为什么我可以返回对本地文字的引用而不是变量? [英] Why can I return a reference to a local literal but not a variable?

问题描述

为什么要编译这段代码?

Why does this code compile?

fn get_iter() -> impl Iterator<Item = i32> {
    [1, 2, 3].iter().map(|&i| i)
}

fn main() {
    let _it = get_iter();
}

[1, 2, 3] 是一个局部变量，iter() 借用了它.此代码不应编译，因为返回值包含对局部变量的引用.

[1, 2, 3] is a local variable and iter() borrows it. This code should not compile because the returned value holds a reference to a local variable.

推荐答案

在您的示例中， [1, 2, 3] 不被视为局部变量，而是视为静态变量！

In your example, [1, 2, 3] is not treated as local variable, but as static one!

我们来看看这段代码:

fn foo() -> &'static [i32] {
    &[1, 2, 3]
}

这有效！

前段时间，RFC 1414: Rvalue Static Promotion 合并:将 constexpr 右值提升为静态内存中的值而不是堆栈槽".这意味着基本上你写的所有文字都可以永远存在.因此，像 let _: &'static i32 = &42; 这样的东西也可以工作！

Some time ago, RFC 1414: Rvalue Static Promotion was merged: "Promote constexpr rvalues to values in static memory instead of stack slots". This means that basically all literals you write can live forever. Thus, things like let _: &'static i32 = &42; also work!

如果我们避免使用文字数组，我们可以看到预期的错误:

If we avoid using a literal array, we can see the expected error:

fn bar() -> impl Iterator<Item = i32> {
    vec![1, 2, 3].iter().map(|&i| i)
}

这里我们得到了v 寿命不够长"错误.

Here we get the "v does not live long enough" error.

这不仅限于整数或数组；它广泛适用于任何仅由文字组成的文字:

This isn't limited to integers or arrays; it applies broadly to any literal that is composed solely of literals:

fn promote_integer() -> &'static i32 {
    &42
}

fn promote_float() -> &'static f64 {
    &42.42
}

fn promote_str() -> &'static str {
    "Hello World!"
}

struct Foo(char);

fn promote_struct() -> &'static Foo {
    &Foo('x')
}

<小时>

除了文字之外，这也适用于标准库中微小数量的函数，但这些很可能是错误.决定任意 const 函数的结果是否可以自动提升为 static 仍然是一个 开放主题.

---

Beyond literals, this also works for a tiny number of functions in the standard library, but these were likely a mistake. Deciding on if the result of arbitrary const functions can be automatically promoted to static is still an open topic.

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