Why can I return a reference to a local literal but not a variable?

问题

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

回答1:

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

Let's take a look at this code:

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

This works!

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!

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)
}

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

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')
}

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.

来源：https://stackoverflow.com/questions/50345139/why-can-i-return-a-reference-to-a-local-literal-but-not-a-variable

标签

reference

rust

borrow-checker