I have this C code:
typedef void (*f_t)(int a);
struct Foo {
f_t f;
};
extern void f(struct Foo *);
bindgen generates the following Rust code (I have removed unimportant details):
#[repr(C)]
#[derive(Copy, Clone)]
#[derive(Debug)]
pub struct Foo {
pub f: ::std::option::Option<extern "C" fn(a: ::std::os::raw::c_int)>,
}
I do not understand why Option is here. Obviously that Rust enum and C pointer are not the same thing on the bit level, so how does the Rust compiler handle this?
When I call the C f function and pass a pointer to a Rust struct Foo, does the compiler convert Foo_rust to Foo_C and then only pass a pointer to Foo_C to f?
From The Rust Programming Language chapter on FFI (emphasis mine):
Certain types are defined to not be null. This includes references (
&T,&mut T), boxes (Box<T>), and function pointers (extern "abi" fn()). When interfacing with C, pointers that might be null are often used. As a special case, a generic enum that contains exactly two variants, one of which contains no data and the other containing a single field, is eligible for the "nullable pointer optimization". When such an enum is instantiated with one of the non-nullable types, it is represented as a single pointer, and the non-data variant is represented as the null pointer. SoOption<extern "C" fn(c_int) -> c_int>is how one represents a nullable function pointer using the C ABI.
Said another way:
Obviously that Rust enum and C pointer are not the same thing on the bit level
They actually are, when the Option contains a specific set of types.
See also:
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