Rust：为什么将闭包绑定到变量会改变类型？

我有这个（最简化的）代码：

fn returns_closure() -> Box<dyn Fn(&u64)> {
    let closure = |_| ();
    Box::new(closure)
}

这不会编译并带有相当无用的错误消息：

error[E0308]: mismatched types
 --> src/main.rs:3:5
  |
3 |     Box::new(closure)
  |     ^^^^^^^^^^^^^^^^^ one type is more general than the other
  |
  = note: expected type `FnOnce<(&u64,)>`
             found type `FnOnce<(&u64,)>`

error[E0308]: mismatched types
 --> src/main.rs:3:5
  |
3 |     Box::new(closure)
  |     ^^^^^^^^^^^^^^^^^ one type is more general than the other
  |
  = note: expected type `FnOnce<(&u64,)>`
             found type `FnOnce<(&u64,)>`

但是，当我不先将闭包绑定到变量，而是直接在 Box 的构造函数中创建它时，它确实会编译：

为什么第一个编译失败，两者有什么区别？

编辑：埃蒙斯的回答似乎是正确的。我用 nightly 工具链 (1.52.0) 编译了完全相同的代码，得到了一个更好的错误：

error: implementation of `FnOnce` is not general enough
 --> src/main.rs:3:5
  |
3 |     Box::new(closure)
  |     ^^^^^^^^^^^^^^^^^ implementation of `FnOnce` is not general enough
  |
  = note: closure with signature `fn(&'2 u64)` must implement `FnOnce<(&'1 u64,)>`, for any lifetime `'1`...
  = note: ...but it actually implements `FnOnce<(&'2 u64,)>`, for some specific lifetime `'2`