Explain more clearly why fn() -> T can't be #[derive(Clone)]

This commit is contained in:
Martin Nordholts 2023-07-24 19:58:13 +02:00
parent 092e4f46be
commit c6566a8037

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@ -86,6 +86,46 @@
/// } /// }
/// ``` /// ```
/// ///
/// If we `derive`:
///
/// ```
/// #[derive(Copy, Clone)]
/// struct Generate<T>(fn() -> T);
/// ```
///
/// the auto-derived implementations will have unnecessary `T: Copy` and `T: Clone` bounds:
///
/// ```
/// # struct Generate<T>(fn() -> T);
///
/// // Automatically derived
/// impl<T: Copy> Copy for Generate<T> { }
///
/// // Automatically derived
/// impl<T: Clone> Clone for Generate<T> {
/// fn clone(&self) -> Generate<T> {
/// Generate(Clone::clone(&self.0))
/// }
/// }
/// ```
///
/// The bounds are unnecessary because clearly the function itself should be
/// copy- and cloneable even if its return type is not:
///
/// ```compile_fail,E0599
/// #[derive(Copy, Clone)]
/// struct Generate<T>(fn() -> T);
///
/// struct NotCloneable;
///
/// fn generate_not_cloneable() -> NotCloneable {
/// NotCloneable
/// }
///
/// Generate(generate_not_cloneable).clone(); // error: trait bounds were not satisfied
/// // Note: With the manual implementations the above line will compile.
/// ```
///
/// ## Additional implementors /// ## Additional implementors
/// ///
/// In addition to the [implementors listed below][impls], /// In addition to the [implementors listed below][impls],