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bbdc5c6039
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
74 lines
2.3 KiB
Rust
74 lines
2.3 KiB
Rust
// cow1.rs
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// This exercise explores the Cow, or Clone-On-Write type.
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// Cow is a clone-on-write smart pointer.
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// It can enclose and provide immutable access to borrowed data, and clone the data lazily when mutation or ownership is required.
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// The type is designed to work with general borrowed data via the Borrow trait.
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//
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// This exercise is meant to show you what to expect when passing data to Cow.
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// Fix the unit tests by checking for Cow::Owned(_) and Cow::Borrowed(_) at the TODO markers.
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// I AM NOT DONE
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use std::borrow::Cow;
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fn abs_all<'a, 'b>(input: &'a mut Cow<'b, [i32]>) -> &'a mut Cow<'b, [i32]> {
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for i in 0..input.len() {
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let v = input[i];
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if v < 0 {
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// Clones into a vector if not already owned.
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input.to_mut()[i] = -v;
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}
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}
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input
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}
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#[cfg(test)]
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mod tests {
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use super::*;
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#[test]
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fn reference_mutation() -> Result<(), &'static str> {
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// Clone occurs because `input` needs to be mutated.
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let slice = [-1, 0, 1];
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let mut input = Cow::from(&slice[..]);
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match abs_all(&mut input) {
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Cow::Owned(_) => Ok(()),
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_ => Err("Expected owned value"),
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}
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}
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#[test]
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fn reference_no_mutation() -> Result<(), &'static str> {
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// No clone occurs because `input` doesn't need to be mutated.
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let slice = [0, 1, 2];
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let mut input = Cow::from(&slice[..]);
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match abs_all(&mut input) {
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// TODO
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}
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}
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#[test]
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fn owned_no_mutation() -> Result<(), &'static str> {
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// We can also pass `slice` without `&` so Cow owns it directly.
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// In this case no mutation occurs and thus also no clone,
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// but the result is still owned because it always was.
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let slice = vec![0, 1, 2];
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let mut input = Cow::from(slice);
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match abs_all(&mut input) {
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// TODO
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}
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}
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#[test]
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fn owned_mutation() -> Result<(), &'static str> {
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// Of course this is also the case if a mutation does occur.
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// In this case the call to `to_mut()` returns a reference to
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// the same data as before.
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let slice = vec![-1, 0, 1];
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let mut input = Cow::from(slice);
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match abs_all(&mut input) {
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// TODO
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}
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}
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}
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