为什么对 const 进行静态引用会返回对临时变量的引用？

Question

In Rust I have the following code:

pub trait Test: Sized {
    const CONST: Self;
    fn static_ref() -> &'static Self {
        &Self::CONST
    }
}

My expectation is that since const is 'static , then I should be able to take a reference to it that is also 'static . However, the compiler gives the following error:

error[E0515]: cannot return reference to temporary value
   --> file.rs:9:9
    |
  9 |         &Self::CONST
    |         ^-----------
    |         ||
    |         |temporary value created here
    |         returns a reference to data owned by the current function

How is a temporary variable being introduced here?

Additionally, it seems that there are some cases where taking a reference to a constant does work. Here is a short concrete example with a slightly different implementation of Test

pub trait Test: Sized {
    fn static_ref() -> &'static Self;
}

struct X;

impl Test for X {
    fn static_ref() -> &'static Self {
        &X
    }
}

Answer 1

A constant in Rust is a compile-time constant, not an actual variable with a memory location. The Rust compiler can substitute the actual value of the constant whereever it is used. If you take the address of such a value, you get the address of a temporary.

Rust also has the concept of a static variable . These variables actually have memory locations that are consistent for the whole program duration, and taking a reference to a static variable indeed results in a reference with 'static lifetime.

See also:

• What is the difference between immutable and const variables in Rust?
• What is the difference between a const variable and a static variable and which should I choose?

Answer 2

When you define a trait, the definition must make sense for all possible implementations.

The problem may not be immediately clear without an example of where it fails. So suppose you had a type like this:

struct MyStruct;
impl MyStruct {
    const fn new() -> Self {
        MyStruct
    }
}

And you attempted to implement the trait like this:

impl Test for MyStruct {
    const CONST: Self = MyStruct::new();
}

This won't work because the implementation of static_ref will now look like this:

fn static_ref() -> &'static Self {
    // &Self::CONST
    &MyStruct::new()
}

It's creating a value inside the function and trying to return it. This value is not static, so the 'static lifetime is invalid.

---

However, with a little re-jigging, you can make something work:

pub trait Test: Sized + 'static {
    // This is now a reference instead of a value:
    const CONST: &'static Self;
    fn static_ref() -> &'static Self {
        Self::CONST
    }
}

struct MyStruct;
impl MyStruct {
    const fn new() -> Self {
        MyStruct
    }
}

impl Test for MyStruct {
    const CONST: &'static Self = &MyStruct::new();
}

This works because CONST is already a 'static reference, so the function can just return it. All possible implementations would have to be able to obtain a 'static reference to Self to implement the trait, so there is no longer an issue with referencing some arbitrary local value.

Answer 3

The mechanism at work here is static promotion. See RFC 1414

Here's a quote:

Inside a function body's block:

• If a shared reference to a constexpr rvalue is taken. (&)
• And the constexpr does not contain a UnsafeCell { ... } constructor.
• And the constexpr does not contain a const fn call returning a type containing a UnsafeCell.
• Then instead of translating the value into a stack slot, translate it into a static memory location and give the resulting reference a 'static lifetime.