//@ `data` public - otherwise, the next parts of this course could not work on `BigInt`s. Of course, in a
//@ real program, one would make the field private to ensure that the invariant (no trailing zeros) is maintained.
pub struct BigInt {
- pub data: Vec<u64>,
+ pub data: Vec<u64>, // least significant digit first, no trailing zeros
}
// Now that we fixed the data representation, we can start implementing methods on it.
//@ fields and initial values assigned to them.
pub fn new(x: u64) -> Self {
if x == 0 {
- BigInt { data: vec![] }
+ BigInt { data: vec![] } /*@*/
} else {
BigInt { data: vec![x] } /*@*/
}
}
// We can convert any vector of digits into a number, by removing trailing zeros. The `mut`
- // declaration for `v` here is just like the one in `let mut ...`, it says that we will locally
- // change the vector `v`.
+ // declaration for `v` here is just like the one in `let mut ...`: We completely own `v`, but Rust
+ // still asks us to make our intention of modifying it explicit. This `mut` is *not* part of the
+ // type of `from_vec` - the caller has to give up ownership of `v` anyway, so they don't care anymore
+ // what you do to it.
//
// **Exercise 05.1**: Implement this function.
//
- // *Hint*: You can use `pop()` to remove the last element of a vector.
+ // *Hint*: You can use `pop` to remove the last element of a vector.
pub fn from_vec(mut v: Vec<u64>) -> Self {
unimplemented!()
}
}
// ## Cloning
-//@ If you have a close look at the type of `BigInt::from_vec`, you will notice that it
-//@ consumes the vector `v`. The caller hence loses access to its vector. There is however something
+//@ If you take a close look at the type of `BigInt::from_vec`, you will notice that it
+//@ consumes the vector `v`. The caller hence loses access to its vector. However, there is something
//@ we can do if we don't want that to happen: We can explicitly `clone` the vector,
//@ which means that a full (or *deep*) copy will be performed. Technically,
-//@ `clone` takes a borrowed vector, and returns a fully owned one.
+//@ `clone` takes a borrowed vector in the form of a shared reference, and returns a fully owned one.
fn clone_demo() {
let v = vec![0,1 << 16];
let b1 = BigInt::from_vec((&v).clone());
let b2 = BigInt::from_vec(v);
}
-//@ Rust has special treatment for methods that borrow its `self` argument (like `clone`, or
+//@ Rust has special treatment for methods that borrow their `self` argument (like `clone`, or
//@ like `test_invariant` above): It is not necessary to explicitly borrow the receiver of the
//@ method. Hence you could replace `(&v).clone()` by `v.clone()` above. Just try it!
match *self { /*@*/
Nothing => Nothing, /*@*/
//@ In the second arm of the match, we need to talk about the value `v`
- //@ that's stored in `self`. However, if we would write the pattern as
+ //@ that's stored in `self`. However, if we were to write the pattern as
//@ `Something(v)`, that would indicate that we *own* `v` in the code
//@ after the arrow. That can't work though, we have to leave `v` owned by
//@ whoever called us - after all, we don't even own `self`, we just borrowed it.
//@ `#[derive(Clone)]` right before the definition of `SomethingOrNothing`.
// **Exercise 05.2**: Write some more functions on `BigInt`. What about a function that returns the number of
-// digits? The number of non-zero digits? The smallest/largest digit?
+// digits? The number of non-zero digits? The smallest/largest digit? Of course, these should all take `self` as a shared reference (i.e., in borrowed form).
// ## Mutation + aliasing considered harmful (part 2)
-//@ Now that we know how to borrow a part of an `enum` (like `v` above), there's another example for why we
+//@ Now that we know how to create references to contents of an `enum` (like `v` above), there's another example we can look at for why we
//@ have to rule out mutation in the presence of aliasing. First, we define an `enum` that can hold either
//@ a number, or a string.
enum Variant {
Number(i32),
Text(String),
}
-//@ Now consider the following piece of code. Like above, `n` will be a borrow of a part of `var`,
-//@ and since we wrote `ref mut`, the borrow will be mutable. In other words, right after the match, `ptr`
+//@ Now consider the following piece of code. Like above, `n` will be a reference to a part of `var`,
+//@ and since we wrote `ref mut`, the reference will be unique and mutable. In other words, right after the match, `ptr`
//@ points to the number that's stored in `var`, where `var` is a `Number`. Remember that `_` means
//@ "we don't care".
fn work_on_variant(mut var: Variant, text: String) {
Variant::Number(ref mut n) => ptr = n,
Variant::Text(_) => return,
}
- /* var = Variant::Text(text); */
+ /* var = Variant::Text(text); */ /* BAD! */
*ptr = 1337;
}
//@ Now, imagine what would happen if we were permitted to also mutate `var`. We could, for example,
//@ I hope this example clarifies why Rust has to rule out mutation in the presence of aliasing *in general*,
//@ not just for the specific case of a buffer being reallocated, and old pointers becoming hence invalid.
-//@ [index](main.html) | [previous](part04.html) | [next](part06.html)
+//@ [index](main.html) | [previous](part04.html) | [raw source](workspace/src/part05.rs) | [next](part06.html)
projects / rust-101.git / blobdiff