How can I make a Vec containing [0,1,2,3,4,3,2,1,0] using ranges?
I tried this:
let mut a = vec![0..len];
let b = vec![len..=0];
a.extend(&b);
But it raises an error:
type mismatch resolving `<&Vec<RangeInclusive<usize>> as IntoIterator>::Item == std::ops::Range<usize>`
There's no direct syntax that I know of to create a Vec directly from a range. A range is an first-class object and implements Iterator, so you can collect a range into a Vec.
Ranges must also be from low to high. To get a reversed range create a regular forward range and call rev to create an iterator that iterates in reverse.
let mut a = (0..len).collect::<Vec<_>>();
let b = (0..=len).rev();
a.extend(&b);
You'll notice that collect requires ::<>, the so-called turbofish operator. That's because it's a generic method that can convert an iterator into any type that implements FromIterator. There are many collection types that implement FromIterator, so we must tell it we want a Vec in particular.
Declaring a's type would have also worked:
let mut a: Vec<_> = (0..len).collect();
let b = (0..=len).rev();
a.extend(&b);
It's also possible to construct the vector in place by chaining the two iterators together and then collecting the result. This way a doesn't need to be mut.
let a = (0..len).chain((0..=len).rev())
.collect::<Vec<_>>();
fn main() {
let len = 4;
let mut a = (0..len).collect::<Vec<_>>();
let b = (0..=len).rev();
a.extend(b);
println!("{:?}", a); // should be [0, 1, 2, 3, 4, 3, 2, 1, 0]
}
Related
I have written a function which takes a generic parameter T with bound AsRef[i32].
Now I want to slice the input further inside my function with get method. But rust compiler would not let me use 1.. range to slice. I can use split_at method to split the slice. That will work. But my question is why can't I use array.as_ref().get([1..]) in this case? Do I need to add any other trait bounds to the generic type to make it work? If I do get with one index like array.as_ref().get(0) that works fine.
Here is my code -
fn find<T>(array: T, key: i32) -> Option<usize>
where
T: AsRef<[i32]>,
{
let arr = array.as_ref().get([1..]);
println!("slicing successful");
None
}
fn main() {
let arr = [1, 2, 3];
find(arr, 1);
}
Playground link.
You are confusing two syntax. The first one is the most commonly used to index a slice:
let arr = array.as_ref()[1..];
This is just syntax sugar for
let arr = array.as_ref().index(1..);
Note that for the second version to work, you need to have the std::ops::Index trait in scope.
This will not work as is because it returns a slice [i32], and [i32]: !Sized. Therefore you need to add a level of indirection:
let arr = &array.as_ref()[1..];
See the playground.
The second possible way is to use the get method of slices:
let arr = array.as_ref().get(1..);
See the playground.
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 have a vector of numbers and use the windows(2) method to create an iterator that gives me neighbouring pairs. For example, the vector [1, 2, 3] is transformed into [1, 2], [2, 3]. I want to use the find method to find a slice that fulfills a specific condition:
fn step(g: u64) -> Option<(u64, u64)> {
let prime_list: Vec<u64> = vec![2, 3, 5, 7]; //For example
if prime_list.len() < 2 {
return None;
}
let res = prime_list.windows(2).find(|&&[a, b]| b - a == g)?;
//...
None
}
I get an error:
error[E0005]: refutable pattern in function argument: `&&[]` not covered
--> src/lib.rs:6:43
|
6 | let res = prime_list.windows(2).find(|&&[a, b]| b - a == g)?;
| ^^^^^^^^ pattern `&&[]` not covered
I don't know what that error means: the list cannot have less than two elements, for example. Maybe the closure parameter is wrong? I tried to vary it but that didn't change anything. a and b are being properly detected as u64 in my IDE too. What is going on here?
You, the programmer, know that each iterated value will have a length of 2, but how do you know that? You can only tell that from the prose documentation of the function:
Returns an iterator over all contiguous windows of length size. The windows overlap. If the slice is shorter than size, the iterator returns no values.
Nowhere does the compiler know this information. The implementation of Windows only states that the iterated value will be a slice:
impl<'a, T> Iterator for Windows<'a, T> {
type Item = &'a [T];
}
I'd convert the slice into an array reference, discarding any slices that were the wrong length (which you know cannot happen):
use std::convert::TryFrom;
fn step(g: u64) -> Option<(u64, u64)> {
let prime_list: Vec<u64> = vec![2, 3, 5, 7]; // For example
if prime_list.len() < 2 {
return None;
}
let res = prime_list
.windows(2)
.flat_map(<&[u64; 2]>::try_from)
.find(|&&[a, b]| b - a == g)?;
//...
None
}
See also:
How to convert a slice into an array reference?
How can I find a subsequence in a &[u8] slice?
How do I imply the type of the value when there are no type parameters or ascriptions?
Alternatively, you could use an iterator of integers and chunk it up.
See also:
Are there equivalents to slice::chunks/windows for iterators to loop over pairs, triplets etc?
At some point in the future, const generics might be stabilized and allow baking the array length into the function call and the return type.
See also:
Is it possible to control the size of an array using the type parameter of a generic?
I want to access the element next to the maximal one in a Vec<i32>. I'm looking for something like this:
let v = vec![1, 3, 2];
let it = v.iter().max_element();
assert_eq!(Some(&2), it.next());
In C++, I would go with std::max_element and then just increase the iterator (with or without bounds checking, depending on how adventurous I feel at the moment). The Rust max only returns a reference to the element, which is not good enough for my use case.
The only solution I came up with is using enumerate to get the index of the item - but this seems manual and cumbersome when compared to the C++ way.
I would prefer something in the standard library.
This example is simplified - I actually want to attach to the highest value and then from that point loop over the whole container (possibly with cycle() or something similar).
C++ iterators are not the same as Rust iterators. Rust iterators are forward-only and can only be traversed once. C++ iterators can be thought of as cursors. See What are the main differences between a Rust Iterator and C++ Iterator? for more details.
In order to accomplish your goal in the most generic way possible, you have to walk through the entire iterator to find the maximum value. Along the way, you have to duplicate the iterator each time you find a new maximum value. At the end, you can return the iterator corresponding to the point after the maximum value.
trait MaxElement {
type Iter;
fn max_element(self) -> Self::Iter;
}
impl<I> MaxElement for I
where
I: Iterator + Clone,
I::Item: PartialOrd,
{
type Iter = Self;
fn max_element(mut self) -> Self::Iter {
let mut max_iter = self.clone();
let mut max_val = None;
while let Some(val) = self.next() {
if max_val.as_ref().map_or(true, |m| &val > m) {
max_iter = self.clone();
max_val = Some(val);
}
}
max_iter
}
}
fn main() {
let v = vec![1, 3, 2];
let mut it = v.iter().max_element();
assert_eq!(Some(&2), it.next());
}
See also:
How can I add new methods to Iterator?
I actually want to attach to the highest value and then from that point loop over the whole container (possibly with cycle() or something similar).
In that case, I'd attempt to be more obvious:
fn index_of_max(values: &[i32]) -> Option<usize> {
values
.iter()
.enumerate()
.max_by_key(|(_idx, &val)| val)
.map(|(idx, _val)| idx)
}
fn main() {
let v = vec![1, 3, 2];
let idx = index_of_max(&v).unwrap_or(0);
let (a, b) = v.split_at(idx);
let mut it = b.iter().chain(a).skip(1);
assert_eq!(Some(&2), it.next());
}
See also:
What's the fastest way of finding the index of the maximum value in an array?
Using max_by_key on a vector of floats
What is the idiomatic way to get the index of a maximum or minimum floating point value in a slice or Vec in Rust?
Find the item in an array with the largest property
a simple solution is to use fold,
the following code produces "largest num is: 99"
let vv:Vec<i32> = (1..100).collect();
let largest = vv.iter().fold(std::i32::MIN, |a,b| a.max(*b));
println!("largest {} ", largest);
If all you want is the value of the item following the maximum, I would do it with a simple call to fold, keeping track of the max found so far and the corresponding next value:
fn main() {
let v = vec![1, 3, 2];
let nxt = v.iter().fold (
(None, None),
|acc, x| {
match acc {
(Some (max), _) if x > max => (Some (x), None),
(Some (max), None) => (Some (max), Some (x)),
(None, _) => (Some (x), None),
_ => acc
}
}
).1;
assert_eq!(Some(&2), nxt);
}
playground
Depending on what you want to do with the items following the max, a similar approach may allow you to do it in a single pass.
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?