I can convert Vec<String> to Vec<&str> this way:
let mut items = Vec::<&str>::new();
for item in &another_items {
items.push(item);
}
Are there better alternatives?
There are quite a few ways to do it, some have disadvantages, others simply are more readable to some people.
This dereferences s (which is of type &String) to a String "right hand side reference", which is then dereferenced through the Deref trait to a str "right hand side reference" and then turned back into a &str. This is something that is very commonly seen in the compiler, and I therefor consider it idiomatic.
let v2: Vec<&str> = v.iter().map(|s| &**s).collect();
Here the deref function of the Deref trait is passed to the map function. It's pretty neat but requires useing the trait or giving the full path.
let v3: Vec<&str> = v.iter().map(std::ops::Deref::deref).collect();
This uses coercion syntax.
let v4: Vec<&str> = v.iter().map(|s| s as &str).collect();
This takes a RangeFull slice of the String (just a slice into the entire String) and takes a reference to it. It's ugly in my opinion.
let v5: Vec<&str> = v.iter().map(|s| &s[..]).collect();
This is uses coercions to convert a &String into a &str. Can also be replaced by a s: &str expression in the future.
let v6: Vec<&str> = v.iter().map(|s| { let s: &str = s; s }).collect();
The following (thanks #huon-dbaupp) uses the AsRef trait, which solely exists to map from owned types to their respective borrowed type. There's two ways to use it, and again, prettiness of either version is entirely subjective.
let v7: Vec<&str> = v.iter().map(|s| s.as_ref()).collect();
and
let v8: Vec<&str> = v.iter().map(AsRef::as_ref).collect();
My bottom line is use the v8 solution since it most explicitly expresses what you want.
The other answers simply work. I just want to point out that if you are trying to convert the Vec<String> into a Vec<&str> only to pass it to a function taking Vec<&str> as argument, consider revising the function signature as:
fn my_func<T: AsRef<str>>(list: &[T]) { ... }
instead of:
fn my_func(list: &Vec<&str>) { ... }
As pointed out by this question: Function taking both owned and non-owned string collections. In this way both vectors simply work without the need of conversions.
All of the answers idiomatically use iterators and collecting instead of a loop, but do not explain why this is better.
In your loop, you first create an empty vector and then push into it. Rust makes no guarantees about the strategy it uses for growing factors, but I believe the current strategy is that whenever the capacity is exceeded, the vector capacity is doubled. If the original vector had a length of 20, that would be one allocation, and 5 reallocations.
Iterating from a vector produces an iterator that has a "size hint". In this case, the iterator implements ExactSizeIterator so it knows exactly how many elements it will return. map retains this and collect takes advantage of this by allocating enough space in one go for an ExactSizeIterator.
You can also manually do this with:
let mut items = Vec::<&str>::with_capacity(another_items.len());
for item in &another_items {
items.push(item);
}
Heap allocations and reallocations are probably the most expensive part of this entire thing by far; far more expensive than taking references or writing or pushing to a vector when no new heap allocation is involved. It wouldn't surprise me if pushing a thousand elements onto a vector allocated for that length in one go were faster than pushing 5 elements that required 2 reallocations and one allocation in the process.
Another unsung advantage is that using the methods with collect do not store in a mutable variable which one should not use if it's unneeded.
another_items.iter().map(|item| item.deref()).collect::<Vec<&str>>()
To use deref() you must add using use std::ops::Deref
This one uses collect:
let strs: Vec<&str> = another_items.iter().map(|s| s as &str).collect();
Here is another option:
use std::iter::FromIterator;
let v = Vec::from_iter(v.iter().map(String::as_str));
Note that String::as_str is stable since Rust 1.7.
Related
I'm trying to create a 2D array from a Vec of 1D arrays using the ndarray crate. In the current implementation, I have Vec<Array1<u32>> as the collection of 1D arrays, and I'm having a hard time figuring out how to convert it to Array2<u32>. I've tried from_vec() on Vec<Array1<u32>> but it yielded Array1<Array1<u32>>. I thought of using the stack! macro, but I'm not sure how to call it on the above Vec. I'm using ndarray 0.12.1 and Rust 1.31.0.
I'm not hugely familiar with ndarray, but it looks like you have to flatten the data as an intermediate step and then rebuild from that. An iterator would probably have been more efficient but I don't see a method to build from an iterator that also lets you specify a shape.
It likely isn't the most performant way to to this, but it does at least work:
fn to_array2<T: Copy>(source: &[Array1<T>]) -> Result<Array2<T>, impl std::error::Error> {
let width = source.len();
let flattened: Array1<T> = source.into_iter().flat_map(|row| row.to_vec()).collect();
let height = flattened.len() / width;
flattened.into_shape((width, height))
}
Note that it can fail if the source arrays has different lengths. This solution is not 100% robust because it won't fail if one array is smaller but compensated by another array being longer. It is probably worth adding a check in there to prevent that, but I'll leave that to you.
Does Rust protect me from iterator invalidation here or am I just lucky with realloc? What guarantees are given for an iterator returned for &'a Vec<T>?
fn main() {
let mut v = vec![0; 2];
println!("capacity: {}", v.capacity());
{
let v_ref = &mut v;
for _each in v_ref.clone() {
for _ in 0..101 {
(*v_ref).push(1); // ?
}
}
}
println!("capacity: {}", v.capacity());
}
In Rust, most methods take an &self - a reference to self. In most circumstances, a call like some_string.len() internally "expands" to something like this:
let a: String = "abc".to_string();
let a_len: usize = String::len(&a); // This is identical to calling `a.len()`.
However, consider a reference to an object: a_ref, which is an &String that references a. Rust is smart enough to determine whether a reference needs to be added or removed, like we saw above (a becomes &a); In this case, a_ref.len() expands to:
let a: String = "abc".to_string();
let a_ref: &String = &a;
let a_len: usize = String::len(a_ref); // This is identical to calling `a_ref.len();`. Since `a_ref` is a reference already, it doesn't need to be altered.
Notice that this is basically equivalent to the original example, except that we're using an explicitly-set reference to a rather than a directly.
This means that v.clone() expands to Vec::clone(&v), and similarly, v_ref.clone() expands to Vec::clone(v_ref), and since v_refis &v (or, specifically, &mut v), we can simplify this back into Vec::clone(&v). In other words, these calls are equivalent - calling clone() on a basic reference (&) to an object does not clone the reference, it clones the referenced object.
In other words, Tamas Hedgeus' comment is correct: You are iterating over a new vector, which contains elements that are clones of the elements in v. The item being iterated over in your for loop is not a &Vec, it's a Vec that is separate from v, and therefore iterator invalidation is not an issue.
As for your question about the guarantees Rust provides, you'll find that Rust's borrow checker handles this rather well without any strings attached.
If you were to remove clone() from the for loop, though, you would receive an error message, use of moved value: '*v_ref', because v_ref is considered 'moved' into the for loop when you iterate over it, and cannot be used for the remainder of the function; to avoid this, the iter function creates an iterator object that only borrows the vector, allowing you to reuse the vector after the loop ends (and the iterator is dropped). And if you were to try iterating over and mutating v without the v_ref abstraction, the error reads cannot borrow 'v' as mutable because it is also borrowed as immutable. v is borrowed immutably within the iterator spawned by v.iter() (which has type signature of fn iter(&self) -> Iter<T> - note, it makes a borrow to the vector), and will not allow you to mutate the vector as a result of Rust's borrow checker, until the iterator is dropped (at the end of the for loop). However, since you can have multiple immutable references to a single object, you can still read from the vector within the for loop, just not write into it.
If you need to mutate an element of a vector while iterating over the vector, you can use iter_mut, which returns mutable references to one element at a time and lets you change that element only. You still cannot mutate the iterated vector itself with iter_mut, because Rust ensures that there is only one mutable reference to an object at a time, as well as ensuring there are no mutable references to an object in the same scope as immutable references to that object.
I'm looking for a method that consumes a Vec and returns one element, without the overhead of restoring Vec's invariants the way remove and swap_remove do:
fn take<T>(vec: Vec<T>, index: usize) -> Option<T>
However, I can't find such a method. Am I missing something? Is this actually unsafe or impossible?
This is a different question from Built in *safe* way to move out of Vec<T>?
There the goal was a remove method that didn't panic on out of bounds access and returned a Result. I'm looking for a method that consumes a Vec and returns one of the elements. None of the answers to the above question address my question.
You can write your function like this:
fn take<T>(mut vec: Vec<T>, index: usize) -> Option<T> {
if vec.get(index).is_none() {
None
} else {
Some(vec.swap_remove(index))
}
}
The code you see here (get and swap_remove) is guaranteed O(1).
However, kind of hidden, vec is dropped at the end of the function and this drop operation is likely not O(1), but O(n) (where n is vec.len()). If T implements Drop, then drop() is called for every element still inside the vector, meaning dropping the vector is guaranteed O(n). If T does not implement Drop, then the Vec only needs to deallocate the memory. The time complexity of the dealloc operation depends on the allocator and is not specified, so we cannot assume it is O(1).
To mention another solution using iterators:
fn take<T>(vec: Vec<T>, index: usize) -> Option<T> {
vec.into_iter().nth(index)
}
I was about to write this:
While Iterator::nth() usually is a linear time operation, the iterator over a vector overrides this method to make it a O(1) operation.
But then I noticed, that this is only true for the iterator which iterates over slices. The std::vec::IntoIter iterator which would be used in the code above, doesn't override nth(). It has been attempted here, but it doesn't seem to be that easy.
So, as of right now, the iterator solution above is a O(n) operation! Not to mention the time needed to drop the vector, as explained above.
The reason fn take<T>(vec: Vec<T>, index: usize) -> Option<T> does not exist in the standard library is that it is not very useful in general. For example, supposing that you have a Vec<String> of length 10, it means throwing away 9 strings and only using 1. This seems wasteful.
In general, the standard library will try to provide an API that is useful in a maximum of scenarios, and in this instance it would be more logical to have a fn take<T>(vec: &mut Vec<T>, index: usize) -> Option<T>.
The only question is how to preserve the invariant, of course:
it can be preserved by exchanging with the last element, which is what Vec::swap_remove does,
it can be preserved by shifting the successor elements in, which is what Vec::drain does.
Those are very flexible, and can be adapted to fill more specific scenarios, such as yours.
Adapting swap_remove:
fn take<T>(mut vec: Vec<T>, index: usize) -> Option<T> {
if index < vec.len() {
Some(vec.swap_remove(index))
} else {
None
}
}
Adapting drain:
fn take<T>(mut vec: Vec<T>, index: usize) -> Option<T> {
if index < vec.len() {
vec.drain(index..index+1).next()
} else {
None
}
}
Noting that the former is more efficient: it's O(1).
I'm looking for a method that consumes the Vec and returns one element, without the overhead of restoring Vec's invariants the way remove and swap_remove do.
This reeks of premature micro-optimization to me.
First of all, note that it is necessary to destroy the elements of the vector; you can accomplish this in two ways:
swap_remove, then iterate over each element to destroy them,
Iterate over each element to destroy them, skipping the specific index.
It is not clear to me that the latter would be faster than the former; if anything it looks more complicated, with more branches (I advise two loops), which may throw off the predictor and may be less amenable to vectorization.
Secondly, before complaining about the overhead of restoring the Vec's invariant, have you properly profiled the solution?
If we look at the swap_remove variant, there are 3 steps:
swap_remove (O(1)),
destroy each remaining element (O(N)),
free the backing memory.
Step 2 may be optimized out if the element has no Drop implementation, but otherwise I would be it's a toss whether (2) or (3) is dominating the cost.
TL;DR: I am afraid that you are fighting ghost issues, profile before trying to optimize.
I have a slice of &[u8] and I'd like to append it to a Vec<u8> with minimal copying. Here are two approaches that I know work:
let s = [0u8, 1u8, 2u8];
let mut v = Vec::new();
v.extend(s.iter().map(|&i| i));
v.extend(s.to_vec().into_iter()); // allocates an extra copy of the slice
Is there a better way to do this in Rust stable? (rustc 1.0.0-beta.2)
There's a method that does exactly this: Vec::extend_from_slice
Example:
let s = [0u8, 1, 2];
let mut v = Vec::new();
v.extend_from_slice(&s);
v.extend(s.iter().cloned());
That is effectively equivalent to using .map(|&i| i) and it does minimal copying.
The problem is that you absolutely cannot avoid copying in this case. You cannot simply move the values because a slice does not own its contents, thus it can only take a copy.
Now, that said, there are two things to consider:
Rust tends to inline rather aggressively; there is enough information in this code for the compiler to just copy the values directly into the destination without any intermediate step.
Closures in Rust aren't like closures in most other languages: they don't require heap allocation and can be directly inlined, thus making them no less efficient than hard-coding the behaviour directly.
Do keep in mind that the above two are dependent on optimisation: they'll generally work out for the best, but aren't guaranteed.
But having said that... what you're actually trying to do here in this specific example is append a stack-allocated array which you do own. I'm not aware of any library code that can actually take advantage of this fact (support for array values is rather weak in Rust at the moment), but theoretically, you could effectively create an into_iter() equivalent using unsafe code... but I don't recommend it, and it's probably not worth the hassle.
I can't speak for the full performance implications, but v + &s will work on beta, which I believe is just similar to pushing each value onto the original Vec.
I need to rewrite these C declarations in Go and Rust for a set of practice problems I am working on. I figured out the Go part, but I am having trouble with the Rust part. Any ideas or help to write these in Rust?
double *a[n];
double (*b)[n];
double (*c[n])();
double (*d())[n];
Assuming n is a constant:
let a: [*mut f64, ..n]; // double *a[n];
let b: *mut [f64, ..n]; // double (*b)[n];
let c: [fn() -> f64, ..n]; // double (*c[n])();
fn d() -> *mut [f64, ..n]; // double (*d())[n];
These are rather awkward and unusual types in any language. Rust's syntax, however, makes these declarations a lot easier to read than C's syntax does.
Note that d in C is a function declaration. In Rust, external function declarations are only allowed in extern blocks (see the FFI guide).
The answer depends on what, exactly, the * is for. For example, is the first one being used as an array of pointers to doubles, or is it an array of arrays of doubles? Are the pointers nullable or not?
Also, is n a constant or not? If it is, then you want an array; if it's not, you want a Vec.
Also also, are these global or local declarations? Are they function arguments? There's different syntax involved for each.
Frankly, without more context, it's impossible to answer this question with any accuracy. Instead, I will give you the following:
The Rust documentation contains all the information you'll need, although it's spread out a bit. Check the reference and any appropriate-looking guides. The FFI Guide is probably worth looking at.
cdecl is a website that will unpick C declarations if that's the part you're having difficulty with. Just note that you'll have to remove the semicolon and the n or it won't parse.
The floating point types in Rust are f32 and f64, depending on whether you're using float or double. Also, don't get caught: int in Rust is not equivalent to int in C. Prefer explicitly-sized types like i32 or u64, or types from libc like c_int. int and uint should only be used with explicitly pointer-sized values.
Normally, you'd write a reference to a T as &T or &mut T, depending on desired mutability (default in C is mutable, default in Rust is immutable).
If you want a nullable reference, use Option<&T>.
If you are trying to use these in a context where you start getting complaints about needing "lifetimes"... well, you're just going to have to learn the language. At that point, simple translation isn't going to work very well.
In Rust, array types are written as brackets around the element type. So an "array of doubles" would be [f64], an array of size n would be [f64, ..n]. Typically, however, the actual equivalent to, say, double[] in C would be &[f64]; that is, a reference to an array, rather then the actual contents of the array.
Use of "raw pointers" is heavily discouraged in Rust, and you cannot use them meaningfully outside of unsafe code. In terms of syntax, a pointer to T is *const T or *mut T, depending on whether it's a pointer to constant or mutable data.
Function pointers are just written as fn (Args...) -> Result. So a function that takes nothing and returns a double would be fn () -> f64.