| A parameter type is missing an explicit lifetime bound and may not live long |
| enough. |
| |
| Erroneous code example: |
| |
| ```compile_fail,E0309 |
| // This won't compile because the applicable impl of |
| // `SomeTrait` (below) requires that `T: 'a`, but the struct does |
| // not have a matching where-clause. |
| struct Foo<'a, T> { |
| foo: <T as SomeTrait<'a>>::Output, |
| } |
| |
| trait SomeTrait<'a> { |
| type Output; |
| } |
| |
| impl<'a, T> SomeTrait<'a> for T |
| where |
| T: 'a, |
| { |
| type Output = u32; |
| } |
| ``` |
| |
| The type definition contains some field whose type requires an outlives |
| annotation. Outlives annotations (e.g., `T: 'a`) are used to guarantee that all |
| the data in `T` is valid for at least the lifetime `'a`. This scenario most |
| commonly arises when the type contains an associated type reference like |
| `<T as SomeTrait<'a>>::Output`, as shown in the previous code. |
| |
| There, the where clause `T: 'a` that appears on the impl is not known to be |
| satisfied on the struct. To make this example compile, you have to add a |
| where-clause like `T: 'a` to the struct definition: |
| |
| ``` |
| struct Foo<'a, T> |
| where |
| T: 'a, |
| { |
| foo: <T as SomeTrait<'a>>::Output |
| } |
| |
| trait SomeTrait<'a> { |
| type Output; |
| } |
| |
| impl<'a, T> SomeTrait<'a> for T |
| where |
| T: 'a, |
| { |
| type Output = u32; |
| } |
| ``` |