Function pointers (fn) as Hashmap values? [duplicate] - rust

Question

I have the following code (Playground):
// Two dummy functions, both with the signature `fn(u32) -> bool`
fn foo(x: u32) -> bool {
x % 2 == 0
}
fn bar(x: u32) -> bool {
x == 27
}
fn call_both<F>(a: F, b: F)
where
F: Fn(u32) -> bool,
{
a(5);
b(5);
}
fn main() {
call_both(foo, bar); // <-- error
}
To me, it seems like this should compile as foo and bar have the same signature: fn(u32) -> bool. Yet, I get the following error:
error[E0308]: mismatched types
--> src/main.rs:20:20
|
20 | call_both(foo, bar);
| ^^^ expected fn item, found a different fn item
|
= note: expected type `fn(u32) -> bool {foo}`
found type `fn(u32) -> bool {bar}`
The same error can be triggered with this code:
let mut x = foo;
x = bar; // <-- error
I also tried to cast bar to the function pointer type again:
let mut x = foo;
x = bar as fn(u32) -> bool; // <-- error
This resulted in a slightly different error:
error[E0308]: mismatched types
--> src/main.rs:20:9
|
20 | x = bar as fn(u32) -> bool;
| ^^^^^^^^^^^^^^^^^^^^^^ expected fn item, found fn pointer
|
= note: expected type `fn(u32) -> bool {foo}`
found type `fn(u32) -> bool`
I don't understand these errors at all. What are fn items vs. fn pointers and why are foo and bar different fn items?

Answer 1

Function item vs. function pointer
When you refer to a function by its name, the type you get is not a function pointer (e.g. fn(u32) -> bool). Instead, you get an a zero-sized value of the function's item type (e.g. fn(u32) -> bool {foo}).
The value is zero-sized because it doesn't store the actual function pointer. The type perfectly identifies the function, so there is no need to store actual data in the type. This has several advantages, mostly about easier optimizations. A function pointer is like you would expect from other languages: it stores an address to the function.
A function pointer refers to the function via stored address; a function item refers to the function via type information.
A function item can be coerced to a function pointer in many situations, for example: as argument to a function and in let _: fn(u32) -> bool = foo; statements. Additionally, you can explicitly cast a function item to a function pointer: foo as fn(u32) -> bool.
You can read more about this topic in the reference on function items, function pointers and coercion.
Solution to your problem
In your case, the compiler isn't smart enough to figure out that you want the function pointers from your foo and bar instead of function items. When you call call_both(foo, bar) the compiler sets the generic type F to be fn(u32) -> bool {foo}, because that's the type of the first argument. And then it complains that the second argument doesn't have the same type.
You can fix that by explicitly specifying the F parameter:
call_both::<fn(u32) -> bool>(foo, bar);
call_both::<fn(_) -> _>(foo, bar); // <-- even this works
After specifying the type, the compiler can correctly coerce the arguments to a function pointer. You could also as-cast the first argument to fn(u32) -> bool explicitly.
You can fix your second example by explicitly stating the function pointer type, too:
let mut x: fn(u32) -> bool = foo;
x = bar;
In general: specifying the function pointer type somewhere to trigger the coercion will work.

Answer 2

Each named function has a distinct type since Rust PR #19891 was merged. You can, however, cast the functions to the corresponding function pointer type with the as operator.
call_both(foo as fn(u32) -> bool, bar as fn(u32) -> bool);
It's also valid to cast only the first of the functions: the cast will be inferred on the second, because both functions must have the same type.
call_both(foo as fn(u32) -> bool, bar);