Error code E0591

Per RFC 401, if you have a function declaration foo:

#![allow(unused)]
fn main() {
struct S;

// For the purposes of this explanation, all of these
// different kinds of `fn` declarations are equivalent:

fn foo(x: S) { /* ... */ }
#[cfg(for_demonstration_only)]
extern "C" {
    fn foo(x: S);
}
#[cfg(for_demonstration_only)]
impl S {
    fn foo(self) { /* ... */ }
}
}

the type of foo is not fn(S), as one might expect. Rather, it is a unique, zero-sized marker type written here as typeof(foo). However, typeof(foo) can be coerced to a function pointer fn(S), so you rarely notice this:

#![allow(unused)]
fn main() {
struct S;
fn foo(_: S) {}
let x: fn(S) = foo; // OK, coerces
}

The reason that this matter is that the type fn(S) is not specific to any particular function: it's a function pointer. So calling x() results in a virtual call, whereas foo() is statically dispatched, because the type of foo tells us precisely what function is being called.

As noted above, coercions mean that most code doesn't have to be concerned with this distinction. However, you can tell the difference when using transmute to convert a fn item into a fn pointer.

This is sometimes done as part of an FFI:

#![allow(unused)]
fn main() {
extern "C" fn foo(userdata: Box<i32>) {
    /* ... */
}

fn callback(_: extern "C" fn(*mut i32)) {}
use std::mem::transmute;
unsafe {
    let f: extern "C" fn(*mut i32) = transmute(foo);
    callback(f);
}
}

Here, transmute is being used to convert the types of the fn arguments. This pattern is incorrect because, because the type of foo is a function item (typeof(foo)), which is zero-sized, and the target type (fn()) is a function pointer, which is not zero-sized. This pattern should be rewritten. There are a few possible ways to do this:

• change the original fn declaration to match the expected signature, and do the cast in the fn body (the preferred option)

• cast the fn item of a fn pointer before calling transmute, as shown here:

  # extern "C" fn foo(_: Box<i32>) {}
  # use std::mem::transmute;
  # unsafe {
  let f: extern "C" fn(*mut i32) = transmute(foo as extern "C" fn(_));
  let f: extern "C" fn(*mut i32) = transmute(foo as usize); // works too
  # }
  

The same applies to transmutes to *mut fn(), which were observed in practice. Note though that use of this type is generally incorrect. The intention is typically to describe a function pointer, but just fn() alone suffices for that. *mut fn() is a pointer to a fn pointer. (Since these values are typically just passed to C code, however, this rarely makes a difference in practice.)