Reputation: 1069
I'm playing around with building a very simple stack based evaluator in Rust and I've come up against an odd situation where I think the borrow checker is being too conservative:
use std::collections::HashMap;
pub type Value = i32;
pub type Result = std::result::Result<(), Error>;
type Op = Box<dyn Fn(&mut Evaluator) -> Result>;
type OpTable = HashMap<String, Op>;
pub struct Evaluator {
stack: Vec<Value>,
ops: OpTable,
}
#[derive(Debug, PartialEq)]
pub enum Error {
DivisionByZero,
StackUnderflow,
UnknownWord,
InvalidWord,
}
impl Evaluator {
fn add(&mut self) -> Result {
if let (Some(x), Some(y)) = (self.stack.pop(), self.stack.pop()) {
self.stack.push(y + x);
Ok(())
} else {
Err(Error::StackUnderflow)
}
}
fn sub(&mut self) -> Result {
if let (Some(x), Some(y)) = (self.stack.pop(), self.stack.pop()) {
self.stack.push(y - x);
Ok(())
} else {
Err(Error::StackUnderflow)
}
}
pub fn new() -> Evaluator {
let stack: Vec<Value> = vec![];
let mut ops: OpTable = HashMap::new();
ops.insert("+".to_string(), Box::new(Evaluator::add));
ops.insert("-".to_string(), Box::new(Evaluator::sub));
Evaluator { stack, ops }
}
pub fn eval(&mut self, input: &str) -> Result {
let symbols = input.split_ascii_whitespace().collect::<Vec<_>>();
// user definition
if let (Some(&":"), Some(&";")) = (symbols.first(), symbols.last()) {
if symbols.len() > 3 {
let statement = symbols[2..symbols.len() - 1].join(" ");
self.ops.insert(
symbols[1].to_string().to_ascii_lowercase(),
Box::new(move |caller: &mut Evaluator| caller.exec(&statement)),
);
return Ok(());
} else {
return Err(Error::InvalidWord);
}
}
self.exec(input)
}
fn exec(&mut self, input: &str) -> Result {
let symbols = input.split_ascii_whitespace().collect::<Vec<_>>();
for sym in symbols {
if let Ok(n) = sym.parse::<i32>() {
self.stack.push(n);
} else {
let s = sym.to_ascii_lowercase();
if let Some(f) = self.ops.get(&s) { // <--------------errors here
f(self)?; // <----------------------------|
} else {
return Err(Error::InvalidWord);
}
}
}
Ok(())
}
}
fn main() {
let mut e = Evaluator::new();
e.eval("1 2 +");
println!("{:?}", e.stack);
e.eval(": plus-1 1 + ;");
e.eval("4 plus-1");
println!("{:?}", e.stack);
}
I'm getting:
error[E0502]: cannot borrow `*self` as mutable because it is also borrowed as immutable
--> src/main.rs:77:21
|
76 | if let Some(f) = self.ops.get(&s) {
| -------- immutable borrow occurs here
77 | f(self)?;
| -^^^^^^
| |
| mutable borrow occurs here
| immutable borrow later used by call
For more information about this error, try `rustc --explain E0502`.
error: could not compile `evaluator` due to previous error
I believe this is because taking part of the hashmap (f
) borrows all of self
immutably, then I'm passing self
mutably to f()
. However, there is no real conflict here (I think).
I'm able to get around this by actually removing and reinserting the value:
fn exec(&mut self, input: &str) -> Result {
let symbols = input.split_ascii_whitespace().collect::<Vec<_>>();
for sym in symbols {
if let Ok(n) = sym.parse::<i32>() {
self.stack.push(n);
} else {
let s = sym.to_ascii_lowercase();
if self.ops.contains_key(&s) {
let f = self.ops.remove(&s).unwrap();
if let Err(e) = f(self) {
self.ops.insert(s, f);
return Err(e);
}
self.ops.insert(s, f);
} else {
return Err(Error::InvalidWord);
}
}
}
Ok(())
}
But this feels hacky and is a lot more verbose and inefficient. Am I missing something? Is there a way to tell the compiler the first version is ok?
Upvotes: 1
Views: 155
Reputation: 16475
The compiler is entirely correct, and so is your explanation: The call to get()
needs a borrow on self.ops
to return an &Op
of the same lifetime. You then try to call that FnMut
with a mutable borrow of self
; this mutable borrow of self
aliases with the immutable borrow on self.ops
, and in theory it would be possible for an implementation of this FnMut
to modify that borrowed Op
through self
, which is not allowed. The compiler prevented a situation where mutation occurs through an aliased pointer.
This situation often occurs when passing &mut self
around, as immutable borrows on members of self
which result in more borrows (&self.ops.get()
has the same lifetime as &self
) "lock" all of self
.
While your second example is cumbersome, it is at least correct as proven by the compiler: By removing Op
from the hashtable, the FnMut
can't reach itself through self
anymore, and mutation while aliasing is prevented.
A better approach is usually to avoid taking &mut self
as an argument (&mut self
as in &mut Executor
).
Upvotes: 2