Rust 编译器不统一在 if let 表达式的分支中都 impl Future 的类型

在try 有条件地创建future 时，为了以后与select_biased!一起使用:

            let mut heartbeat_timer = if let ServerState::Leader(_, _, _) = server_state {
                // if we're the Leader, we want to send out AppendEntries RPC at a
                // minimum interval as required to maintain our leadership
                task::sleep(HEARTBEAT_DURATION)
            } else {
                // if we're not the leader, no need to send those
                pending::<()>()
            };

// ...

            select_biased! {
                h = heartbeat_timer => self.heartbeat().await,
                // ... some others declared in priority order
                default => task::yield_now().await,
            }

我很难理解为什么会出现编译错误:

error[E0308]: `if` and `else` have incompatible types
   --> src/raft.rs:185:17
    |
182 |               let heartbeat_timer = if let ServerState::Leader(_, _, _) = server_state {
    |  ___________________________________-
183 | |                 task::sleep(HEARTBEAT_DURATION)
    | |                 ------------------------------- expected because of this
184 | |             } else {
185 | |                 pending::<()>()
    | |                 ^^^^^^^^^^^^^^^ expected future, found `Pending<()>`
186 | |             };
    | |_____________- `if` and `else` have incompatible types
    |
    = note: expected opaque type `impl futures::Future<Output = ()>`
                    found struct `futures::future::Pending<()>`

单击(在我的IDE中)pending会显示它的类型是Pending<T>，我还可以进一步看到Pending<T>实现了Future<Output=T>:

impl<T> Future for Pending<T> {
    type Output = T;

    fn poll(self: Pin<&mut Self>, _: &mut Context<'_>) -> Poll<T> {
        Poll::Pending
    }
}

出于某种原因，我还不能理解，编译器没有遵循相同的面包屑.编译器知道我遗漏了什么秘密？我的代码有没有一些直接的转换可以满足编译器的要求？我应该怎么想才能提供编译器所需要的东西呢？

rustup --version包括输出:

info: The currently active `rustc` version is `rustc 1.69.0 (84c898d65 2023-04-16)`