How to create a slice from a single element without copying that element?

Question

I have a really big struct (on heap and on stack) that I use in a function. Most of the time, I want a simple variable of that struct, because I'm doing stuff with the big struct directly. At one point, however, I'm forced (by a function signature) to pass this struct inside of a slice.

struct VeryBig(Vec<String>, [u64; 50]);

fn takes_slice(_: &[VeryBig]) {}

fn main() {
    let foo = VeryBig(vec!["Ferris".to_string(); 100], [27; 50]);

    // Use `foo` directly a bunch of times

    takes_slice(&foo); // <-- mismatched type

    // Use `foo` directly a bunch of times
}

This obviously and understandably leads to this error:

error[E0308]: mismatched types
  --> src/main.rs:10:17
   |
10 |     takes_slice(&foo); // <-- mismatched type
   |                 ^^^^ expected slice, found struct `VeryBig`
   |
   = note: expected type `&[VeryBig]`
              found type `&VeryBig`

So I wonder: what is the best way to solve this? I could just make foo a [VeryBig; 1], but that would mean that I have to use foo[0] everywhere where I want to use the big struct directly -- annoying. Alternatively, I could put the big struct into an array temporarily, call takes_slice and move it out of the array again. But that's also not very nice.

As far as I know, &T and &[T] should have the same memory layout (the pointee, not pointer), so I would expect there to be a way to reinterpret one as the other easily, but I haven't found such a feature.

Answer 1

You can use slice::from_ref.

fn from_ref<T>(s: &T) -> &[T]

In your code (Playground):

use std::slice;

takes_slice(slice::from_ref(&foo));

---

For mutable slices, you can use slice::from_mut. Also see this question about that topic.

See How can I convert a reference to any type to a reference to an array of length 1? for arrays.