acc.push(*j);
}
}
acc
});
}
I have the following:
enum SomeType {
VariantA(String),
VariantB(String, i32),
}
fn transform(x: SomeType) -> SomeType {
// very complicated transformation, reusing parts of x in order to produce result:
match x {
SomeType::VariantA(s) => SomeType::VariantB(s, 0),
SomeType::VariantB(s, i) => SomeType::VariantB(s, 2 * i),
}
}
fn main() {
let mut data = vec![
SomeType::VariantA("hello".to_string()),
SomeType::VariantA("bye".to_string()),
SomeType::VariantB("asdf".to_string(), 34),
];
}
I would now like to call transform on each element of data and store the resulting value back in data. I could do something like data.into_iter().map(transform).collect(), but this will allocate a new Vec. Is there a way to do this in-place, reusing the allocated memory of data? There once was Vec::map_in_place in Rust but it has been removed some time ago.
As a work-around, I've added a Dummy variant to SomeType and then do the following:
for x in &mut data {
let original = ::std::mem::replace(x, SomeType::Dummy);
*x = transform(original);
}
This does not feel right, and I have to deal with SomeType::Dummy everywhere else in the code, although it should never be visible outside of this loop. Is there a better way of doing this?
Your first problem is not map, it's transform.
transform takes ownership of its argument, while Vec has ownership of its arguments. Either one has to give, and poking a hole in the Vec would be a bad idea: what if transform panics?
The best fix, thus, is to change the signature of transform to:
fn transform(x: &mut SomeType) { ... }
then you can just do:
for x in &mut data { transform(x) }
Other solutions will be clunky, as they will need to deal with the fact that transform might panic.
No, it is not possible in general because the size of each element might change as the mapping is performed (fn transform(u8) -> u32).
Even when the sizes are the same, it's non-trivial.
In this case, you don't need to create a Dummy variant because creating an empty String is cheap; only 3 pointer-sized values and no heap allocation:
impl SomeType {
fn transform(&mut self) {
use SomeType::*;
let old = std::mem::replace(self, VariantA(String::new()));
// Note this line for the detailed explanation
*self = match old {
VariantA(s) => VariantB(s, 0),
VariantB(s, i) => VariantB(s, 2 * i),
};
}
}
for x in &mut data {
x.transform();
}
An alternate implementation that just replaces the String:
impl SomeType {
fn transform(&mut self) {
use SomeType::*;
*self = match self {
VariantA(s) => {
let s = std::mem::replace(s, String::new());
VariantB(s, 0)
}
VariantB(s, i) => {
let s = std::mem::replace(s, String::new());
VariantB(s, 2 * *i)
}
};
}
}
In general, yes, you have to create some dummy value to do this generically and with safe code. Many times, you can wrap your whole element in Option and call Option::take to achieve the same effect .
See also:
Change enum variant while moving the field to the new variant
Why is it so complicated?
See this proposed and now-closed RFC for lots of related discussion. My understanding of that RFC (and the complexities behind it) is that there's an time period where your value would have an undefined value, which is not safe. If a panic were to happen at that exact second, then when your value is dropped, you might trigger undefined behavior, a bad thing.
If your code were to panic at the commented line, then the value of self is a concrete, known value. If it were some unknown value, dropping that string would try to drop that unknown value, and we are back in C. This is the purpose of the Dummy value - to always have a known-good value stored.
You even hinted at this (emphasis mine):
I have to deal with SomeType::Dummy everywhere else in the code, although it should never be visible outside of this loop
That "should" is the problem. During a panic, that dummy value is visible.
See also:
How can I swap in a new value for a field in a mutable reference to a structure?
Temporarily move out of borrowed content
How do I move out of a struct field that is an Option?
The now-removed implementation of Vec::map_in_place spans almost 175 lines of code, most of having to deal with unsafe code and reasoning why it is actually safe! Some crates have re-implemented this concept and attempted to make it safe; you can see an example in Sebastian Redl's answer.
You can write a map_in_place in terms of the take_mut or replace_with crates:
fn map_in_place<T, F>(v: &mut [T], f: F)
where
F: Fn(T) -> T,
{
for e in v {
take_mut::take(e, f);
}
}
However, if this panics in the supplied function, the program aborts completely; you cannot recover from the panic.
Alternatively, you could supply a placeholder element that sits in the empty spot while the inner function executes:
use std::mem;
fn map_in_place_with_placeholder<T, F>(v: &mut [T], f: F, mut placeholder: T)
where
F: Fn(T) -> T,
{
for e in v {
let mut tmp = mem::replace(e, placeholder);
tmp = f(tmp);
placeholder = mem::replace(e, tmp);
}
}
If this panics, the placeholder you supplied will sit in the panicked slot.
Finally, you could produce the placeholder on-demand; basically replace take_mut::take with take_mut::take_or_recover in the first version.
I'm trying to sort an array with a map() over an iterator.
struct A {
b: Vec<B>,
}
#[derive(PartialEq, Eq, PartialOrd, Ord)]
struct B {
c: Vec<i32>,
}
fn main() {
let mut a = A { b: Vec::new() };
let b = B { c: vec![5, 2, 3] };
a.b.push(b);
a.b.iter_mut().map(|b| b.c.sort());
}
Gives the warning:
warning: unused `std::iter::Map` that must be used
--> src/main.rs:16:5
|
16 | a.b.iter_mut().map(|b| b.c.sort());
| ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
|
= note: #[warn(unused_must_use)] on by default
= note: iterators are lazy and do nothing unless consumed
Which is true, sort() isn't actually called here. This warning is described in the book, but I don't understand why this variation with iter_mut() works fine:
a.b.iter_mut().find(|b| b == b).map(|b| b.c.sort());
As the book you linked to says:
If you are trying to execute a closure on an iterator for its side effects, use for instead.
That way it works, and it's much clearer to anyone reading the code. You should use map when you want to transform a vector to a different one.
I don't understand why this variation with iter_mut() works fine:
a.b.iter_mut().find(|b| b == b).map(|b| b.c.sort());
It works because find is not lazy; it's an iterator consumer. It returns an Option not an Iterator. This might be why it is confusing you, because Option also has a map method, which is what you are using here.
As others have said, map is intended for transforming data, without modifying it and without any other side-effects. If you really want to use map, you can map over the collection and assign it back:
fn main() {
let mut a = A { b: Vec::new() };
let mut b = B { c: vec![5, 2, 3] };
a.b.push(b);
a.b =
a.b.into_iter()
.map(|mut b| {
b.c.sort();
b
})
.collect();
}
Note that vector's sort method returns (), so you have to explicitly return the sorted vector from the mapping function.
I use for_each.
According to the doc:
It is equivalent to using a for loop on the iterator, although break and continue are not possible from a closure. It's generally more idiomatic to use a for loop, but for_each may be more legible when processing items at the end of longer iterator chains. In some cases for_each may also be faster than a loop, because it will use internal iteration on adaptors like Chain.