Question

我遇到了section talking about lifetimes，一开始有点困惑。因此，我决定通过编写一些示例来尝试一下。（Playground）

void limit_capabilities(void)
{
        ...
        if (setuid(getuid()) < 0) {
                perror("setuid");
                exit(-1);
        }

据我了解，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"; }函数将生存期lifetime绑定到返回的参考值。因此，通常我希望行'a在编译时中断，并出现一个错误，指出生存期println!("{}", d);超出范围，这是我错的。

我理解错了什么？为什么要编译此代码？

我看到this和that基本上使我更加困惑，因为他们首先说了我期望的结果。

Answer 1

找到上面评论中指出的关于contravariance的信息后，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);
}

现在给出实际的预期编译时错误。

它在我最初的问题中发现生命周期被推断为'static，这是str引用中最常见的生命周期。

当其中一个字符串文字超出范围时，为什么带有字符串文字的代码会编译？

1 个答案: