Hide private fields of a Rust lib when generating its C header

Question

I am making a Rust library that contains the following code:

pub mod my_module{

    use std::os::raw::{c_int, c_double};
    use std::collections::HashMap;

    struct MyPrivateClass {
        my_parameter:c_int
    }

    (...)

    #[repr(C)]
    pub struct MyPublicClass {
        my_private_parameter:HashMap<String,MyPrivateClass>,
        pub my_public_parameter:c_int,
        pub my_other_public_parameter:c_double
    }

    (...)

}

I want to use that library from Objective-C. So I’m using cbindgen to generate a C header. From my Objective-C project, I only need to access the public fields of MyPublicClass. But the C header generated contains all the fields of my public structs, including the private field.

This is what the generated C header looks like:

#include <stdarg.h>
#include <stdbool.h>
#include <stdint.h>
#include <stdlib.h>

typedef struct HashMap_String__MyPrivateClass HashMap_String__MyPrivateClass;

typedef struct {
  HashMap_String__MyPrivateClass my_private_parameter;
  int my_public_parameter;
  double my_other_public_parameter;
} MyPublicClass;

In this particular case, the HashMap type doesn’t have a straight C equivalent, so I can’t use this C header.

I only need to use this HashMap inside of my Rust library. I don’t need to use it from Objective-C.

How can I generate a C header that doesn’t expose it, without breaking anything?

You can't have a `HashMap` inside a C struct. C doesn't know its ABI, C doesn't know it can't make bit-wise copies of it. C doesn't know it has to drop it. My guess would be that you don't actually want `MyPublicClass` to be known by C at all, and merely want pointers to it in C. In that case you'd just declare `typedef struct MyPublicClass MyPublicClass;`. — mcarton, Dec 05 '19 at 21:53
C and C++ cannot ignore private members because they need to know the **size** of the type. The only way around that is with pointer members like forward declared structs or void*. — Zan Lynx, Dec 06 '19 at 01:23

score 0 · Answer 1 · answered Dec 09 '19 at 20:59

I ended up following mcarton's answer and got rid of the need to expose MyPublicClass to my C header.

As a consequence, I rely on functions to access the parameters, instead of accessing them directly as instance parameters. This has the drawback of not being able to check the values of MyPublicClass fields from the debugger when running my Objective-C project from Xcode, but at least it compiles and works as expected.

My Rust code:

// This struct is no longer exposed to C:
pub struct MyPublicClass {
    my_private_parameter:HashMap<String,MyPrivateClass>,
    pub my_public_parameter:c_int,
    pub my_other_public_parameter:c_double
}


impl MyPublicClass {
    pub fn new() -> Self {
        (…)
        MyPublicClass {
            my_private_parameter: … ,
            my_public_parameter: … ,
            my_other_public_parameter: … ,
        }
    }

    // Necessary to access my_public_parameter outside of Rust:
    pub fn get_my_public_parameter_value(&self) -> c_int {
        return self.my_public_parameter;
    }
}

// Exposed to C:
#[no_mangle]
pub unsafe extern fn my_public_class_new() -> *mut MyPublicClass {
    Box::into_raw(Box::new(MyPublicClass::new()))
}

// Exposed to C and necessary to access my_public_parameter outside of Rust:
#[no_mangle]
pub unsafe extern fn my_public_class_get_my_public_parameter_value(ptr: *mut MyPublicClass) -> c_int {
    let mpc = unsafe {
        assert!(!ptr.is_null());
        &mut *ptr
    };
    mpc.get_my_public_parameter_value()
}

The C header:

#include <stdarg.h>
#include <stdbool.h>
#include <stdint.h>
#include <stdlib.h>

typedef struct MyPublicClass MyPublicClass;
MyPublicClass *my_public_class_new();
int my_public_class_get_my_public_parameter_value(MyPublicClass *ptr);

Hide private fields of a Rust lib when generating its C header

1 Answers1