Instantiating a struct with stdin data in Rust - closures

I am very, very new to Rust and trying to implement some simple things to get the feel for the language. Right now, I'm stumbling over the best way to implement a class-like struct that involves casting a string to an int. I'm using a global-namespaced function and it feels wrong to my Ruby-addled brain.
What's the Rustic way of doing this?
use std::io;
struct Person {
name: ~str,
age: int
}
impl Person {
fn new(input_name: ~str) -> Person {
Person {
name: input_name,
age: get_int_from_input(~"Please enter a number for age.")
}
}
fn print_info(&self) {
println(fmt!("%s is %i years old.", self.name, self.age));
}
}
fn get_int_from_input(prompt_message: ~str) -> int {
println(prompt_message);
let my_input = io::stdin().read_line();
let my_val =
match from_str::<int>(my_input) {
Some(number_string) => number_string,
_ => fail!("got to put in a number.")
};
return my_val;
}
fn main() {
let first_person = Person::new(~"Ohai");
first_person.print_info();
}
This compiles and has the desired behaviour, but I am at a loss for what to do here--it's obvious I don't understand the best practices or how to implement them.
Edit: this is 0.8

Here is my version of the code, which I have made more idiomatic:
use std::io;
struct Person {
name: ~str,
age: int
}
impl Person {
fn print_info(&self) {
println!("{} is {} years old.", self.name, self.age);
}
}
fn get_int_from_input(prompt_message: &str) -> int {
println(prompt_message);
let my_input = io::stdin().read_line();
from_str::<int>(my_input).expect("got to put in a number.")
}
fn main() {
let first_person = Person {
name: ~"Ohai",
age: get_int_from_input("Please enter a number for age.")
};
first_person.print_info();
}
fmt!/format!
First, Rust is deprecating the fmt! macro, with printf-based syntax, in favor of format!, which uses syntax similar to Python format strings. The new version, Rust 0.9, will complain about the use of fmt!. Therefore, you should replace fmt!("%s is %i years old.", self.name, self.age) with format!("{} is {} years old.", self.name, self.age). However, we have a convenience macro println!(...) that means exactly the same thing as println(format!(...)), so the most idiomatic way to write your code in Rust would be
println!("{} is {} years old.", self.name, self.age);
Initializing structs
For a simple type like Person, it is idiomatic in Rust to create instances of the type by using the struct literal syntax:
let first_person = Person {
name: ~"Ohai",
age: get_int_from_input("Please enter a number for age.")
};
In cases where you do want a constructor, Person::new is the idiomatic name for a 'default' constructor (by which I mean the most commonly used constructor) for a type Person. However, it would seem strange for the default constructor to require initialization from user input. Usually, I think you would have a person module, for example (with person::Person exported by the module). In this case, I think it would be most idiomatic to use a module-level function fn person::prompt_for_age(name: ~str) -> person::Person. Alternatively, you could use a static method on Person -- Person::prompt_for_age(name: ~str).
&str vs. ~str in function parameters
I've changed the signature of get_int_from_input to take a &str instead of ~str. ~str denotes a string allocated on the exchange heap -- in other words, the heap that malloc/free in C, or new/delete in C++ operate on. Unlike in C/C++, however, Rust enforces the requirement that values on the exchange heap can only be owned by one variable at a time. Therefore, taking a ~str as a function parameter means that the caller of the function can't reuse the ~str argument that it passed in -- it would have to make a copy of the ~str using the .clone method.
On the other hand, &str is a slice into the string, which is just a reference to a range of characters in the string, so it doesn't require a new copy of the string to be allocated when a function with a &str parameter is called.
The reason to use &str rather than ~str for prompt_message in get_int_from_input is that the function doesn't need to hold onto the message past the end of the function. It only uses the prompt message in order to print it (and println takes a &str, not a ~str). Once you change the function to take &str, you can call it like get_int_from_input("Prompt") instead of get_int_from_input(~"Prompt"), which avoids the unnecessary allocation of "Prompt" on the heap (and similarly, you can avoid having to clone s in the code below):
let s: ~str = ~"Prompt";
let i = get_int_from_input(s.clone());
println(s); // Would complain that `s` is no longer valid without cloning it above
// if `get_int_from_input` takes `~str`, but not if it takes `&str`.
Option<T>::expect
The Option<T>::expect method is the idiomatic shortcut for the match statement you have, where you want to either return x if you get Some(x) or fail with a message if you get None.
Returning without return
In Rust, it is idiomatic (following the example of functional languages like Haskell and OCaml) to return a value without explicitly writing a return statement. In fact, the return value of a function is the result of the last expression in the function, unless the expression is followed by a semicolon (in which case it returns (), a.k.a. unit, which is essentially an empty placeholder value -- () is also what is returned by functions without an explicit return type, such as main or print_info).
Conclusion
I'm not a great expert on Rust by any means. If you want help on anything related to Rust, you can try, in addition to Stack Overflow, the #rust IRC channel on irc.mozilla.org or the Rust subreddit.

This isn't really rust-specifc, but try to split functionality into discrete units. Don't mix the low-level tasks of putting strings on the terminal and getting strings from the terminal with the more directly relevant (and largely implementation dependent) tasks of requesting a value, and verify it. When you do that, the design decisions you should make start to arise on their own.
For instance, you could write something like this (I haven't compiled it, and I'm new to rust myself, so they're probably at LEAST one thing wrong with this :) ).
fn validated_input_prompt<T>(prompt: ~str) {
println(prompt);
let res = io::stdin().read_line();
loop {
match res.len() {
s if s == 0 => { continue; }
s if s > 0 {
match T::from_str(res) {
Some(t) -> {
return t
},
None -> {
println("ERROR. Please try again.");
println(prompt);
}
}
}
}
}
}
And then use it as:
validated_input_prompt<int>("Enter a number:")
or:
validated_input_prompt<char>("Enter a Character:")
BUT, to make the latter work, you'd need to implement FromStr for chars, because (sadly) rust doesn't seem to do it by default. Something LIKE this, but again, I'm not really sure of the rust syntax for this.
use std::from_str::*;
impl FromStr for char {
fn from_str(s: &str) -> Option<Self> {
match len(s) {
x if x >= 1 => {
Option<char>.None
},
x if x == 0 => {
None,
},
}
return s[0];
}
}

A variation of telotortium's input reading function that doesn't fail on bad input. The loop { ... } keyword is preferred over writing while true { ... }. In this case using return is fine since the function is returning early.
fn int_from_input(prompt: &str) -> int {
println(prompt);
loop {
match from_str::<int>(io::stdin().read_line()) {
Some(x) => return x,
None => println("Oops, that was invalid input. Try again.")
};
}
}

Related

Peeking at stdin using match

I'm trying to port a translator/parser example from an old compiler textbook from C into Rust.
I have the following code:
use std::io::Read;
fn lexan() {
let mut input = std::io::stdin().bytes().peekable();
loop {
match input.peek() {
Some(&ch) => {
match ch {
_ => println!("{:?}", input.next()),
}
}
None => break,
}
}
}
At this point I'm not actively trying to parse the input, just get my head around how match works. The aim is to add parse branches to the inner match. Unfortunately this fails to compile because I appear to fail in understanding the semantics of match:
error[E0507]: cannot move out of borrowed content
--> src/main.rs:7:18
|
7 | Some(&ch) => {
| ^--
| ||
| |hint: to prevent move, use `ref ch` or `ref mut ch`
| cannot move out of borrowed content
From what I understand, this error is because I don't own the return value of the match. The thing is, I don't believe that I'm using the return value of either match. I thought perhaps input.next() may have been the issue, but the same error occurs with or without this part (or indeed, the entire println! call).
What am I missing here? It's been some time since I looked at Rust (and never in a serious level of effort), and most of the search results for things of this nature appear to be out of date.
It's got nothing to do with the return value of match, or even match itself::
use std::io::Read;
fn lexan() {
let mut input = std::io::stdin().bytes().peekable();
if let Some(&ch) = input.peek() {}
}
The issue is that you are attempting to bind the result of Peekable::peek while dereferencing it (that's what the & in &ch does). In this case, the return type is an Option<&Result<u8, std::io::Error>> because the Bytes iterator returns errors from the underlying stream. Since this type does not implement Copy, trying to dereference the type requires that you transfer ownership of the value. You cannot do so as you don't own the original value — thus the error message.
The piece that causes the inability to copy is the error type of the Result. Because of that, you can match one level deeper:
match input.peek() {
Some(&Ok(ch)) => {
match ch {
_ => println!("{:?}", input.next()),
}
}
Some(&Err(_)) => panic!(),
None => break,
}
Be aware that this code is pretty close to being uncompilable though. The result of peek will be invalidated when next is called, so many small changes to this code will trigger the borrow checker to fail the code. I'm actually a bit surprised the above worked on the first go.
If you didn't care about errors at all, you could do
while let Some(&Ok(ch)) = input.peek() {
match ch {
_ => println!("{:?}", input.next()),
}
}
Unfortunately, you can't split the middle, as this would cause the borrow of input to last during the call to next:
while let Some(x) = input.peek() {
match *x {
Ok(ch) => {
match ch {
_ => println!("{:?}", input.next()),
}
}
Err(_) => {}
}
// Could still use `x` here, compiler doesn't currently see that we don't
}

What is a macro for concatenating an arbitrary number of components to build a path in Rust?

In Python, a function called os.path.join() allows concatenating multiple strings into one path using the path separator of the operating system. In Rust, there is only a function join() that appends a string or a path to an existing path. This problem can't be solved with a normal function as a normal function needs to have a fixed number of arguments.
I'm looking for a macro that takes an arbitrary number of strings and paths and returns the joined path.
There's a reasonably simple example in the documentation for PathBuf:
use std::path::PathBuf;
let path: PathBuf = [r"C:\", "windows", "system32.dll"].iter().collect();
Once you read past the macro syntax, it's not too bad. Basically, we take require at least two arguments, and the first one needs to be convertible to a PathBuf via Into. Each subsequent argument is pushed on the end, which accepts anything that can be turned into a reference to a Path.
macro_rules! build_from_paths {
($base:expr, $($segment:expr),+) => {{
let mut base: ::std::path::PathBuf = $base.into();
$(
base.push($segment);
)*
base
}}
}
fn main() {
use std::{
ffi::OsStr,
path::{Path, PathBuf},
};
let a = build_from_paths!("a", "b", "c");
println!("{:?}", a);
let b = build_from_paths!(PathBuf::from("z"), OsStr::new("x"), Path::new("y"));
println!("{:?}", b);
}
A normal function which takes an iterable (e.g. a slice) can solve the problem in many contexts:
use std::path::{Path, PathBuf};
fn join_all<P, Ps>(parts: Ps) -> PathBuf
where
Ps: IntoIterator<Item = P>,
P: AsRef<Path>,
{
parts.into_iter().fold(PathBuf::new(), |mut acc, p| {
acc.push(p);
acc
})
}
fn main() {
let parts = vec!["/usr", "bin", "man"];
println!("{:?}", join_all(&parts));
println!("{:?}", join_all(&["/etc", "passwd"]));
}
Playground

Why does returning early not finish outstanding borrows?

I'm trying to write a function which pushes an element onto the end of a sorted vector only if the element is larger than the last element already in the vector, otherwise returns an error with a ref to the largest element. This doesn't seem to violate any borrowing rules as far as I cant tell, but the borrow checker doesn't like it. I don't understand why.
struct MyArray<K, V>(Vec<(K, V)>);
impl<K: Ord, V> MyArray<K, V> {
pub fn insert_largest(&mut self, k: K, v: V) -> Result<(), &K> {
{
match self.0.iter().next_back() {
None => (),
Some(&(ref lk, _)) => {
if lk > &k {
return Err(lk);
}
}
};
}
self.0.push((k, v));
Ok(())
}
}
error[E0502]: cannot borrow `self.0` as mutable because it is also borrowed as immutable
--> src/main.rs:15:9
|
6 | match self.0.iter().next_back() {
| ------ immutable borrow occurs here
...
15 | self.0.push((k, v));
| ^^^^^^ mutable borrow occurs here
16 | Ok(())
17 | }
| - immutable borrow ends here
Why doesn't this work?
In response to Paolo Falabella's answer.
We can translate any function with a return statement into one without a return statement as follows:
fn my_func() -> &MyType {
'inner: {
// Do some stuff
return &x;
}
// And some more stuff
}
Into
fn my_func() -> &MyType {
let res;
'outer: {
'inner: {
// Do some stuff
res = &x;
break 'outer;
}
// And some more stuff
}
res
}
From this, it becomes clear that the borrow outlives the scope of 'inner.
Is there any problem with instead using the following rewrite for the purpose of borrow-checking?
fn my_func() -> &MyType {
'outer: {
'inner: {
// Do some stuff
break 'outer;
}
// And some more stuff
}
panic!()
}
Considering that return statements preclude anything from happening afterwards which might otherwise violate the borrowing rules.
If we name lifetimes explicitly, the signature of insert_largest becomes fn insert_largest<'a>(&'a mut self, k: K, v: V) -> Result<(), &'a K>. So, when you create your return type &K, its lifetime will be the same as the &mut self.
And, in fact, you are taking and returning lk from inside self.
The compiler is seeing that the reference to lk escapes the scope of the match (as it is assigned to the return value of the function, so it must outlive the function itself) and it can't let the borrow end when the match is over.
I think you're saying that the compiler should be smarter and realize that the self.0.push can only ever be reached if lk was not returned. But it is not. And I'm not even sure how hard it would be to teach it that sort of analysis, as it's a bit more sophisticated than the way I understand the borrow checker reasons today.
Today, the compiler sees a reference and basically tries to answer one question ("how long does this live?"). When it sees that your return value is lk, it assigns lk the lifetime it expects for the return value from the fn's signature ('a with the explicit name we gave it above) and calls it a day.
So, in short:
should an early return end the mutable borrow on self? No. As said the borrow should extend outside of the function and follow its return value
is the borrow checker a bit too strict in the code that goes from the early return to the end of the function? Yes, I think so. The part after the early return and before the end of the function is only reachable if the function has NOT returned early, so I think you have a point that the borrow checked might be less strict with borrows in that specific area of code
do I think it's feasible/desirable to change the compiler to enable that pattern? I have no clue. The borrow checker is one of the most complex pieces of the Rust compiler and I'm not qualified to give you an answer on that. This seems related to (and might even be a subset of) the discussion on non-lexical borrow scopes, so I encourage you to look into it and possibly contribute if you're interested in this topic.
For the time being I'd suggest just returning a clone instead of a reference, if possible. I assume returning an Err is not the typical case, so performance should not be a particular worry, but I'm not sure how the K:Clone bound might work with the types you're using.
impl <K, V> MyArray<K, V> where K:Clone + Ord { // 1. now K is also Clone
pub fn insert_largest(&mut self, k: K, v: V) ->
Result<(), K> { // 2. returning K (not &K)
match self.0.iter().next_back() {
None => (),
Some(&(ref lk, _)) => {
if lk > &k {
return Err(lk.clone()); // 3. returning a clone
}
}
};
self.0.push((k, v));
Ok(())
}
}
Why does returning early not finish outstanding borrows?
Because the current implementation of the borrow checker is overly conservative.
Your code works as-is once non-lexical lifetimes are enabled, but only with the experimental "Polonius" implementation. Polonius is what enables conditional tracking of borrows.
I've also simplified your code a bit:
#![feature(nll)]
struct MyArray<K, V>(Vec<(K, V)>);
impl<K: Ord, V> MyArray<K, V> {
pub fn insert_largest(&mut self, k: K, v: V) -> Result<(), &K> {
if let Some((lk, _)) = self.0.iter().next_back() {
if lk > &k {
return Err(lk);
}
}
self.0.push((k, v));
Ok(())
}
}

Return something that's allocated on the stack

I have the following simplified code, where a struct A contains a certain attribute. I'd like to create new instances of A from an existing version of that attribute, but how do I make the lifetime of the attribute's new value last past the function call?
pub struct A<'a> {
some_attr: &'a str,
}
impl<'a> A<'a> {
fn combine(orig: &'a str) -> A<'a> {
let attr = &*(orig.to_string() + "suffix");
A { some_attr: attr }
}
}
fn main() {
println!("{}", A::combine("blah").some_attr);
}
The above code produces
error[E0597]: borrowed value does not live long enough
--> src/main.rs:7:22
|
7 | let attr = &*(orig.to_string() + "suffix");
| ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ does not live long enough
8 | A { some_attr: attr }
9 | }
| - temporary value only lives until here
|
note: borrowed value must be valid for the lifetime 'a as defined on the impl at 5:1...
--> src/main.rs:5:1
|
5 | / impl<'a> A<'a> {
6 | | fn combine(orig: &'a str) -> A<'a> {
7 | | let attr = &*(orig.to_string() + "suffix");
8 | | A { some_attr: attr }
9 | | }
10| | }
| |_^
This question most certainly was answered before, but I'm not closing it as a duplicate because the code here is somewhat different and I think it is important.
Note how you defined your function:
fn combine(orig: &'a str) -> A<'a>
It says that it will return a value of type A whose insides live exactly as long as the provided string. However, the body of the function violates this declaration:
let attr = &*(orig.to_string() + "suffix");
A {
some_attr: attr
}
Here you construct a new String obtained from orig, take a slice of it and try to return it inside A. However, the lifetime of the implicit variable created for orig.to_string() + "suffix" is strictly smaller than the lifetime of the input parameter. Therefore, your program is rejected.
Another, more practical way to look at this is consider that the string created by to_string() and concatenation has to live somewhere. However, you only return a borrowed slice of it. Thus when the function exits, the string is destroyed, and the returned slice becomes invalid. This is exactly the situation which Rust prevents.
To overcome this you can either store a String inside A:
pub struct A {
some_attr: String
}
or you can use std::borrow::Cow to store either a slice or an owned string:
pub struct A<'a> {
some_attr: Cow<'a, str>
}
In the last case your function could look like this:
fn combine(orig: &str) -> A<'static> {
let attr = orig.to_owned() + "suffix";
A {
some_attr: attr.into()
}
}
Note that because you construct the string inside the function, it is represented as an owned variant of Cow and so you can use 'static lifetime parameter for the resulting value. Tying it to orig is also possible but there is no reason to do so.
With Cow it is also possible to create values of A directly out of slices without allocations:
fn new(orig: &str) -> A {
A { some_attr: orig.into() }
}
Here the lifetime parameter of A will be tied (through lifetime elision) to the lifetime of the input string slice. In this case the borrowed variant of Cow is used, and no allocation is done.
Also note that it is better to use to_owned() or into() to convert string slices to Strings because these methods do not require formatting code to run and so they are more efficient.
how can you return an A of lifetime 'static when you're creating it on the fly? Not sure what "owned variant of Cow" means and why that makes 'static possible.
Here is the definition of Cow:
pub enum Cow<'a, B> where B: 'a + ToOwned + ?Sized {
Borrowed(&'a B),
Owned(B::Owned),
}
It looks complex but it is in fact simple. An instance of Cow may either contain a reference to some type B or an owned value which could be derived from B via the ToOwned trait. Because str implements ToOwned where Owned associated type equals to String (written as ToOwned<Owned = String>, when this enum is specialized for str, it looks like this:
pub enum Cow<'a, str> {
Borrowed(&'a str),
Owned(String)
}
Therefore, Cow<str> may represent either a string slice or an owned string - and while Cow does indeed provide methods for clone-on-write functionality, it is just as often used to hold a value which can be either borrowed or owned in order to avoid extra allocations. Because Cow<'a, B> implements Deref<Target = B>, you can get &B from Cow<'a, B> with simple reborrowing: if x is Cow<str>, then &*x is &str, regardless of what is contained inside of x - naturally, you can get a slice out of both variants of Cow.
You can see that the Cow::Owned variant does not contain any references inside it, only String. Therefore, when a value of Cow is created using Owned variant, you can choose any lifetime you want (remember, lifetime parameters are much like generic type parameters; in particular, it is the caller who gets to choose them) - there are no restrictions on it. So it makes sense to choose 'static as the greatest lifetime possible.
Does orig.to_owned remove ownership from whoever's calling this function? That sounds like it would be inconvenient.
The to_owned() method belongs to ToOwned trait:
pub trait ToOwned {
type Owned: Borrow<Self>;
fn to_owned(&self) -> Self::Owned;
}
This trait is implemented by str with Owned equal to String. to_owned() method returns an owned variant of whatever value it is called on. In this particular case, it creates a String out of &str, effectively copying contents of the string slice into a new allocation. Therefore no, to_owned() does not imply ownership transfer, it's more like it implies a "smart" clone.
As far as I can tell String implements Into<Vec<u8>> but not str, so how can we call into() in the 2nd example?
The Into trait is very versatile and it is implemented for lots of types in the standard library. Into is usually implemented through the From trait: if T: From<U>, then U: Into<T>. There are two important implementations of From in the standard library:
impl<'a> From<&'a str> for Cow<'a, str>
impl<'a> From<String> for Cow<'a, str>
These implementations are very simple - they just return Cow::Borrowed(value) if value is &str and Cow::Owned(value) if value is String.
This means that &'a str and String implement Into<Cow<'a, str>>, and so they can be converted to Cow with into() method. That's exactly what happens in my example - I'm using into() to convert String or &str to Cow<str>. Without this explicit conversion you will get an error about mismatched types.

Rust: using traits/typeclasses to implement a generic numeric function

I have a function which works to make a linked list of integers:
enum List<T> { Cons(T, ~List<T>), End }
fn range(start: int, end: int) -> ~List<int> {
if start >= end { ~End }
else { ~Cons(start, range(start+1, end)) }
}
However, I want to make a range of any numeric type, including uints, doubles and the like. But this, for example, doesn't work:
fn range<T: ord>(start: T, end: T) -> ~List<T> {
if start >= end { ~End }
else { ~Cons(start, range(start+1, end)) }
}
which produces:
> rustc list.rs
list.rs:3:12: 3:15 error: use of undeclared type name `ord`
list.rs:3 fn range<T: ord>(start: T, end: T) -> ~List<T> {
^~~
error: aborting due to previous error
How can I make a generic function in rust which restricts itself to be callable by "numeric" types? Without having to specifically write the interface myself? I had assumed that there were a number of standard-library traits (such as those listed in section 6.2.1.1 of the manual like eq, ord, etc, though now I'm wondering if those are proper "traits" at all) that I could use when declaring generic functions?
The traits are usually uppercase. In this case it is Ord. See if that helps.
In the current master, there is a trait named 'Num' which serves as a general trait for all numeric types. Work has been done recently to unify many of the common math functions to work on this trait rather than u8, f32, etc , specifically.
See https://github.com/mozilla/rust/blob/master/src/libstd/num/num.rs#L26 for the aforementioned Num trait.
Hope this helps!

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