Getting pointer to a mut f32

I have:

let mut x = 0.0_f32;

I am dealing with C FFI, so I need to construct a

let y: *mut f32 = ...

How do I construct this y that points to x? (The C FFI function that takes a pointer to a f32 and writes the result there.)

You can coerce a mutable reference (&mut f32) into a raw pointer (*mut f32):

let y: *mut f32 = &mut x;
1 Like

let y = &mut x as *mut f32 should do the trick

As general rule, try to minimize the number of "as" casts in your code. So mbrubeck solution seems better :slight_smile:

That's fair. Though if it were me I would just let the reference coerce at the function call site instead of making a separate variable at all. i.e.

unsafe extern "C" fn takes_a_raw_pointer(ptr: *mut f32) {
    *ptr = 42.;

fn main() {
    let mut x = 0.0_f32;
    unsafe {
        takes_a_raw_pointer(&mut x); // automagically coerces here
    println!("{}", x); // prints 42

Thanks @mbrubeck , @FenrirWolf , @leonardo ...

after reading the solutions, I figured out what I was doing wrong:

(& x) as *mut f32

I was taking a read-only ref instead of a mutable ref, and trying to cast it to a *mut f32. LOL.

PS: I also agree with @FenrirWolf 's appraoch of "use as *mut f32 directly at call site" -- seems to be how FFI is often handled.