| 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; | |
| } | |
| ``` |