为什么在返回类型选项时返回 None<T> 当 T 不是引用时不返回空指针

Question

The FFI section in the nomicon states that

The most common type that takes advantage of the nullable pointer optimization is Option<T> , where None corresponds to null . So Option<extern "C" fn(c_int) -> c_int> is a correct way to represent a nullable function pointer using the C ABI (corresponding to the C type int (*)(int) ).

However, this function does not return null instead returning an address.

#[no_mangle]
pub extern "C" fn test() -> Option<u8> {
    None
}

Rust also issues this warning :

warning: `extern` fn uses type `Option<u8>`, which is not FFI-safe
 --> src/lib.rs:2:29
  |
2 | pub extern "C" fn test() -> Option<u8> {
  |                             ^^^^^^^^^^ not FFI-safe
  |
  = note: `#[warn(improper_ctypes_definitions)]` on by default
  = help: consider adding a `#[repr(C)]`, `#[repr(transparent)]`, or integer `#[repr(...)]` attribute to this enum
  = note: enum has no representation hint

Changing the return type to Option<Box<u8>> makes the function return null as expected,

#[no_mangle]
pub extern "C" fn test2() -> Option<Box<u8>> {
    None
}

The functions are called from c++ using the following header file

extern "C" {
    void * test();
    void * test2();
}

The following c++ code

int main()
{
    cout << test() << endl;
    cout << test2() << endl;
}

prints :

0x5555a09d1200
0

Answer 1

nullable pointer optimization.

If the Option contains a pointer type, such as Box then the optimisation can be applied. A u8 cannot be null and is not a pointer, but a Box<u8> is an owned pointer to a u8 , so the optimisation works there.

More generally, this optimisation is called "niche-filling" which also applies to a few non-pointer Rust types, notably enums . In the general case this is not safe for FFI though, and therefore doesn't apply to data in #[repr(C)] structs except for pointer types.