rust - 为什么当其中一个似乎超出范围时，带有字符串文字的代码会编译？

Question

我遇到了section talking about lifetimes起初有点困惑。所以我决定通过编写一个小例子来尝试一下。 (Playground)

fn main() {
    let b = "hello";
    let mut d = "Unassigned";
    {
        let a = "hi";
        let c = lifetime(&a, &b);
        d = &c;
    }

    // My confusion happens here, I was expecting a compile-time error
    // since the lifetime of 'c' is the same as 'a' which is in the previous block
    // from my understanding. But this compiles just fine
    println!("{}", d);
}

fn lifetime<'a, 'b>(test: &'a str, test2: &'b str) -> &'a str {
    println!("{}, {}", test, test2);
    return "returned value of lifetime";
}

据我了解， lifetime函数绑定(bind)生命周期 'a到返回的引用值。
所以通常我会期待这条线 println!("{}", d);在编译时因引用生命周期的错误而中断 'a超出范围，我错了。

我理解错了什么？
为什么要编译这段代码？

我看过 this和 that这基本上只是让我更加困惑，因为他们首先说出了我所期望的结果。

Answer 1

了解contravariance后正如上面评论中所指出的，我写了这个小片段(类似于问题)

struct Potato(i32);

fn doesnt_work() {
    let b = Potato(1);
    let d;
    {
        let a = Potato(2);
        let c = lifetime2(&b, &a);
        d = c;
    }

    // Compile-time error ! Borrowed value does not live long enough.
    println!("{}", d.0);
}

fn lifetime2<'a, 'b>(test: &'a Potato, test2: &'b Potato) -> &'a Potato {
    return &Potato(test.0);
}

现在给出了实际的预期编译时错误。

它在我原来的问题中发现生命周期被推断为 'static这是 str 的最高共同生命周期引用。