refactor(cow1): replace main with tests

Following the discussion in #1195 this is the best I could come up with.
The issue for me (and apparently a few other learners) was that the code
needed to complete the exercise was not _missing_, but was rather there
but wrong.

In the end, what made the difference between this exercise and others
(for me) was that in this exercise I was supposed to learn what to
*expect* of an output. So I think it makes sense here to let the learner
modify the tests and not the code itself.

Fixes #1195

Signed-off-by: Daan Wynen <black.puppydog@gmx.de>

# Conflicts:
#	info.toml
pull/1244/head
Daan Wynen 1 year ago
parent 149e0c8ac2
commit bbdc5c6039

@ -4,6 +4,9 @@
// Cow is a clone-on-write smart pointer.
// It can enclose and provide immutable access to borrowed data, and clone the data lazily when mutation or ownership is required.
// The type is designed to work with general borrowed data via the Borrow trait.
//
// This exercise is meant to show you what to expect when passing data to Cow.
// Fix the unit tests by checking for Cow::Owned(_) and Cow::Borrowed(_) at the TODO markers.
// I AM NOT DONE
@ -20,29 +23,52 @@ fn abs_all<'a, 'b>(input: &'a mut Cow<'b, [i32]>) -> &'a mut Cow<'b, [i32]> {
input
}
fn main() {
// No clone occurs because `input` doesn't need to be mutated.
let slice = [0, 1, 2];
let mut input = Cow::from(&slice[..]);
match abs_all(&mut input) {
Cow::Borrowed(_) => println!("I borrowed the slice!"),
_ => panic!("expected borrowed value"),
#[cfg(test)]
mod tests {
use super::*;
#[test]
fn reference_mutation() -> Result<(), &'static str> {
// Clone occurs because `input` needs to be mutated.
let slice = [-1, 0, 1];
let mut input = Cow::from(&slice[..]);
match abs_all(&mut input) {
Cow::Owned(_) => Ok(()),
_ => Err("Expected owned value"),
}
}
// Clone occurs because `input` needs to be mutated.
let slice = [-1, 0, 1];
let mut input = Cow::from(&slice[..]);
match abs_all(&mut input) {
Cow::Owned(_) => println!("I modified the slice and now own it!"),
_ => panic!("expected owned value"),
#[test]
fn reference_no_mutation() -> Result<(), &'static str> {
// No clone occurs because `input` doesn't need to be mutated.
let slice = [0, 1, 2];
let mut input = Cow::from(&slice[..]);
match abs_all(&mut input) {
// TODO
}
}
// No clone occurs because `input` is already owned.
let slice = vec![-1, 0, 1];
let mut input = Cow::from(slice);
match abs_all(&mut input) {
// TODO
Cow::Borrowed(_) => println!("I own this slice!"),
_ => panic!("expected borrowed value"),
#[test]
fn owned_no_mutation() -> Result<(), &'static str> {
// We can also pass `slice` without `&` so Cow owns it directly.
// In this case no mutation occurs and thus also no clone,
// but the result is still owned because it always was.
let slice = vec![0, 1, 2];
let mut input = Cow::from(slice);
match abs_all(&mut input) {
// TODO
}
}
#[test]
fn owned_mutation() -> Result<(), &'static str> {
// Of course this is also the case if a mutation does occur.
// In this case the call to `to_mut()` returns a reference to
// the same data as before.
let slice = vec![-1, 0, 1];
let mut input = Cow::from(slice);
match abs_all(&mut input) {
// TODO
}
}
}

@ -1010,11 +1010,11 @@ https://doc.rust-lang.org/stable/book/ch16-00-concurrency.html
[[exercises]]
name = "cow1"
path = "exercises/smart_pointers/cow1.rs"
mode = "compile"
mode = "test"
hint = """
Since the vector is already owned, the `Cow` type doesn't need to clone it.
If Cow already owns the data it doesn't need to clone it when to_mut() is called.
Checkout https://doc.rust-lang.org/std/borrow/enum.Cow.html for documentation
Check out https://doc.rust-lang.org/std/borrow/enum.Cow.html for documentation
on the `Cow` type.
"""

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