Rust 函数指针似乎被borrow 判断器视为有状态的

以下示例代码无法编译:

fn invoke(i: i32, mut f: impl FnMut(i32)) {
    f(i)
}

fn main() {
    let f: fn(i32, _) = invoke;

    let mut sum: i32 = 0;
    for i in 0..10 {
        _ = f(i, |x| sum += x);
    }

    println!("{:?}", sum);
}

编译器返回以下错误:

   Compiling playground v0.0.1 (/playground)
error[E0499]: cannot borrow `sum` as mutable more than once at a time
  --> src/main.rs:10:18
   |
10 |         _ = f(i, |x| sum += x);
   |             -    ^^^ --- borrows occur due to use of `sum` in closure
   |             |    |
   |             |    `sum` was mutably borrowed here in the previous iteration of the loop
   |             first borrow used here, in later iteration of loop

For more information about this error, try `rustc --explain E0499`.
error: could not compile `playground` due to previous error

如果我将f赋值移到for循环，代码编译:

fn invoke(i: i32, mut f: impl FnMut(i32)) {
    f(i)
}

fn main() {
    let mut sum: i32 = 0;
    for i in 0..10 {
        let f: fn(i32, _) = invoke;
        _ = f(i, |x| sum += x);
    }

    println!("{:?}", sum);
}

我搞不懂为什么第一个代码不能编译.变量f的类型为fn，这意味着它是无状态的.变量f也是不可变的，因此即使它的类型是有状态的，它也不能存储闭包.因此，编译器应该能够在for循环的下一次迭代之前得出闭包将被删除的结论.然而，编译器的行为就像f是可变的，并且它可以存储闭包.你能解释一下为什么编译器会这样做吗.

Rustc版本:稳定v1.68.2