Moving and returning a mutable pointer - memory

I am working through this Rust tutorial, and I'm trying to solve this problem:
Implement a function, incrementMut that takes as input a vector of integers and modifies the values of the original list by incrementing each value by one.
This seems like a fairly simple problem, yes?
I have been trying to get a solution to compile for a while now, and I'm beginning to lose hope. This is what I have so far:
fn main() {
let mut p = vec![1i, 2i, 3i];
increment_mut(p);
for &x in p.iter() {
print!("{} ", x);
}
println!("");
}
fn increment_mut(mut x: Vec<int>) {
for &mut i in x.iter() {
i += 1;
}
}
This is what the compiler says when I try to compile:
Compiling tut2 v0.0.1 (file:///home/nate/git/rust/tut2)
/home/nate/git/rust/tut2/src/main.rs:5:12: 5:13 error: use of moved value: `p`
/home/nate/git/rust/tut2/src/main.rs:5 for &x in p.iter() {
^
/home/nate/git/rust/tut2/src/main.rs:3:16: 3:17 note: `p` moved here because it has type `collections::vec::Vec<int>`, which is non-copyable
/home/nate/git/rust/tut2/src/main.rs:3 increment_mut(p);
^
error: aborting due to previous error
Could not compile `tut2`.
To learn more, run the command again with --verbose.
I also tried a version with references:
fn main() {
let mut p = vec![1i, 2i, 3i];
increment_mut(&p);
for &x in p.iter() {
print!("{} ", x);
}
println!("");
}
fn increment_mut(x: &mut Vec<int>) {
for &mut i in x.iter() {
i += 1i;
}
}
And the error:
Compiling tut2 v0.0.1 (file:///home/nate/git/rust/tut2)
/home/nate/git/rust/tut2/src/main.rs:3:16: 3:18 error: cannot borrow immutable dereference of `&`-pointer as mutable
/home/nate/git/rust/tut2/src/main.rs:3 increment_mut(&p);
^~
error: aborting due to previous error
Could not compile `tut2`.
To learn more, run the command again with --verbose.
I feel like I'm missing some core idea about memory ownership in Rust, and it's making solving trivial problems like this very difficult, could someone shed some light on this?

There are a few mistakes in your code.
increment_mut(&p), given a p that is Vec<int>, would require the function increment_mut(&Vec<int>); &-references and &mut-references are completely distinct things syntactically, and if you want a &mut-reference you must write &mut p, not &p.
You need to understand patterns and how they operate; for &mut i in x.iter() will not do what you intend it to: what it will do is take the &int that each iteration of x.iter() produces, dereference it (the &), copying the value (because int satisfies Copy, if you tried it with a non-Copy type like String it would not compile), and place it in the mutable variable i (mut i). That is, it is equivalent to for i in x.iter() { let mut i = *i; … }. The effect of this is that i += 1 is actually just incrementing a local variable and has no effect on the vector. You can fix this by using iter_mut, which produces &mut int rather than &int, and changing the &mut i pattern to just i and the i += 1 to *i += 1, meaning “change the int inside the &mut int.
You can also switch from using &mut Vec<int> to using &mut [int] by calling .as_mut_slice() on your vector. This is a better practice; you should practically never need a reference to a vector as that is taking two levels of indirection where only one is needed. Ditto for &String—it’s exceedingly rare, you should in such cases work with &str.
So then:
fn main() {
let mut p = vec![1i, 2i, 3i];
increment_mut(p.as_mut_slice());
for &x in p.iter() {
print!("{} ", x);
}
println!("");
}
fn increment_mut(x: &mut [int]) {
for i in x.iter_mut() {
*i += 1;
}
}

Related

With closures as parameter and return values, is Fn or FnMut more idiomatic?

Continuing from How do I write combinators for my own parsers in Rust?, I stumbled into this question concerning bounds of functions that consume and/or yield functions/closures.
From these slides, I learned that to be convenient for consumers, you should try to take functions as FnOnce and return as Fn where possible. This gives the caller most freedom what to pass and what to do with the returned function.
In my example, FnOnce is not possible because I need to call that function multiple times. While trying to make it compile I arrived at two possibilities:
pub enum Parsed<'a, T> {
Some(T, &'a str),
None(&'a str),
}
impl<'a, T> Parsed<'a, T> {
pub fn unwrap(self) -> (T, &'a str) {
match self {
Parsed::Some(head, tail) => (head, &tail),
_ => panic!("Called unwrap on nothing."),
}
}
pub fn is_none(&self) -> bool {
match self {
Parsed::None(_) => true,
_ => false,
}
}
}
pub fn achar(character: char) -> impl Fn(&str) -> Parsed<char> {
move |input|
match input.chars().next() {
Some(c) if c == character => Parsed::Some(c, &input[1..]),
_ => Parsed::None(input),
}
}
pub fn some_v1<T>(parser: impl Fn(&str) -> Parsed<T>) -> impl Fn(&str) -> Parsed<Vec<T>> {
move |input| {
let mut re = Vec::new();
let mut pos = input;
loop {
match parser(pos) {
Parsed::Some(head, tail) => {
re.push(head);
pos = tail;
}
Parsed::None(_) => break,
}
}
Parsed::Some(re, pos)
}
}
pub fn some_v2<T>(mut parser: impl FnMut(&str) -> Parsed<T>) -> impl FnMut(&str) -> Parsed<Vec<T>> {
move |input| {
let mut re = Vec::new();
let mut pos = input;
loop {
match parser(pos) {
Parsed::Some(head, tail) => {
re.push(head);
pos = tail;
}
Parsed::None(_) => break,
}
}
Parsed::Some(re, pos)
}
}
#[test]
fn try_it() {
assert_eq!(some_v1(achar('#'))("##comment").unwrap(), (vec!['#', '#'], "comment"));
assert_eq!(some_v2(achar('#'))("##comment").unwrap(), (vec!['#', '#'], "comment"));
}
playground
Now I don't know which version is to be preferred. Version 1 takes Fn which is less general, but version 2 needs its parameter mutable.
Which one is more idiomatic/should be used and what is the rationale behind?
Update: Thanks jplatte for the suggestion on version one. I updated the code here, that case I find even more interesting.
Comparing some_v1 and some_v2 as you wrote them I would say version 2 should definitely be preferred because it is more general. I can't think of a good example for a parsing closure that would implement FnMut but not Fn, but there's really no disadvantage to parser being mut - as noted in the first comment on your question this doesn't constrain the caller in any way.
However, there is a way in which you can make version 1 more general (not strictly more general, just partially) than version 2, and that is by returning impl Fn(&str) -> … instead of impl FnMut(&str) -> …. By doing that, you get two functions that each are less constrained than the other in some way, so it might even make sense to keep both:
Version 1 with the return type change would be more restrictive in its argument (the callable can't mutate its associated data) but less restrictive in its return type (you guarantee that the returned callable doesn't mutate its associated data)
Version 2 would be less restrictive in its argument (the callable is allowed to mutate its associated data) but more restrictive in its return type (the returned callable might mutate its associated data)

Returning closures which capture outer variables in Rust

As the title states I am looking to return a closure from a function which has some initial, mutable state. In the following examples CowRow is a struct with a time field. It also has a String field, so is thus not copyable. Concretely, I would like a function that looks something like:
pub fn agg1() -> Box<Fn(&CowRow)> {
let res = 0;
Box::new(move |r| { res += r.time; })
}
Of course, this produces the error:
src/queries.rs:9:25: 9:38 error: cannot assign to captured outer variable in an `Fn` closure
src/queries.rs:9 Box::new(move |r| { res += r.time; })
^~~~~~~~~~~~~
src/queries.rs:9:14: 9:41 help: consider changing this closure to take self by mutable reference
src/queries.rs:9 Box::new(move |r| { res += r.time; })
^~~~~~~~
It is my understanding that Rust needs to know about the size of returned values and because closures borrow their stack frame from their environment we need to introduce the Box and move to get a size for the return and put the closure on the heap.
Is there some way to also put res on the heap in this closures environment? Or otherwise allow for this behaviour? Of course I have looked at: Cannot borrow captured outer variable in an `Fn` closure as mutable but this seems overly complicated and it's not clear to me how this would perform in the case of multiple threads running this function simultaneously.
Another technique I tried was to change the closure to take a mutable reference to an i32 which I can initialise outside of the agg function. Example:
pub fn agg0() -> Box<Fn(&CowRow, &mut i32)> {
Box::new(move |r, &mut acc| { acc += r.time; })
}
However, this produces the error:
src/queries.rs:4:35: 4:48 error: re-assignment of immutable variable `acc` [E0384]
src/queries.rs:4 Box::new(move |r, &mut acc| { acc += r.time; })
^~~~~~~~~~~~~
src/queries.rs:4:35: 4:48 help: run `rustc --explain E0384` to see a detailed explanation
src/queries.rs:4:28: 4:31 note: prior assignment occurs here
src/queries.rs:4 Box::new(move |r, &mut acc| { acc += r.time; })
This one is a total mystery to me.
You need to do two things here: make res mutable, and return an FnMut closure, not an Fn one:
pub struct CowRow {
time: u64,
}
pub fn agg1() -> Box<FnMut(&CowRow) -> u64> {
let mut res = 0;
Box::new(move |r| { res += r.time; res })
}
fn main() {
let mut c = agg1();
let moo = CowRow { time: 2 };
println!("{:?}", c(&moo));
println!("{:?}", c(&moo));
println!("{:?}", c(&moo));
}
The Fn trait forbids the implementor from changing itself when invoked. Since this closure is modifying its own state, this means it cannot be Fn [1]. Instead, you need to use FnMut which does allow mutation of the closure's captured environment.
[1]: Unless you involve interior mutability, of course.
DK. already said how to fix agg1, but I wanted to explain what's wrong with agg0, for completeness.
pub fn agg0() -> Box<Fn(&CowRow, &mut i32)> {
Box::new(move |r, &mut acc| { acc += r.time; })
}
We can infer from agg0's return type that the type of the closure's second parameter is &mut i32. &mut acc is a pattern that deconstructs the mutable reference, defining acc as an i32, initialized to a copy of the referenced value. You can't mutate it, because you didn't define acc to be mutable (you'd need to write &mut mut acc instead of &mut acc), but that's not what you want anyway, because then you'd be mutating a copy. What you want is to mutate the pointed-to integer, so you need to define your closure like this:
pub fn agg0() -> Box<Fn(&CowRow, &mut i32)> {
Box::new(move |r, acc| { *acc += r.time; })
}
Here, the type of acc is &mut i32, so in order to mutate the i32, we need to dereference the pointer first (this yields an lvalue that refers to the i32 behind the pointer; it's not a copy!).

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(())
}
}

Why is function argument lifetime different to the lifetime of a binding inside a function?

Gave VisualRust another try and to see how far they got, I wrote a few lines of code. And as usual, the code causes me to write a question on stackoverflow...
See first, read my question later:
fn make_counter( state : &mut u32 ) -> Box<Fn()->u32>
{
Box::new(move || {let ret = *state; *state = *state + 1; ret })
}
fn test_make_counter() {
let mut cnt : u32 = 0;
{
let counter = make_counter( & mut cnt );
let x1 = counter();
let x2 = counter();
println!("x1 = {} x2 = {}",x1,x2);
}
}
fn alt_make_counter ( init : u32 ) -> Box<Fn()->u32> {
let mut state = init;
Box::new(move || {let ret = state; state = state + 1; ret })
}
fn test_alt_make_counter() {
let counter = alt_make_counter( 0u32 );
let x1 = counter();
let x2 = counter();
println!("x1 = {} x2 = {}",x1,x2);
}
fn main() {
test_make_counter();
test_alt_make_counter();
}
The difference between make_counter() and alt_make_counter() is, that in one case, the state is a pointer to a mutable u32 passed to the function and in the other case, it is a mutable u32 defined inside the function. As the test_make_counter() function shows clearly, there is no way, that the closure lives longer than the variable cnt. Even if I removed the block inside test_make_counter() they would still have the identical lifetime. With the block, the counter will die before cnt. And yet, Rust complains:
src\main.rs(4,2): error : captured variable state does not outlive the enclosing closure
src\main.rs(3,1): warning : note: captured variable is valid for the anonymous lifetime #1 defined on the block at 3:0
If you look at the alt_make_counter() function now, the lifetime of state should basically cause the same error message, right? If the code captures the state for the closure, it should not matter if the pointer is passed in or if the variable is bound inside the function, right? But obviously, those 2 cases are magically different.
Who can explain, why they are different (bug, feature, deep insight, ...?) and if there is a simple rule one can adopt which prevents wasting time over such issues now and then?
The difference is not in using a local variable vs. using a parameter. Parameters are perfectly ordinary locals. In fact, this version of alt_make_counter works1:
fn alt_make_counter (mut state: u32) -> Box<FnMut() -> u32> {
Box::new(move || {let ret = state; state = state + 1; ret })
}
The problem is that the closure in make_counter closes over a &mut u32 instead of u32. It doesn't have its own state, it uses an integer somewhere else as its scratch space. And thus it needs to worry about the lifetime of that location. The function signature needs to communicate that the closure can only work while it can still use the reference that was passed in. This can be expressed with a lifetime parameter:
fn make_counter<'a>(state: &'a mut u32) -> Box<FnMut() -> u32 + 'a> {
Box::new(move || {let ret = *state; *state = *state + 1; ret })
}
Note that 'a is also attached to the FnMut() -> u32 (though with a different syntax because it's a trait).
The simplest rule to avoid such trouble is to not use references when they cause problems. There is no good reason for this closure to borrow its state, so don't do it. I don't know whether you fall under this, but I've seen a bunch of people that were under the impression that &mut is the primary or only way to mutate something. That is wrong. You can just store it by value and then just mutate that directly by storing it, or the larger structure in which it is contained, in a local variable that is tagged as mut. A mutable reference is only useful if the results of the mutation needs to be shared with some other code and you can't just pass the new value to that code.
Of course, sometimes juggling references in complicated ways is necessary. Unfortunately there doesn't seem to be a quick and easy way to learn to deal with those confidently. It's a big pedagogic challenge, but so far it appears everyone just struggled for a while and then progressively had fewer problems as they get more experienced. No, there is no single simple rule that solves all lifetime woes.
1 The return type has to be FnMut in all cases. You just didn't get an error about that yet because your current error happens at an earlier stage in the compilation.

Instantiating a struct with stdin data in Rust

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.")
};
}
}

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