How to represent a pointer to a C array in Rust?

Question

I need an extern "C" FFI function in Rust and want to accept an array of fixed size. The C code passes something like:

// C code
extern int(*)[4] call_rust_funct(unsigned char (*)[3]);
....
unsigned char a[] = { 11, 255, 212 };
int(*p)[4] = call_rust_funct(&a);

How do I write my Rust function for it ?

// Pseudo code - DOESN'T COMPILE
pub unsafe extern "C" fn call_rust_funct(_p: *mut u8[3]) -> *mut i32[4] {
    Box::into_raw(Box::new([99i32; 4]))
}

It's a pointer, so you can just use *mut std::os::raw::c_void in the extern function signature and transmute it to the correct type. — Pavel Strakhov
– Pavel Strakhov, Commented Aug 29, 2016 at 14:44
@PavelStrakhov Using the correct type gives some type safety than using void*, so i would use it only as a last resort. Doesn't rust have a solution for this you mean ? — ustulation
– ustulation, Commented Aug 29, 2016 at 14:48

Shepmaster · Accepted Answer · 2020-11-15 17:23:13Z

8

You need to use Rust's syntax for fixed size arrays:

pub unsafe extern "C" fn call_rust_funct(_p: *mut [u8; 3]) -> *mut [i32; 4] {
    Box::into_raw(Box::new([99i32; 4]))
}

You can also always use *mut std::os::raw::c_void and transmute it to the correct type.

Shepmaster

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answered Aug 29, 2016 at 15:03

Pavel Strakhov

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