A parameter type is missing an explicit lifetime bound and may not live long enough.
Erroneous code example:
// 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;
}
RunThe 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;
}
Run